Chapter 3: Laws of Motion
On 2 August 1971, an astronaut stood on the surface of the Moon with a small television camera watching him. His name was David Scott, and he was the commander of the Apollo 15 mission. In one hand he held a geologist’s hammer. In the other he held a feather from a falcon.
He held them out at the same height, side by side, and let them go together.
On Earth, everybody knows what happens next. The hammer thuds down at once and the feather drifts, tips, floats and takes its time. That is what we grow up seeing. But the Moon has almost no air at all, and on that grey, airless ground the hammer and the feather fell side by side, keeping pace with each other the whole way down, and struck the dust at the same instant.
They also fell slowly. Anyone watching the film notices it — the fall looks lazy, as though the two objects were sinking through water. The Moon pulls on things far more gently than the Earth does.
That short demonstration, lasting only a few seconds, brings together almost everything this chapter is about: what a force is, why objects need a push or a pull to change what they are doing, and why the Earth and the Moon pull on things at all.
Diagram
Fig. 3.0-1 — An Astronaut Drops A Hammer And A Feather On The Moon Illustration of an astronaut in a spacesuit standing on the grey lunar surface with a black star-filled sky behind. He holds one arm out to each side at shoulder height; a geologist’s hammer falls from the right hand and a feather falls from the left hand, both drawn at the same height above the ground with short motion lines behind them. Label the hammer, the feather, the equal heights above the surface with a dashed horizontal line, and the lunar surface. Add a small caption arrow noting ‘no air on the Moon’.
3.1 Effect Of A Force
You use forces all day without naming them. You pull a door, you press a switch, you kick a stone out of your path. In each case something happens because you pushed or pulled.
In Physics we are interested in what that push or pull actually does to the body it acts on. When a force acts on a body, it can produce two main effects. It can change the way the body is moving, and it can change the body’s size or shape.
3.1.1 Change In The State Of Rest Or Of Motion
The first effect of a force is that it can change the state of rest or of motion of a body. In other words, it can start motion in a body that was standing still, and it can alter the motion of a body that was already moving.
Think of a broom being pushed across a floor. The dust and litter lying still on the floor are set moving by the push. A ball lying on the ground stays where it is until somebody kicks it, and then it moves off. A railway engine pulls a whole train into motion. In each case a body at rest has been made to move.
A force can also alter a motion that is already going on. It can do this in three different ways, shown in Fig. 3.1-1.
Three Ways Of Altering A Motion
The first way is to make the body faster. A ball rolling down a slope keeps gaining speed, because the Earth’s pull acts along its direction of motion the whole way down.
The second way is to make the body slower. When the brakes of a moving vehicle are applied, a force acts on the wheels against the motion and the vehicle slows down and finally stops.
The third way is to change the direction in which the body is travelling. Tie a stone to one end of a string and whirl it in a horizontal circle at a steady speed. The stone is not getting any faster, yet its direction keeps changing every instant. The force doing this is the pull of the string, and it acts along the string, at right angles to the direction in which the stone is moving at that instant.
A single force can do two of these things at once. When a batsman hits a moving ball with a bat, or a player strikes a ball with a hockey stick in a direction other than the one it was travelling in, both the speed and the direction of the ball change together.
Diagram
Fig. 3.1-1 — Three Ways In Which A Force Alters A Motion Three panels side by side, each showing the body, its velocity arrow (thin, labelled ‘motion’) and the applied force arrow (thick, labelled ‘force’). Panel (a) speeding up: a ball on a slope with the force arrow pointing down the slope, along the direction of motion; label ‘speed increases’. Panel (b) slowing down: a vehicle wheel moving right with the force arrow pointing left, opposite to the motion; label ‘speed decreases’. Panel (c) change of direction: a stone whirling in a horizontal circle, with the velocity arrow drawn as a tangent to the circle and the force arrow along the string pointing towards the centre, at right angles to the velocity; label ‘direction changes, speed the same’.
3.1.2 Change In Size Or Shape
The second effect of a force is that it can change the size or the shape of a body, that is, it can change the dimensions of the body. Some examples are drawn in Fig. 3.1-2.
Hang a spring from a rigid support and load it. The spring becomes longer than it was. Take a small piece of silver sheet and hammer it: the sheet spreads out into a large thin foil, so here the force has increased the surface area. Press a piece of rubber between your fingers and it changes shape while you press.
A force can also squeeze a body into a smaller volume, or let it spread into a larger one. When the piston of a cycle pump is pushed down, the air trapped in the barrel is compressed and comes to occupy a smaller volume. The steam pushing out from a pressure cooker does the opposite — once it escapes into the atmosphere it occupies a much larger volume.
Diagram
Fig. 3.1-2 — A Force Changing The Size Or Shape Of A Body Three panels. Panel (a): a spring hanging from a hatched rigid support, drawn twice — unloaded on the left, and stretched longer with a weight hanging at its lower end on the right; mark the increase in length. Panel (b): a piece of rubber between finger and thumb, drawn twice — normal shape and squashed shape while pressed, with force arrows pointing inwards from both sides. Panel (c): a cycle pump in section with the piston up and the air column tall, and beside it the same pump with the piston pushed down and the air column short; label ‘air compressed into a smaller volume’ and draw the downward force arrow on the piston.
Rigid And Non-Rigid Bodies
Whether a force changes a body’s dimensions depends on the kind of body it acts on. A rigid body is one whose particles hold their spacing firmly, such as a steel ball or a brick. A non-rigid body, such as a spring, a rubber band or a lump of dough, allows its particles to be pushed closer together or pulled further apart.
When a force is applied on a free rigid object, it does not change the spacing between the particles of the object, so the dimensions of the object stay the same and only its motion changes. When a force is applied on a non-rigid object, the spacing between its particles does change, so the object changes its dimensions, and it may be set moving as well.
3.1.3 Force Defined From Its Effects
We now know a force only by what it does. It changes the size, or the shape, or the state of rest or of motion of a body. That is enough to define it.
Notice the words ‘or tends to change’ in the definition below. If you push hard against a heavy almirah and it does not budge, your push is still a force. It tended to move the almirah, but some other force acting on the almirah cancelled it.
Forces are of many kinds, and it helps to sort them out. From the point of view of how they are applied, forces are classified into two categories: those that act only when one body is physically touching another, and those that act across a gap without any touching at all. The next two sections take these two categories in turn.
3.2 Contact Forces
Most of the forces you meet in a day are forces you can trace to a touch. Your hand touches the door, the tyre touches the road, the string touches the stone. These are called contact forces.
There are five common contact forces at this level: the force of friction, the normal reaction force, the force of tension in a string, the force exerted by a spring, and the force exerted on two bodies during a collision. We take them one by one.
3.2.1 Frictional Force
When a body slides, or rolls, over a rough surface, a force starts acting on the body in a direction opposite to its motion, along the surface in contact. This force is called the frictional force, or the force of friction.
Fig. 3.2-1 shows a book resting on a table top. If the book is pushed towards the right, the force of friction on the book acts towards the left. Notice the direction carefully: friction never chooses a direction of its own, it simply opposes whichever way the body is moving or tending to move.
Friction does not wait for the body to actually move. It comes into play as soon as the book tends to move, and it resists the motion of the book on the table top. That is why a heavy carton on a floor can be pushed hard without moving at all — friction has grown to match your push.
Diagram
Fig. 3.2-1 — Force Of Friction On A Sliding Body Side view of a book resting on a horizontal table top. A thick arrow labelled ‘push’ points to the right on the book. A second thick arrow labelled ‘force of friction’ points to the left, drawn at the surface where the book and the table meet. Label the book and the table, and mark the contact surface with a short hatched line.
3.2.2 Normal Reaction Force
When a body is placed on a surface, the body exerts a force downwards on the surface, equal to its weight. Yet the body does not sink or fall. It cannot, because the surface pushes back on the body with an equal and opposite force. This push of the surface on the body, acting at right angles to the surface, is called the normal reaction force.
The word ‘normal’ here has nothing to do with ‘ordinary’. In Physics, normal means perpendicular. So the normal reaction is always perpendicular to the surface in contact, whichever way that surface is turned.
Hold a block on your open palm, as in Fig. 3.2-2. The block presses down on your palm because of its weight, and you have to push upwards on the block, normal to your palm, to keep it in position. The same thing happens with the book of Fig. 3.2-1: the book exerts a downward force equal to its weight on the table top, and the table top exerts an equal reaction force upwards, normal to the top of the table.
Diagram
Fig. 3.2-2 — Normal Reaction On A Block Held On The Palm An open hand, palm upwards, with a rectangular block resting on it. Draw an upward arrow starting at the palm and passing through the block, labelled ‘reaction force on block due to palm’. Draw a downward arrow from the centre of the block onto the palm, labelled ‘force due to weight of block on palm’. Label the palm and the block. Show both arrows as equal in length and perpendicular to the palm surface.
3.2.3 Tension Force Through A String
When a body is suspended by a string attached to a rigid support, the body pulls the string vertically downwards because of its weight \(W\). The string, now stretched, pulls the body upwards by a force that balances the weight of the body. This force developed in the string is called tension, or the force of tension, written \(T\).
Fig. 3.2-3 shows the two forces. For a body hanging at rest, the two must balance, so
\[T = W\]
where \(T\) is the tension in the string in newton (N) and \(W\) is the weight of the body in newton (N). This equality holds only while the body hangs in equilibrium, that is, at rest with nothing else pulling on it. If the support is jerked upwards, the tension is no longer equal to the weight.
That the string really is pulling upwards can be seen with one cut. If the string is cut slightly above the body, the free piece of string above the cut flicks slightly upwards with a jerk, because the tension in it was acting upwards, and only then does it fall down under its own weight.
Diagram
Fig. 3.2-3 — Force Of Tension In A String A hatched horizontal rigid support at the top. A vertical string hangs from it with a small rectangular block tied at its lower end. Draw an upward arrow along the string labelled ‘tension T (= W) developed in the string’ and a downward arrow from the block labelled ‘weight W of body pulling the string’. Both arrows equal in length. Label the rigid support, the string and the body.
3.2.4 Force Exerted By A Spring
Take a spring with one end kept fixed, as in Fig. 3.2-4(a). Now stretch its free end, as in Fig. 3.2-4(b), or compress it, as in Fig. 3.2-4(c). In both cases the spring exerts a force \(F\) on whatever is holding its free end, and this force is opposite to the direction in which the free end was displaced. Pull the end out and the spring pulls back in; push the end in and the spring pushes back out.
The size of this force is not fixed. It depends on how far the free end has been moved: the magnitude of the force is directly proportional to the magnitude of the displacement of the free end, that is, to the elongation or the compression of the spring. Stretch it twice as far and it pulls back twice as hard. This force is called the restoring force, because the spring is always trying to come back to its original form or shape. A spring balance works on this principle.
Now attach two objects \(A\) and \(B\), one at each end of a horizontal spring, as in Fig. 3.2-5. When the spring is in its normal form, neither stretched nor compressed, it exerts no force on either object. Compress the spring, and it pushes away each object with a restoring force \(F\) at its ends. Stretch the spring, and it pulls in each object with a restoring force \(F\) at its ends. In each case the spring has a tendency to come back to its original form or shape.
Diagram
Fig. 3.2-4 — Restoring Force Exerted By A Spring With One End Fixed Three panels, one below the other, each with the right-hand end of the spring attached to a hatched fixed wall labelled ‘fixed’. Panel (a): spring at its natural length, labelled ‘spring in normal position’, no force arrows. Panel (b): spring drawn longer, with an arrow at the free end pointing left labelled ‘(pull) F’ and an arrow on the spring pointing right labelled ‘restoring force F’; caption label ‘restoring force on a stretched spring’. Panel (c): spring drawn shorter and squeezed, with an arrow at the free end pointing right labelled ‘(push) F’ and an arrow on the spring pointing left labelled ‘restoring force F’; caption label ‘restoring force on a compressed spring’.
Diagram
Fig. 3.2-5 — Restoring Force Of A Horizontal Spring Carrying Two Objects Three panels, one below the other. Panel (a): a horizontal spring at natural length with hatched block A at its left end and hatched block B at its right end; label ‘normal form of spring — it exerts no force on either object’. Panel (b): the same spring drawn compressed, with an arrow on A pointing left and an arrow on B pointing right, each labelled ‘restoring force F’; label ‘compressed spring pushes away the objects’. Panel (c): the same spring drawn stretched, with an arrow on A pointing right and an arrow on B pointing left, each labelled ‘restoring force F’; label ‘stretched spring pulls in the objects’.
3.2.5 Force Exerted During A Collision
When two bodies collide, they push each other. As a result, equal and opposite forces act on each body. These two forces are called the force of action and the force of reaction.
Fig. 3.2-6 shows two bodies \(A\) and \(B\) moving towards each other. On meeting, body \(B\) exerts a force \(F\_{AB}\) on body \(A\), which is called the force of action. At the same instant, body \(A\) also exerts an equal and opposite force of reaction \(F\_{BA}\) on body \(B\). Because of these forces, the two bodies move apart after the collision.
Two points are worth noticing now, and both are taken up properly in section 3.13. The two forces act on different bodies — one on \(A\) and one on \(B\), never both on the same body. And they act at the same instant; neither comes first.
Diagram
Fig. 3.2-6 — Forces Exerted During A Collision Between Two Bodies Three panels, one below the other. Panel (a): circle A on the left with a velocity arrow pointing right labelled ‘motion of A’, circle B on the right with a velocity arrow pointing left labelled ‘motion of B’; caption label ‘two bodies moving towards each other’. Panel (b): the two circles touching, with an arrow on A pointing left labelled ‘F(AB) = force on A due to B’ and an arrow on B pointing right labelled ‘F(BA) = force on B due to A’; caption label ‘forces exerted during collision’. Panel (c): circle A on the left with its velocity arrow now pointing left and circle B on the right with its velocity arrow now pointing right; caption label ‘bodies moving away from each other after collision’.
3.3 Non-Contact Forces
Not every force needs a touch. A coin released above the floor starts moving downwards although nothing is holding it, and a pin jumps to a magnet before the two ever meet. Forces of this kind reach across a gap.
There are three non-contact forces you need at this level: the gravitational force, the electrostatic force and the magnetic force.
3.3.1 Gravitational Force
In the universe, every particle attracts every other particle because of its mass. This force of attraction between two masses is called the gravitational force.
The Earth has an enormous mass, so it attracts every mass around it. The force on a body due to the Earth’s attraction is called the force of gravity on the body, and it is the same thing as the weight of the body. It acts towards the Earth, that is, downwards, and it causes motion in that direction whenever the body is free to move.
So it is the force of gravity that makes a body fall when it is released from a height. Three ordinary examples: a ball placed on a table starts rolling down when the table is tilted; a body thrown up in the air goes up, reaches some height and then returns to the ground; a coin falls down when it is released at a height.
Here is a fact that surprises most students. The body attracts the Earth by an equal force at the same time. We never notice it, because the Earth’s mass is so huge that this force produces no perceptible motion in it. The pull is mutual; only the response is one-sided.
3.3.2 Electrostatic Force
A body can carry an electric charge, and charges come in two kinds, which we call positive and negative. Two like charges repel each other, while two unlike charges attract each other. The force between charges is called the electrostatic force, and it acts between charged objects even when they are separated.
You have seen it work. When a comb is rubbed on dry hair, it gets charged. Bring this comb near small bits of paper, and opposite charges are induced on the bits of paper, so they begin to move towards the comb. The motion of the paper bits is due to the electrostatic force of attraction between the unlike charges on the comb and on the paper bits.
3.3.3 Magnetic Force
A magnet has two poles, a north pole and a south pole. Two like magnetic poles repel each other, while two unlike magnetic poles attract each other. The force between the magnetic poles is called the magnetic force, and it acts even when the poles are at a separation.
Bring one pole of a magnet near a small iron nail without touching it. An opposite polarity is induced on the nail, and the nail moves towards the magnet. The motion of the nail is due to the magnetic force of attraction exerted between the unlike poles on the magnet and on the nail.
The attraction and repulsion in these last two cases are drawn together in Fig. 3.3-1.
Diagram
Fig. 3.3-1 — Attraction And Repulsion Between Charges And Between Magnetic Poles Four small panels in a two-by-two arrangement, each showing two objects with a gap between them and a pair of force arrows. Top left: two spheres each marked ‘+’, with arrows on both pointing away from each other, labelled ‘like charges repel’. Top right: one sphere marked ‘+’ and one marked ‘–’, with arrows on both pointing towards each other, labelled ‘unlike charges attract’. Bottom left: two bar magnets with their N poles facing each other, arrows pointing away, labelled ‘like poles repel’. Bottom right: two bar magnets with the N pole of one facing the S pole of the other, arrows pointing towards each other, labelled ‘unlike poles attract’. Mark the gap between the objects in every panel to show that no contact is needed.
3.3.4 General Properties Of Non-Contact Forces
The three forces above are different in origin, but they share a set of properties.
- They act at a distance. That is, these forces act even when the bodies are physically not in touch.
- They can be attractive or repulsive. The gravitational force is always attractive, while the electrostatic force and the magnetic force can be either attractive or repulsive.
- Their magnitude depends on the distance of separation between the two bodies. The force decreases with an increase in separation and increases as the separation decreases.
- They are long range forces, that is, they act over large distances.
- They can act even in vacuum, that is, they do not require any medium. For example, the Earth attracts the Moon by the gravitational force across empty space, which clearly shows that the gravitational force is a long range force and does not require a medium.
How Fast The Force Falls With Distance
The third property can be made exact. The magnitude of a non-contact force varies inversely as the square of the distance of separation:
\[F \\propto \\frac{1}{r^{2}}\]
where \(F\) is the magnitude of the force in newton (N) and \(r\) is the distance of separation between the two bodies in metre (m). This is called the inverse square relation.
Read the relation carefully, because the squaring is what students forget. It does not say that doubling the separation halves the force. It says that on doubling the separation, the force becomes one-fourth of what it was, since \(2^{2} = 4\). Triple the separation and the force falls to one-ninth.
The comparison in Table 3.1 collects the difference between the two families of forces you have now met.
3.4 Newton’s First Law Of Motion
In chapter 2 you studied how a body moves — its distance, its speed, its velocity, its acceleration. Now comes the harder question. What is the cause of motion? What produces motion in an object?
It is our common experience that a force has to be applied on an object to produce motion in it. To move a cycle, the cyclist applies a force on its pedal. In a car, the petrol engine provides the force needed to move the car. To move a horse-cart, the horse applies a force by stretching its muscles. To move a boat ahead, the oar applies a force on the water, pushing it backward, and the water pushes the boat ahead.
So far, so obvious. The difficulty begins when we ask whether that force has to keep acting.
3.4.1 The Older View Of Motion
Before Galileo, scientists held that a force is needed not only to start a motion but also to keep an object moving even with uniform velocity. In their view, an object remains in motion only so long as the external force that produced the motion remains present. A force, they said, must act continuously to keep a body in motion.
They had good reason for thinking so, because that is exactly what the world looks like. A cycle keeps moving as long as force is applied on its pedal; stop pushing the pedal and the cycle stops. Turn off the engine of a car and the car stops. The horse-cart stops after the horse stops pulling, and the boat stops after we stop pushing the oar.
Every one of these observations is correct. What is wrong is the conclusion drawn from them.
3.4.2 Galileo’s Conclusion And The Law Of Inertia
Galileo did not accept the older view. From his experiments he found that no force is needed to continue the motion of a moving body. Once a body is set in motion, it will remain in motion even when the force that set it moving is withdrawn — provided that there is no other force, such as friction, present to oppose the motion.
That last condition is the whole point, and everyday experience supports it. When a ball is kicked, it moves a certain distance before coming to rest. It does not stop by itself; it stops because of the force of friction between the ball and the ground. Reduce that friction by kicking the ball on a smooth road, and it travels a much longer distance before coming to rest.
Now carry the argument one step further. If it were somehow possible to remove the force of friction completely, the ball would have remained in motion forever after being kicked once. So in the absence of the force of friction, no force is required to keep an object moving after it has been set in motion. In other words, an object, once set in motion, moves with uniform velocity if no force acts on it.
A body therefore continues to be in the state of rest, or in the state of uniform motion, unless an external force is applied on it. This is called Galileo’s law of inertia.
From the above discussion we conclude that:
- If a body is at rest, it remains at rest unless a force is applied on it.
- If a body is moving, it will continue to move with the same speed in the same direction unless a force is applied on it.
Diagram
Fig. 3.4-1 — Galileo’s Inclined Planes Three panels one below the other, each showing a ball released from a marked height on a left-hand inclined plane and travelling up a right-hand plane. Draw a dashed horizontal line across all three panels at the release height. Panel (a): the right-hand plane at the same steep angle; the ball rises to the dashed line after a short distance. Panel (b): the right-hand plane at a gentler angle; the ball rises to the same dashed line but travels a much longer distance along the slope. Panel (c): the right-hand plane laid horizontal; the ball rolls along it with a long arrow and the label ‘the ball never reaches the dashed line, so it keeps rolling’. Label the release height, the dashed line as ‘same height’, and the ball in each panel.
3.4.3 Statement Of Newton’s First Law
Newton put the above observations in the form of a law, which is called Newton’s first law of motion.
The law can be understood in two parts, because it does two jobs at once. Its first part gives us the definition of inertia, and its second part gives us the definition of force.
3.4.4 The First Law Defines Inertia
In its first part, Newton’s first law tells us that an object cannot change its state by itself. If the object is in the state of rest, it will remain in the state of rest; and if it is moving in some direction, it will continue to move with the same speed in the same direction, unless an external force is applied on it.
A book lying on a table top will remain placed where it is unless it is displaced. Similarly, a ball rolling on a horizontal plane keeps on rolling unless the force of friction between the ball and the plane stops it.
3.4.5 The First Law Defines Force
The second part of Newton’s first law defines force. According to it, force is that external cause which can move a stationary object, or which can change the state of motion of a moving object.
A book lying on a table gets displaced from its place when it is pushed — a stationary object has been moved. A rolling marble comes to rest when you hold your palm in its path — the state of motion of a moving object has been changed. In both cases the cause came from outside the object itself.
Three Points To Note About Force
Force is a vector quantity, which means it has a direction as well as a magnitude. Saying that a force of ten units acts on a body tells only half the story until you say which way it acts.
Because forces have direction, they can cancel. The sum of two equal and opposite forces acting on a body is zero. More generally, a body acted upon by several forces can have the resultant net force on it equal to zero.
This is worth holding on to, because it settles a common doubt. When the first law speaks of no force acting, it includes the case of several forces acting whose resultant is zero. A book lying on a table has the Earth pulling it down and the table pushing it up, and it stays at rest because these add to nothing.
3.5 Mass And Inertia
Inertia is an inherent property of each body, by virtue of which it has a tendency to resist a change in its state of rest or state of motion. But bodies do not all resist equally. A rolling marble is easy to stop; a rolling drum is not.
What decides how much a body resists? The property of inertia is because of the mass of the body. The greater the mass, the greater is the inertia of the body. Thus a lighter body has less inertia than a heavier body.
Put the same idea the other way round. The more the mass of a body, the more difficult it is to move the body from rest, and the more difficult it is to stop the body if it is initially in motion. Since the mass of a body decides how strongly it resists, mass is a measure of inertia.
Two Everyday Comparisons
Take a football and a table tennis ball, and push each of them with an equal force for the same duration. The football, having much more mass, acquires a much smaller velocity than the table tennis ball. And if the two are moving with the same velocity, it is far harder to stop the football than the light table tennis ball.
The same shows up with a trolley, as in Fig. 3.5-1. A larger force is required to set a loaded trolley, which has more mass, in motion than to set an unloaded trolley, which has less mass, in motion. Similarly it is harder to stop a loaded trolley than an unloaded one, if both are moving initially with the same velocity.
Diagram
Fig. 3.5-1 — Equal Pushes On A Loaded And An Unloaded Trolley Two panels side by side. Panel (a): an empty four-wheeled trolley with a hand pushing it, the push arrow labelled ‘same force F’, and a long velocity arrow in front labelled ‘large velocity’. Panel (b): an identical trolley stacked with boxes, with a hand pushing it, the push arrow drawn the same length and labelled ‘same force F’, and a short velocity arrow in front labelled ‘small velocity’. Label panel (a) ‘less mass, less inertia’ and panel (b) ‘more mass, more inertia’.
3.6 Kinds Of Inertia And Examples Of The First Law
A body at rest resists being moved, and a body in motion resists being stopped. Both are inertia, but it is convenient to name them separately. Inertia is of two kinds: inertia of rest, and inertia of motion.
Each of the examples below is really Newton’s first law caught in the act, so read each one for its reason and not for its story.
3.6.1 Inertia Of Rest
If a body is at rest, it will remain at rest unless an external force is applied to change its state of rest. This property of a body is called the inertia of rest, or static inertia.
The examples that follow all share one pattern: one part of a system is forced into motion, while another part, touched by no such force, stays where it was.
- When a train suddenly starts moving forward, a passenger standing in the compartment tends to fall backwards. The reason is that the lower part of the passenger’s body is in close contact with the train. As the train starts moving, his lower part shares the motion at once, but the upper part, due to inertia of rest, cannot share the motion at the same instant and so tends to remain at the same place. The lower part of the body moves ahead while the upper part is left behind, so the passenger tends to fall backwards.
- When a hanging carpet is beaten with a stick, dust particles start falling out of it. The part of the carpet where the stick strikes comes into motion at once, while the dust particles settled in its fur remain in position due to inertia of rest. The carpet moves ahead with the stick, leaving the dust particles behind, and they then fall down due to the Earth’s pull.
- On shaking, or giving jerks to, the branches of a tree, the fruits fall down. When the stem or branches of the tree are shaken, they come into motion, while the fruits, due to inertia, remain in the state of rest. The massive and weakly attached fruits therefore get detached from the branches and fall down due to the pull of gravity.
- On striking the coin at the bottom of a pile of carrom coins with a striker, only the lowest coin moves away, while the rest of the pile remains intact. As the striker hits the lowest coin, that coin moves, that is, it changes its state of rest. The remaining pile, due to inertia of rest, remains where it is, and ultimately takes the place of the original pile due to the force of gravity.
- When a smooth card placed over the mouth of a tumbler is flicked sharply in the horizontal direction, the card flies away, but the coin kept over the card falls into the tumbler. When the card is flicked, a momentary force acts on the card, so it moves away, as shown in Fig. 3.6-1. But the coin kept on it does not share the motion at once and remains at its place due to inertia of rest. The coin then falls down into the tumbler due to the pull of gravity.
- When a corridor train suddenly starts, the sliding doors of some compartments may open. The frame of a sliding door is fixed on the floor of the train, so it comes into motion as soon as the train starts. The sliding door itself remains in its position due to inertia. Thus the frame moves ahead with the train while the door slides opposite to the direction of motion of the train, and so the door may open.
Diagram
Fig. 3.6-1 — Falling Of A Coin Kept On A Card Two panels side by side. Panel (a): a tumbler standing on a table with a smooth card lying flat across its mouth and a coin resting at the centre of the card; a finger is shown at the edge of the card with a horizontal arrow labelled ‘sharp flick’. Panel (b): the same tumbler with the card shown flying away sideways, labelled ‘card moves away’, and the coin drawn falling straight down into the tumbler with a short downward arrow labelled ‘coin stays in place, then falls’. Label the tumbler, the card and the coin in both panels.
3.6.2 Inertia Of Motion
A body in a state of motion continues to be in the state of motion, with the same speed in the same direction in a straight line, unless an external force is applied on it to change its state. This property of a body is called the inertia of motion, or dynamic inertia.
The pattern now reverses: something that was moving keeps moving, even after the cause of its motion is taken away, or after the body it was riding on has stopped.
- A cyclist riding along a level road does not come to rest immediately after he stops pedalling. The bicycle continues to move due to inertia of motion even after the cyclist stops applying force on the pedal. It comes to rest after some time as a result of the retarding force of friction between the tyres of the bicycle and the ground.
- When a passenger jumps out of a moving train, he falls down. Inside the train, his whole body was in a state of motion with the train. On jumping out, as soon as his feet touch the ground, the lower part of his body comes to rest, while the upper part still remains in motion due to inertia of motion. As a result, he falls in the direction of motion of the train and gets hurt. To avoid falling, as his feet touch the ground he should start running on the ground in the direction of motion of the train for some distance.
- When a running car stops suddenly, the passenger tends to lean forward. In a running car, the whole body of the passenger is in the state of motion. When the car stops suddenly, the lower part of his body, being in contact with the car, comes to rest immediately, but his upper part remains in the state of motion due to inertia. Thus his body leans forward.
- An athlete often runs before taking a long jump. By running, he brings his entire body into the state of motion. When the body is already in motion, it becomes easier to take a long jump.
- A ball thrown vertically upwards by a person in a moving train comes back to his hand. At the moment when the ball was thrown, it was in motion along with the person and the train, and it remains in that same state of forward motion during the whole time it is in the air. The person, the air inside the compartment and the ball all move ahead through the same distance due to inertia, and so the ball falls back into his palm on its return.
3.7 Linear Momentum
It is our common experience that if two bodies of different masses are moving with the same velocity, and they are to be brought to rest in the same time, the force needed to stop the heavier body is more than that needed for the lighter body. A loaded truck rolling at walking pace is a serious matter; a cycle rolling at the same pace is not.
Now keep the mass the same and change the velocity. If two bodies of the same mass are moving with different velocities, then to stop them in the same time, the force needed for the faster moving body is more than that for the slower one.
So the force needed to stop a moving body in a definite time depends both on the mass of the body and on its velocity. Neither by itself is enough to describe how hard the body is to stop. What the force actually depends on is the product of the mass and the velocity, and this product is called the linear momentum of the moving body. It is shown in Fig. 3.7-1.
Diagram
Fig. 3.7-1 — Momentum Depends On Both Mass And Velocity Three panels in a row, each showing a body with a velocity arrow and, below it, a shaded horizontal bar whose length represents its momentum. Panel (a): a cycle with a short velocity arrow labelled ‘small mass, small velocity’ and a short momentum bar. Panel (b): a cycle with a long velocity arrow labelled ‘small mass, large velocity’ and a medium momentum bar. Panel (c): a loaded truck with a short velocity arrow labelled ‘large mass, small velocity’ and a long momentum bar. Label the bars ‘momentum = mass × velocity’.
The linear momentum of a body is denoted by the letter \(p\). The word momentum by itself is generally used to mean linear momentum.
For a body of mass \(m\) moving with velocity \(v\), the linear momentum \(p\) is expressed as
\[p = mv\]
where \(p\) is the linear momentum in \(\\text{kg m s}^{-1}\), \(m\) is the mass of the body in kilogram (kg), and \(v\) is its velocity in \(\\text{m s}^{-1}\).
Momentum Has A Direction
Momentum is a vector quantity, having the direction of motion of the body, that is, along the velocity of the body. Mass has no direction of its own, so multiplying by mass cannot change the direction; it can only change how large the momentum is. A ball of momentum \(3\\ \\text{kg m s}^{-1}\) moving east and an identical ball moving west with the same speed do not have the same momentum, and a full answer always names the direction.
The Unit Of Momentum
The unit of momentum is not one you have to remember separately, because the relation gives it to you. From \(p = mv\),
unit of momentum = unit of mass × unit of velocity
In the S.I. system, mass is in kilogram and velocity in \(\\text{m s}^{-1}\), so momentum is in \(\\text{kg m s}^{-1}\). In the C.G.S. system, mass is in gram and velocity in \(\\text{cm s}^{-1}\), so momentum is in \(\\text{g cm s}^{-1}\).
3.8 Change In Momentum
A body’s momentum is not fixed for ever. A ball speeds up when it is hit, and slows down when it is caught, so its momentum changes. We now need a way of writing that change down.
From the relation \(p = mv\), the change in momentum is
\[\\Delta p = \\Delta(mv)\]
Here the symbol \(\\Delta\), called del or delta, written before a quantity, denotes a small change in that quantity. So \(\\Delta p\) means a small change in the momentum, and \(\\Delta(mv)\) means a small change in the product of mass and velocity.
When Only The Velocity Changes
Look at the product \(mv\). A change in it can come about in three ways: due to a change in the mass \(m\), due to a change in the velocity \(v\), or due to a change in both the mass \(m\) and the velocity \(v\) together.
For almost everything you will handle in this chapter, the mass stays the same. If the mass remains constant, then a change in momentum can only be due to a change in velocity. In that case the constant mass may be taken outside the bracket, and the relation becomes
\[\\Delta p = \\Delta(mv) = m,\\Delta v\]
where \(\\Delta p\) is the change in momentum in \(\\text{kg m s}^{-1}\), \(m\) is the constant mass of the body in kilogram (kg), and \(\\Delta v\) is the change in velocity in \(\\text{m s}^{-1}\). This form holds only while the mass of the body remains constant.
The reasoning behind taking \(m\) outside is worth stating once. The symbol \(\\Delta\) before \(mv\) denotes a small change in the product of \(m\) and \(v\). If the mass \(m\) does not change, the product \(mv\) will change only because of a change in \(v\), and so \(m\) can be written before the symbol \(\\Delta\). The quantity \(\\Delta v\) now represents a small change in \(v\) only.
When The Mass Itself Changes
There are cases in which the mass does not stay constant, and then the shorter form fails.
In the case of atomic particles moving with a velocity comparable to the velocity of light \(c\), where \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\), it was observed that the mass of the particle does not remain constant, but increases with an increase in velocity, according to the relation
\[m = \\frac{m\_0}{\\sqrt{1 - (v/c)^{2}}}\]
where \(m\) is the mass of the particle in kilogram (kg) when it is moving with velocity \(v\) in \(\\text{m s}^{-1}\), \(c\) is the speed of light in \(\\text{m s}^{-1}\), and \(m\_0\) is the mass of the particle when it is at rest, that is, when \(v = 0\).
In such a case, where an object or a particle is moving at extremely high speed, we cannot write \(\\Delta(mv) = m,\\Delta v\). The relation \(\\Delta p = m,\\Delta v\) is true only if the velocity \(v\) of the moving particle is much smaller than the velocity of light \(c\), that is, \(v \\ll c\). This happens when the velocity of the particle is of the order of \(10^{6}\\ \\text{m s}^{-1}\) or less than this; then the variation in mass with velocity is small enough, and the mass can be considered to be constant.
However, the general relation \(\\Delta p = \\Delta(mv)\) is always true, whether the mass \(m\) varies or it remains constant.
Mass need not change only at enormous speeds. In rocket motion, the mass of the rocket decreases as the burnt gases are expelled out of the nozzle, so the mass does not remain constant there either, even though the speeds involved are ordinary.
3.9 Rate Of Change Of Momentum
When a force is applied on a moving body, its velocity changes. Because the velocity changes, the momentum of the body also changes.
But knowing that the momentum changed is not enough. The same change spread over a long time and the same change crammed into an instant are very different things — as anyone who has caught a hard ball knows. So we now ask how fast the momentum changes, that is, we work out the rate of change of momentum.
Setting Up The Derivation
Let a force \(F\) be applied on a body of mass \(m\) for a time \(t\), due to which its velocity changes from \(u\) to \(v\). Here \(u\) is the initial velocity in \(\\text{m s}^{-1}\), \(v\) is the final velocity in \(\\text{m s}^{-1}\), \(m\) is the mass in kilogram (kg) and \(t\) is the time in second (s). The mass is taken to remain constant throughout.
Initial momentum of the body \(= mu\)
Final momentum of the body \(= mv\)
Change in momentum of the body in time \(t\) is the final momentum minus the initial momentum:
\[mv - mu = m(v - u)\]
The route from here is drawn in FC 3.1.
Flowchart
FC 3.1 — Route To The Rate Of Change Of Momentum Initial momentum mu → final momentum mv → change in momentum m(v − u) → divide by the time t taken → rate of change of momentum m(v − u)/t → replace (v − u)/t by the acceleration a → rate of change of momentum = ma
Working It Through
Dividing the change in momentum by the time taken,
\[\\text{rate of change of momentum} = \\frac{\\text{change in momentum}}{\\text{time}} = \\frac{m(v - u)}{t}\]
Now look at the fraction that has been left standing next to \(m\). You met it in chapter 2, because acceleration is defined as
\[a = \\frac{\\text{change in velocity}}{\\text{time}} = \\frac{v - u}{t}\]
where \(a\) is the acceleration in \(\\text{m s}^{-2}\). Substituting this in the line above,
\[\\text{rate of change of momentum} = m,a = \\text{mass} \\times \\text{acceleration}\]
This relation holds true when the mass of the body remains constant.
Thus, when a force acts on a body, the rate of change in momentum of the body is equal to the product of the mass of the body and the acceleration produced in it due to that force, provided that the mass of the body remains constant.
Alternative Method
The same result follows from the delta notation of the last section, and it is worth seeing both routes.
For a body of mass \(m\) moving with velocity \(v\), its linear momentum is \(p = mv\). In a time \(\\Delta t\), if its linear momentum changes by \(\\Delta p\) on applying a force on it, then the rate of change of linear momentum is
\[\\frac{\\Delta p}{\\Delta t} = \\frac{\\Delta(mv)}{\\Delta t} = m,\\frac{\\Delta v}{\\Delta t}\]
where the last step is allowed only if the mass \(m\) remains constant.
But \(\\dfrac{\\Delta v}{\\Delta t} = a\), the acceleration. So the rate of change of momentum is
\[\\frac{\\Delta p}{\\Delta t} = ma = \\text{mass} \\times \\text{acceleration}\]
which is the same result as before.
3.10 Newton’s Second Law Of Motion
Newton’s first law of motion defines force only qualitatively. A force is that physical cause which changes the state of motion of a body when it is applied on it. This means that the force produces acceleration in the body, that is, the force is the cause of acceleration.
That tells us what a force does, but not how much of it we have. Newton’s second law of motion gives the quantitative value of force, that is, it relates force to the measurable quantities acceleration and mass. Two separate experimental results go into it, and we take them one at a time.
3.10.1 Acceleration Depends On The Force Applied
It is our common experience that if we push a tennis ball gently, a small acceleration is produced in it, and it acquires a small velocity in a certain time. If the same tennis ball is pushed hard, a larger acceleration is produced in it and it acquires a large velocity in the same time interval.
This gives us a way of comparing two forces. Their magnitudes can be compared by measuring the accelerations they produce when they are applied one by one on the same body. If a force \(F\_1\) produces an acceleration of \(5\\ \\text{m s}^{-2}\) and a force \(F\_2\) produces an acceleration of \(10\\ \\text{m s}^{-2}\) on the same body, then the magnitude of the force \(F\_2\) is two times the magnitude of the force \(F\_1\).
Experimentally, Newton found that the acceleration produced in a body is directly proportional to the force applied on it, that is,
\[a \\propto F \\qquad \\text{(if mass remains constant)}\]
where \(a\) is the acceleration in \(\\text{m s}^{-2}\) and \(F\) is the applied force. A graph plotted for acceleration \(a\) against force \(F\), for a body of fixed mass, is a straight line as shown in Fig. 3.10-1.
Diagram
Fig. 3.10-1 — Variation Of Acceleration With Force, For A Constant Mass Graph with the horizontal axis labelled ‘force F’ and the vertical axis labelled ‘acceleration a (m s⁻²)’. Draw a single straight line starting exactly at the origin and rising steadily to the top right. Label the line ‘mass constant’. Mark the origin O. Show two dashed guide lines from a point on the line down to each axis to indicate a reading being taken.
3.10.2 Force Depends On The Mass Of The Body
Now try to produce the same change in velocity in the same time, that is, to produce the same acceleration, in a football and in a tennis ball, both initially at rest. You will need to apply a much larger force on the football than on the tennis ball.
So the force needed to produce the same acceleration in two bodies of different masses is not the same. It is more for the body of larger mass and less for the body of smaller mass. If a force \(F\) is needed to produce an acceleration of \(5\\ \\text{m s}^{-2}\) in a body of mass \(2\\ \\text{kg}\), then to produce the same acceleration of \(5\\ \\text{m s}^{-2}\) in a body of mass \(4\\ \\text{kg}\), the force needed is of double the magnitude, that is, \(2F\).
Experimentally, Newton found that the force needed to produce the same acceleration in different bodies is proportional to their masses, that is,
\[F \\propto m \\qquad \\text{(if the acceleration remains the same)}\]
where \(m\) is the mass of the body in kilogram (kg). A graph plotted for force \(F\) against mass \(m\) is a straight line as shown in Fig. 3.10-2.
Diagram
Fig. 3.10-2 — Variation Of Force With Mass, For A Constant Acceleration Graph with the horizontal axis labelled ‘mass m (kg)’ and the vertical axis labelled ‘force F’. Draw a single straight line starting at the origin and rising steadily to the top right. Label the line ‘acceleration constant’. Mark the origin O.
3.10.3 Combining The Two Results To Get F = ma
On the basis of his experiments, Newton concluded two things:
- The acceleration produced in a body of given mass is directly proportional to the force applied on it, \(a \\propto F\).
- The force needed to produce a given acceleration in a body is directly proportional to the mass of the body, \(F \\propto m\).
Combining the two relations,
\[F \\propto m \\times a \\qquad \\text{or} \\qquad F = K,m,a\]
Here \(K\) is a constant of proportionality. Now comes a step that looks like a trick but is really a choice. The unit of force has not yet been fixed, and we are free to fix it in whichever way is most convenient. The unit of force is so chosen that \(K\) becomes \(1\), when \(m = 1\) and \(a = 1\). That is, that amount of force is taken as one unit of force which, when applied on a body of unit mass, produces a unit acceleration in the body.
With the unit of force so chosen, the relation takes the following form:
\[F = m \\times a \\qquad \\text{or} \\qquad \\text{force} = \\text{mass} \\times \\text{acceleration}\]
where \(F\) is the force applied on the body, \(m\) is the mass of the body in kilogram (kg) and \(a\) is the acceleration produced in it in \(\\text{m s}^{-2}\). This is the mathematical expression of Newton’s second law of motion, and it holds while the mass of the body remains constant. The unit that \(F\) carries is fixed in the next section.
Same Force On Different Masses
Turn the relation around. If a given force is applied on bodies of different masses, the acceleration produced in them is inversely proportional to their masses, that is,
\[a \\propto \\frac{1}{m} \\qquad \\text{(for a given force } F)\]
A graph plotted for acceleration \(a\) against mass \(m\), for a fixed force, is therefore not a straight line but a curve, called a hyperbola, as shown in Fig. 3.10-3. The curve falls steeply at first and then flattens, but it never touches either axis: however large the mass, some acceleration is still produced.
Diagram
Fig. 3.10-3 — Variation Of Acceleration With Mass, For A Constant Force Graph with the horizontal axis labelled ‘mass m (kg)’ and the vertical axis labelled ‘acceleration a (m s⁻²)’. Draw a smooth curve starting high on the left near the vertical axis, falling steeply and then flattening as it moves right, approaching but never touching the horizontal axis. Label the curve ‘F = constant’. Mark the origin O.
3.11 C.G.S. And S.I. Units Of Force
The force is related to mass and acceleration as
\[\\text{force} = \\text{mass} \\times \\text{acceleration} \\qquad \\text{or} \\qquad F = ma\]
This relation is what fixes the unit of force. Since we agreed in the last section that the unit of force is the one for which the constant becomes \(1\), a unit of force is simply a unit of mass multiplied by a unit of acceleration. Choose the units of mass and acceleration, and the unit of force follows.
3.11.1 The Newton
In the S.I. system, mass is measured in kilogram and acceleration in \(\\text{m s}^{-2}\). On this basis, the S.I. unit of force is the newton.
The standard symbol of newton is N. Notice how it is written: the name of the unit is written with a small letter, newton, while its symbol is the capital letter N. This is the rule for every unit named after a scientist.
3.11.2 The Dyne
In the C.G.S. system, mass is measured in gram and acceleration in \(\\text{cm s}^{-2}\). On this basis, the C.G.S. unit of force is the dyne.
A dyne is a very small force. That is not a defect of the unit; it simply belongs to a system built on grams and centimetres rather than on kilograms and metres.
3.11.3 Relation Between Newton And Dyne
The two units measure the same physical quantity, so one must be convertible into the other. All we do is convert the kilogram into grams and the metre into centimetres, using \(1\\ \\text{kg} = 1000\\ \\text{g}\) and \(1\\ \\text{m} = 100\\ \\text{cm}\).
\[1\\ \\text{newton} = 1\\ \\text{kg} \\times 1\\ \\text{m s}^{-2}\] \[= 1000\\ \\text{g} \\times 100\\ \\text{cm s}^{-2} = 10^{5}\\ \\text{g cm s}^{-2}\] \[= 10^{5}\\ \\text{dyne}\]
Thus
\[1\\ \\text{newton} = 10^{5}\\ \\text{dyne}\]
To go the other way, divide instead of multiplying: \(1\\ \\text{dyne} = 10^{-5}\\ \\text{N}\).
3.12 The Second Law In Terms Of Rate Of Change Of Momentum
When a force is applied on a body, it produces acceleration in the body, due to which the velocity, and hence the momentum, of the body changes. So the second law can be said in a second way — in terms of momentum rather than acceleration — and this second way turns out to be the more powerful of the two.
From the result of section 3.9, the rate of change of momentum is equal to the product of mass and acceleration:
\[\\frac{\\Delta p}{\\Delta t} = ma \\qquad \\text{(if the mass remains constant)}\]
And from Newton’s second law, force \(F = ma\). The two right-hand sides are the same, so the two left-hand sides must be equal:
\[F = \\frac{\\Delta p}{\\Delta t} = \\frac{\\Delta(mv)}{\\Delta t} = \\frac{m,\\Delta v}{\\Delta t} = ma\]
where \(F\) is the applied force in newton (N), \(\\Delta p\) is the change in momentum in \(\\text{kg m s}^{-1}\), \(\\Delta t\) is the time in second (s) over which that change takes place, \(m\) is the mass in kilogram (kg) and \(\\Delta v\) is the change in velocity in \(\\text{m s}^{-1}\). The last two steps hold only if the mass remains constant.
Force is therefore the rate of change of momentum.
3.12.1 Statement And Explanation
Read the second half of the statement as carefully as the first. It fixes a direction, and direction is physical information, not decoration. Push a trolley forward and its momentum grows forward; push it backward and its momentum changes backward, which for a forward-moving trolley means slowing down.
According to this form of the law,
\[F = \\frac{\\Delta p}{\\Delta t} = \\frac{\\Delta(mv)}{\\Delta t}\]
This is the general form of Newton’s second law of motion, and it covers every case: when the momentum changes due to a change in mass, when it changes due to a change in velocity, and when it changes due to a change in both the mass and the velocity together.
When Does It Reduce To F = ma?
It is observed that the mass of a particle increases with an increase in its velocity, but this becomes perceptible only when the velocity \(v\) of the particle is comparable with the speed of light \(c\), where \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\). At velocities \(v \\ll c\), that is, when \(v \\ll 10^{6}\\ \\text{m s}^{-1}\), the change in mass is not perceptible. At such velocities the mass \(m\) can be considered to be constant, and Newton’s second law then takes the form
\[F = m,\\frac{\\Delta v}{\\Delta t} = ma\]
Thus, for the relation \(F = m\\dfrac{\\Delta v}{\\Delta t} = ma\) to hold, two conditions are needed: first, that the velocities are much smaller than the velocity of light, and second, that the mass remains constant.
Newton’s second law is the fundamental law of motion. The first law of motion is also included in the second law, as we can now see.
3.12.2 Obtaining The First Law From The Second Law
From Newton’s second law, \(F = ma\).
If \(F = 0\), then \(a = 0\), since the mass of a body cannot be zero.
This means that if no force is applied on the body, its acceleration will be zero. And zero acceleration means no change in velocity at all. So if the body is at rest, it will remain at rest, and if it is moving, it will remain moving in the same direction with the same speed.
Thus a body not acted upon by any external force does not change its state of rest or of motion. This is exactly the statement of Newton’s first law of motion.
3.12.3 Applications Of The Second Law In Daily Life
The mathematical expression of the second law is \(F = \\dfrac{\\Delta p}{\\Delta t}\). Rearranged, it gives
\[\\Delta p = F \\times \\Delta t\]
Read this as a bargain. A given change in momentum of a body can be brought about either by applying a large force for a small duration, or by applying a small force for a long duration. You choose which side of the bargain you want.
Sometimes we want the large force. In hitting a cricket ball with a bat, in hammering a nail, or in hitting a ping-pong ball with a racket, a required momentum is imparted to the body by applying a force of large magnitude for a very short duration.
More often we want the opposite. In the three examples that follow, the momentum of a moving body is brought to zero over a longer duration, so as to reduce the force exerted by, or on, the moving body.
Catching A Ball With Withdrawn Hands
Let \(u\) be the velocity of a ball of mass \(m\) when it reaches the hands of the player catching it.
The initial momentum of the ball \(= mu\).
When the cricketer stops the ball, its final velocity \(v = 0\), so its final momentum \(= m \\times 0 = 0\).
Change in momentum = final momentum − initial momentum \(= 0 - mu = -mu\), that is, \(mu\) numerically. The negative sign shows that the change in momentum is in a direction opposite to the initial direction of motion, which is only to be expected, since the ball is being stopped.
If the cricketer does not pull back his hands and stops the ball as soon as it touches his hands, he uses a very short time \(t\_1\) to stop the ball. The force exerted by the ball on his hands is then
\[F\_1 = \\frac{\\text{change in momentum}}{\\text{time interval}} = \\frac{mu}{t\_1}\]
But if he pulls back his hands along with the ball, he takes a much longer time \(t\_2\) to stop it. The force exerted by the ball on his hands is now
\[F\_2 = \\frac{mu}{t\_2}\]
Since \(t\_2 > t\_1\), and the numerator \(mu\) is the same in both cases, therefore \(F\_2 < F\_1\). The force exerted on the hands of the cricketer by the fast moving ball is less when he withdraws his hands. This is how the cricketer avoids the chance of injury to his palms, as shown in Fig. 3.12-1.
Diagram
Fig. 3.12-1 — Catching A Ball With Rigid And With Withdrawn Hands Two panels side by side, each showing a ball arriving at a pair of cupped hands with a velocity arrow labelled ‘u’. Panel (a): the hands held stiff and still at one position, with a short time label ‘stopped in a short time t₁’ and a long force arrow on the hands labelled ‘large force F₁’. Panel (b): the hands drawn in two positions, the second further back, joined by a dashed arrow labelled ‘hands withdrawn’, with a time label ‘stopped in a longer time t₂’ and a short force arrow on the hands labelled ‘small force F₂’. Keep the ball identical in both panels to show that the change in momentum is the same.
Landing On Sand Or Foam After A High Jump
When an athlete lands from a height on a hard floor, he may hurt his feet, because his feet come to rest almost instantaneously, that is, in a very short interval of time, so a very large force is exerted by the floor on his feet.
When he lands on sand or on foam instead, his feet push the sand or the foam for some distance before stopping. The time duration in which his feet come to rest therefore increases, and as a result the force exerted on his feet decreases, and he is saved from getting hurt.
Glass Vessels On A Hard Floor And On A Carpet
When a glass vessel falls from a height on a hard floor, it comes to rest almost instantaneously, that is, in a very short time, so the floor exerts a large force on the vessel and it breaks.
But if the same vessel falls on a carpet or on sand, the time duration in which the vessel comes to rest increases, and so the carpet or the sand exerts a smaller force on the vessel, and it does not break.
3.13 Newton’s Third Law Of Motion
Newton’s first law tells us that to bring a change in the state of rest or in the state of motion of an object, a force is needed, that is, a force produces a change in the velocity of the object. Newton’s second law tells us about the magnitude of the acceleration produced by that force when it is applied on the object.
Neither of these two laws explains how the force acts on the object — where it comes from, and what happens to the body that supplied it. This question is answered by Newton’s third law.
3.13.1 Statement And Explanation
Two examples make the meaning clear.
Consider a book lying on a table. The weight of the book acts vertically downwards, but the book does not move downwards. This implies that the resultant force on the book is zero, which is possible only if the table exerts an equal upward force on the book, equal to the weight of the book. This force on the book balances the weight of the book.
Next, consider yourself walking. While moving on the ground, we exert a force by our feet to push the ground backwards. The ground exerts a force of the same magnitude on our feet in the forward direction, and it is this force that makes us move forward.
In both of these examples there are two objects and two forces. In the first example, the weight of the book acting on the table downwards is the force of action, and the force exerted by the table on the book upwards is the force of reaction. In the second example, the force exerted by our feet on the ground is the force of action, and the force exerted by the ground on our feet is the force of reaction.
The law can now be stated more exactly, using the labels introduced in section 3.2.5.
Two points must never be dropped. The action and the reaction never act on the same body; they always act on two different bodies. And they act simultaneously — neither one comes first, and neither is the cause of the other. The forces of interaction are always present in a pair.
3.13.2 Experimental Demonstration Of The Third Law
The law can be shown with two spring balances, as in Fig. 3.13-1.
The ring of a spring balance \(B\) is attached to a hook fixed in a wall, and then the hook of another spring balance \(A\) is attached to the hook of the spring balance \(B\). Now the ring of the spring balance \(A\) is pulled. We find that both the balances show the same reading.
The reading on \(B\) is easy to expect: the spring of balance \(A\) pulls the spring of balance \(B\), so balance \(B\) registers that pull. The reading on \(A\) is the interesting one. The same reading is seen in balance \(A\) because the spring of balance \(B\) also pulls the spring of balance \(A\) by the same force.
Here the pull on the spring \(B\) by the spring \(A\) is the action \(F\_{BA}\), and the pull on the spring \(A\) by the spring \(B\) is the reaction \(F\_{AB}\). The experiment demonstrates that to every action there is an equal and opposite reaction, that is, in magnitude \(F\_{AB} = F\_{BA}\), but they are in opposite directions, which is written as
\[\\vec{F}*{AB} = -,\\vec{F}*{BA}\]
where the minus sign carries the physical information that the two forces point opposite ways.
Diagram
Fig. 3.13-1 — Demonstration Of Action And Reaction With Two Spring Balances Horizontal arrangement drawn in a line. At the right, a hatched wall with a hook fixed in it; the ring of spring balance B is hooked to it. To the left of B, the hook of spring balance A is attached to the hook of B. At the far left, a hand pulls the ring of A, with an arrow labelled ‘pull’. Draw an arrow above balance A pointing right labelled ‘reaction F(AB)’ and an arrow above balance B pointing left labelled ‘action F(BA)’. Show the dial of each balance with an identical reading, and label both ‘reading = 40 N’. Label the ring and hook of each balance and the wall.
3.13.3 Examples Of Action And Reaction
- A book on a table. When a book is placed on a table top, the book exerts a force equal to its weight \(W\), the action, on the table in the downward direction, and the table balances it by an equal force called the reaction \(R\), where \(R = W\), acting upwards on the book. These are shown in Fig. 3.13-2.
- Pushing a wall. When you exert a force on a wall by pushing the palm of your hand against it, the action, you experience a force exerted by the wall on your palm, the reaction.
- Motion of a boat moving away from the shore. When a boatman wants to move the boat away from the shore, he pushes the shore with a bamboo or with his oar, the action. The reaction force of the shore pushes the boatman, along with the boat, away with an equal and opposite force.
- Motion of a boat in water. To move a boat ahead in water, the boatman pushes the water backwards with his oar, the action, and the water exerts an equal and opposite force, the reaction, in the forward direction on the boat, due to which the boat moves ahead.
- Firing a bullet from a gun. When a man fires a bullet from a gun, a force \(F\) is exerted on the bullet, the action, and the gun experiences an equal recoil \(R\), the reaction, as shown in Fig. 3.13-3.
- Rocket motion. In a rocket, fuel is burnt inside the rocket, and the burnt gases at high pressure and high temperature are expelled out of the rocket through a nozzle. Thus the rocket exerts a force \(F\) on the gases, the action, to expel them through the nozzle backwards. The outgoing gases exert an equal and opposite force \(R\), the reaction, on the rocket, due to which it moves in the forward direction, as shown in Fig. 3.13-4.
- Motion of a man on the ground. When a man applies a force \(F\) backward by his foot on the ground against the force of friction, the action, the ground exerts an equal and opposite force \(R\), the reaction, forward on his foot. The horizontal component of this force of reaction \(R\) enables the man to move forward, as shown in Fig. 3.13-5. Obviously it will be difficult for a man to move on a slippery road, where friction is less.
- Motion of a boat away from the shore while stepping down from it. In order to get out of the boat, a man exerts a force on the board of the boat, the action. This creates a force of reaction which enables him to step out of the boat. At the same instant, the boat tends to leave the shore due to the force exerted by the man. For the safety of passengers, the boatman ties the boat to a pole on the shore so that it does not move away.
- Catching a ball. While catching a ball, the ball exerts a force on the palm of the cricketer, the action, and the cricketer exerts an equal force on the ball, the reaction, to stop it.
Diagram
Fig. 3.13-2 — Action And Reaction On A Book Lying On A Table A book drawn as a flat rectangular block resting on a table top shown in perspective. From the centre of the book draw a downward arrow onto the table labelled ‘W (action)’. From the same point draw an upward arrow labelled ‘R (reaction)’, equal in length to the downward arrow. Label the book and the table top, and mark the point of contact at the centre of the lower face of the book.
Diagram
Fig. 3.13-3 — Action And Reaction On Firing A Bullet From A Gun Side view of a gun pointing to the right, with a bullet drawn just leaving the muzzle. Draw an arrow on the bullet pointing right labelled ‘force on bullet (action) F’. Draw an arrow on the gun pointing left labelled ‘recoil on gun (reaction) R’. Label the gun, the barrel and the bullet.
Diagram
Fig. 3.13-4 — Action And Reaction In Rocket Motion A rocket drawn vertically, nose upwards, with a nozzle at its base and burnt gases streaming downwards out of the nozzle. Draw an upward arrow on the body of the rocket labelled ‘force on rocket (reaction) R’. Draw a downward arrow on the escaping gases labelled ‘force on gases (action) F’. Label the rocket, the nozzle and the gases.
Diagram
Fig. 3.13-5 — Action And Reaction While A Man Moves On The Ground Side view of a person’s legs in a walking stride, with the rear foot pressing on the ground. Draw an arrow from the foot pointing backwards and downwards onto the ground labelled ‘force exerted by man on ground (action) F’. Draw an arrow from the ground onto the foot pointing forwards and upwards labelled ‘force exerted by ground on foot (reaction) R’. Show the ground as a hatched horizontal line.
3.14 Universal Law Of Gravitation
Each mass particle of the universe attracts every other mass particle. In section 3.3 you met this force of attraction between two masses and gave it a name — the gravitational force of attraction. What you did not have then was a way of working out how large it is.
For the magnitude of this force, Sir Isaac Newton gave a law, known as Newton’s law of gravitation.
3.14.1 Statement And Equation Of The Law
In Fig. 3.14-1, let there be two bodies of masses \(m\_1\) and \(m\_2\) at a separation \(r\). The magnitude of the force of attraction \(F\) acting between them is governed by two statements:
\[F \\propto m\_1 m\_2 \\qquad \\text{(directly, as the product of the masses)}\] \[F \\propto \\frac{1}{r^{2}} \\qquad \\text{(inversely, as the square of the distance)}\]
Combining the two relations,
\[F \\propto \\frac{m\_1 m\_2}{r^{2}} \\qquad \\text{or} \\qquad F = G,\\frac{m\_1 m\_2}{r^{2}}\]
where \(F\) is the force of attraction in newton (N), \(m\_1\) and \(m\_2\) are the two masses in kilogram (kg), \(r\) is the distance between their centres in metre (m), and \(G\) is a constant of proportionality known as the gravitational constant.
Diagram
Fig. 3.14-1 — Gravitational Force Between Two Bodies Two small filled circles drawn on a horizontal line, well separated. Label the left circle m₁ and the right circle m₂. Draw a long double-headed arrow between them labelled r. Draw a short arrow on m₁ pointing right, towards m₂, labelled F₁₂. Draw a short arrow on m₂ pointing left, towards m₁, labelled F₂₁. Draw the two force arrows equal in length and mark the dotted line joining the two centres.
The Direction Of The Two Forces
Force is a vector quantity, so it is necessary to indicate its direction as well. The direction of the force \(\\vec{F}*{12}\) on the mass \(m\_1\) is towards \(m\_2\), along the line joining the masses \(m\_1\) and \(m\_2\), whereas the force \(\\vec{F}*{21}\) on the mass \(m\_2\) is towards the mass \(m\_1\) along the same line.
Both these forces are equal in magnitude but opposite in direction, that is, \(\\vec{F}*{12} = -,\\vec{F}*{21}\). So they form an action–reaction pair in the sense of section 3.13: a particle exerts a force on the other particle equal and opposite to the force that the second particle exerts on the first.
3.14.2 The Universal Gravitational Constant
The value of \(G\) remains the same at all places, and it is independent of the nature of the particles, the temperature, the medium between them, and so on. Therefore it is a universal constant, and is known as the universal gravitational constant.
Its unit follows from the relation itself. Rearranging,
\[G = \\frac{F \\times r^{2}}{m\_1 m\_2}\]
so the S.I. unit of \(G\) is
\[\\frac{\\text{newton} \\times \\text{metre}^{2}}{\\text{kilogram} \\times \\text{kilogram}} = \\text{N m}^{2},\\text{kg}^{-2}\]
The value of \(G\) is
\[G = 6.67 \\times 10^{-11}\\ \\text{N m}^{2},\\text{kg}^{-2}\]
Now put \(m\_1 = 1\\ \\text{kg}\), \(m\_2 = 1\\ \\text{kg}\) and \(r = 1\\ \\text{m}\) in the relation. Every factor except \(G\) becomes one, so \(G = F\). This gives \(G\) a physical meaning rather than leaving it as a number in a table.
How Big Is This Force?
Two examples settle the question.
Take two ordinary bodies of masses \(m\_1 = 1\\ \\text{kg}\) and \(m\_2 = 1\\ \\text{kg}\) kept at a separation of \(r = 1\\ \\text{m}\). The force between them is
\[F = \\frac{(6.67 \\times 10^{-11}) \\times 1 \\times 1}{(1)^{2}} = 6.67 \\times 10^{-11}\\ \\text{N}\]
which is far too small to notice, let alone measure by hand.
Now take the Moon, of mass \(7.36 \\times 10^{22}\\ \\text{kg}\), and the Earth, of mass \(5.96 \\times 10^{24}\\ \\text{kg}\), with the distance of the Moon from the Earth taken to be \(3.80 \\times 10^{8}\\ \\text{m}\). Then
\[F = \\frac{(6.67 \\times 10^{-11}) \\times (5.96 \\times 10^{24}) \\times (7.36 \\times 10^{22})}{(3.80 \\times 10^{8})^{2}} = 2 \\times 10^{20}\\ \\text{N}\]
From these two examples it is clear that the gravitational force of attraction is highly significant between heavenly bodies, but insignificant between ordinary bodies.
3.14.3 Properties Of The Gravitational Force
Regarding the gravitational force, it is important to note that the gravitational force between two masses:
- is always attractive. Unlike the electrostatic and magnetic forces, it never repels.
- is directly proportional to the product of the masses. Doubling either mass doubles the force.
- is inversely proportional to the square of the separation between them, that is, it obeys the inverse square law. Thus, on doubling the separation between the two masses, the force of attraction between them is reduced to one-fourth.
- is significant between heavenly bodies, but is insignificant between ordinary bodies, because of the small magnitude of \(G\).
3.14.4 Applications Of The Universal Law Of Gravitation
Newton used this law to explain the motion of the planets around the Sun, the motion of the Moon, which is the Earth’s satellite, around the Earth, and the motion of a freely falling body. Since then it has been put to work in many ways.
- Determining the mass and size of celestial bodies. Scientists use this law to calculate the mass and dimensions of planets, stars and other celestial bodies. For example, the mass of the Earth, the mass of the Moon, and even the masses of distant planets such as Saturn and Neptune, have been determined only through gravitational calculations.
- Formation of ocean tides. The Moon’s gravitational pull causes the Earth’s oceans to bulge, producing the regular cycle of high and low tides in coastal regions. In India, tidal patterns along the Gujarat and Konkan coasts influence fishing activities, port operations and even tidal energy projects.
- Launch and motion of satellites and space stations. Satellites are used for weather prediction, global positioning, communication, remote sensing and much else. For all this, satellites have to orbit around the Earth, and the determination of their orbital velocity and of how to keep them in continuous motion is done using this law. Indian space missions such as Chandrayaan-3, the Mars Orbiter Mission, and the placing of GSAT communication satellites are successful applications of these gravitational principles. India’s own navigation system, NavIC, also works entirely on carefully calculated satellite orbits.
- Launch of rockets. The law is used to calculate the escape velocity required for rockets to overcome the Earth’s gravitational pull. Every rocket launch by ISRO, including the missions under the Gaganyaan human spaceflight programme, requires a precise calculation of escape velocity and of gravitational forces, in order to leave the Earth and safely reach space.
- Discovery and prediction in astronomy. The law aids in the discovery of new planets and in predicting stellar phenomena. For instance, the planet Neptune was discovered by studying the gravitational effects on Uranus. Scientists also predict the motion of comets and asteroids, the occurrence of solar and lunar eclipses, and the orbits of planets and moons, using this law.
3.15 Force Due To Gravity
According to the law of gravitation, the Earth attracts every object around it towards its centre. The force with which the Earth attracts a body is called the force due to gravity on the body.
Two practical points go with it. This force can be taken to act vertically downwards at the centre of gravity of the body, so that a single arrow drawn from one point is enough to represent it, however oddly the body is shaped. And in vertical motion near the Earth’s surface, for a height much smaller than the radius of the Earth, the force of gravity on the body is assumed to be the same throughout the motion.
Working Out The Force
The force due to gravity on a body of mass \(m\) kept on the surface of the Earth is nothing other than the force of attraction between the Earth and that body, so the law of gravitation gives it directly. Let the Earth have mass \(M\) and radius \(R\). Since the body sits on the surface, the distance between the centre of the Earth and the body is \(R\) itself, as shown in Fig. 3.15-1. Then
\[F = \\frac{GMm}{R^{2}}\]
where \(F\) is the force due to gravity in newton (N), \(G\) is the universal gravitational constant in \(\\text{N m}^{2},\\text{kg}^{-2}\), \(M\) is the mass of the Earth in kilogram (kg), \(m\) is the mass of the body in kilogram (kg) and \(R\) is the radius of the Earth in metre (m).
Diagram
Fig. 3.15-1 — Force Due To Gravity On A Body At The Earth’s Surface A large circle representing the Earth with its centre marked O. A small block is drawn resting on the surface at the top of the circle. Draw a straight line from the block to the centre O, labelled R, marked with arrowheads at both ends and the label ‘radius of the earth’. Draw a thick arrow from the block pointing along this line towards O, labelled ‘force due to gravity F’. Label the Earth with ‘mass M’ and the block with ‘mass m’.
Now put in the numbers. Taking the mass of the Earth \(M = 5.96 \\times 10^{24}\\ \\text{kg}\) and the radius of the Earth \(R = 6.37 \\times 10^{6}\\ \\text{m}\), the force of gravity on a body of mass \(m = 1\\ \\text{kg}\) on the surface of the Earth will be
\[F = \\frac{(6.67 \\times 10^{-11}) \\times (5.96 \\times 10^{24}) \\times 1}{(6.37 \\times 10^{6})^{2}} = 9.8\\ \\text{N}\]
Thus the Earth attracts a body of mass \(1\\ \\text{kg}\) by a force of \(9.8\\ \\text{N}\) towards its centre. Compare this with the \(6.67 \\times 10^{-11}\\ \\text{N}\) of the last section, and you can see what an enormous mass does to a gravitational pull.
The Earth Is Pulled Too
As the Earth attracts an object towards it, the object also attracts the Earth towards it by an equal force — the third law again, and the law of gravitation says the same. Since the object is free to move, it starts moving towards the Earth. The Earth, because of its very large inertia, does not move perceptibly towards the object.
3.16 Acceleration Due To Gravity
The Earth pulls a falling body, and the last section told us how strong that pull is. A force produces acceleration, so the next question follows at once: how much acceleration does the pull of the Earth produce, and is it the same for every body?
3.16.1 Galileo’s Finding On Falling Bodies
Galileo was the first scientist to study the motion of different bodies under the force of attraction of the Earth, that is, under gravity. From his experiments he found that if bodies of different masses and different sizes or shapes are simultaneously made to fall in vacuum, that is, in the absence of air, from the same height, they all reach the Earth’s surface simultaneously and with the same velocity. In other words, all such bodies travel the same distance in the same interval of time.
He also found something about the fall itself. The velocity of a freely falling body does not remain constant, but it increases at a constant rate. That is, the motion of a freely falling body is a uniformly accelerated motion, and this acceleration is the same for all bodies.
In other words, the acceleration of a freely falling body does not depend on the mass of the body, its size or its shape. This acceleration is called the acceleration due to gravity.
Why Bodies Do Not Fall Together In Air
Galileo’s result is stated for a vacuum, and that condition is not a formality. In the presence of air, two extra forces act on a falling body in the upward direction: the viscous force of the air, which resists motion through it, and the force due to buoyancy of the air.
These two forces depend on the size, the shape and the velocity of the body, so they are different in magnitude for different bodies. Therefore the downward acceleration will not remain the same, equal to \(g\), for all bodies, and so they will not reach the ground simultaneously. A feather is affected far more than a coin of the same mass, simply because it presents so much more surface to the air. This is drawn in Fig. 3.16-1.
Diagram
Fig. 3.16-1 — Falling Bodies In Vacuum And In Air Two panels side by side, each showing two bodies released at the same instant from a marked release line at the top. Panel (a) labelled ‘in vacuum’: a heavy metal ball on the left and a light plastic ball on the right, drawn at the same lower level with a dashed horizontal line through both, labelled ‘they fall together’. Panel (b) labelled ‘in air’: the same two bodies, the metal ball drawn lower than the plastic ball; add small upward arrows on each body labelled ‘air resistance’ with a longer arrow on the lighter body, and label ‘the lighter body lags behind’. Mark the release line in both panels.
The acceleration due to gravity is denoted by the letter \(g\). It is a vector quantity, directed vertically downwards towards the centre of the Earth.
3.16.2 Variation In The Value Of g
Experimentally, it is found that the value of the acceleration due to gravity \(g\) does not remain constant.
On the Earth’s surface, the value of \(g\) varies from place to place. At the equator it is slightly less as compared to that at the poles. The mean value of \(g\) on the Earth’s surface is
\[g = 9.8\\ \\text{m s}^{-2}\]
and this is the value used throughout this chapter unless a question states otherwise.
At altitudes above the Earth’s surface, or at depths below the Earth’s surface, the value of \(g\) decreases. The value of \(g\) is zero at the centre of the Earth.
The value of \(g\) is also different on different planets and satellites. The value of \(g\) on the Moon’s surface is nearly one-sixth of the value of \(g\) on the Earth’s surface.
3.16.3 Relationship Between g And G
Let \(g\) be the acceleration due to gravity at a planet, or a satellite, of mass \(M\) and radius \(R\). There are now two ways of writing the force on a body of mass \(m\) resting on its surface, and setting them equal is what links \(g\) to \(G\).
By Newton’s second law of motion, the force due to gravity on the body is
\[F = \\text{mass} \\times \\text{acceleration due to gravity} \\qquad \\text{or} \\qquad F = mg\]
By Newton’s gravitational law, the same attractive force is given by
\[F = \\frac{GMm}{R^{2}}\]
From these two relations,
\[\\frac{GMm}{R^{2}} = mg\]
The mass \(m\) of the body appears on both sides, so it cancels — and that cancellation is the whole reason a hammer and a feather fall alike. This gives
\[g = \\frac{GM}{R^{2}}\]
where \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\), \(G\) is the universal gravitational constant in \(\\text{N m}^{2},\\text{kg}^{-2}\), \(M\) is the mass of the planet or satellite in kilogram (kg) and \(R\) is its radius in metre (m).
The relation above relates the acceleration due to gravity \(g\) with the gravitational constant \(G\). Obviously, the value of \(g\) on a planet, or a satellite, depends on the mass and the radius of that planet or satellite — and on nothing about the falling body.
Checking The Relation With Two Worlds
Taking the mass of the Earth \(M = 5.96 \\times 10^{24}\\ \\text{kg}\) and the radius of the Earth \(R = 6.37 \\times 10^{6}\\ \\text{m}\), the acceleration due to gravity at a place on the surface of the Earth comes out to be
\[g\_{\\text{earth}} = \\frac{(6.67 \\times 10^{-11}) \\times (5.96 \\times 10^{24})}{(6.37 \\times 10^{6})^{2}} = 9.8\\ \\text{m s}^{-2}\]
On the surface of the Moon, of mass \(M = 7.36 \\times 10^{22}\\ \\text{kg}\) and radius \(R = 1.75 \\times 10^{6}\\ \\text{m}\), the acceleration due to gravity is
\[g\_{\\text{moon}} = \\frac{(6.67 \\times 10^{-11}) \\times (7.36 \\times 10^{22})}{(1.75 \\times 10^{6})^{2}} = 1.6\\ \\text{m s}^{-2}\]
Obviously, \(g\_{\\text{moon}} \\cong \\dfrac{1}{6},g\_{\\text{earth}}\).
3.17 Free Fall
In chapter 2 you studied one dimensional motion and the three equations of motion that go with uniform acceleration. The motion of a freely falling body from a height, and the motion of a body thrown vertically upwards from the surface of the Earth, are both one dimensional motions under gravity. So those same equations apply here — with one change.
The change is that the acceleration is no longer some unknown \(a\). It is the acceleration due to gravity, whose value we already know. The acceleration of a vertically falling body is \(a = +g\), and that of a body going vertically upwards is \(a = -g\).
The sign is doing real physical work here. Gravity always acts downwards, whichever way the body happens to be travelling. When the body is going down, the acceleration is along the motion and the body speeds up. When the body is going up, the acceleration is opposite to the motion and the body slows down; this is why the upward case is written with a minus sign and is called retardation.
One thing to remember about all three sets of equations below: they hold in the absence of air. In the presence of air, or of any other medium, there are retarding forces — buoyancy and viscosity — which reduce the downward acceleration from the ideal value of \(g\) in vacuum.
A Body Falling From Rest
If a body falls from rest freely from a height \(h\) under gravity, then \(u = 0\) and the acceleration \(a\) is replaced by \(g\). The equations of motion become
\[v = gt, \\qquad h = \\tfrac{1}{2}gt^{2}, \\qquad v^{2} = 2gh\]
where \(v\) is the velocity gained in \(\\text{m s}^{-1}\), \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\), \(t\) is the time of fall in second (s) and \(h\) is the height fallen through in metre (m).
A Body Thrown Downwards
But if the initial velocity of fall of the body is \(u\), that is, if the body is thrown downwards rather than simply released, then the equations of motion are
\[v = u + gt, \\qquad h = ut + \\tfrac{1}{2}gt^{2}, \\qquad v^{2} = u^{2} + 2gh\]
with the symbols carrying the same meanings and units as above, and \(u\) the initial downward velocity in \(\\text{m s}^{-1}\).
A Body Thrown Vertically Upwards
If a body is thrown vertically up with an initial velocity \(u\) to a height \(h\), there will be retardation, so \(a = -g\), and the equations of motion are
\[v = u - gt, \\qquad h = ut - \\tfrac{1}{2}gt^{2}, \\qquad v^{2} = u^{2} - 2gh\]
At the highest point of its reach, the body is momentarily not moving, so the final velocity \(v = 0\). Putting \(v = 0\) in \(v^{2} = u^{2} - 2gh\) gives the maximum height reached:
\[h\_{\\text{max}} = \\frac{u^{2}}{2g}\]
and putting \(v = 0\) in \(v = u - gt\) gives the time taken by the body to rise to the highest point:
\[t = \\frac{u}{g}\]
The same will be the time it takes to come back to the initial point after reaching the highest point. So the total time of journey is
\[t' = 2t = \\frac{2u}{g}\]
and the total distance travelled by the body, up and then down again, is
\[h' = 2h\_{\\text{max}} = \\frac{u^{2}}{g}\]
The journey is drawn in Fig. 3.17-1.
Diagram
Fig. 3.17-1 — A Body Thrown Vertically Upwards A vertical line from the ground to the highest point of the flight, with the upward and downward paths drawn as two curved arrows side by side for clarity. At the ground, mark the upward launch with an arrow labelled ‘u’ and the return with a downward arrow labelled ‘v = u’. At the top, mark a dot labelled ‘highest point, v = 0’. Mark the vertical distance from the ground to the top as h(max) with a double-headed arrow. Label the upward path ‘a = −g, slowing down, time = u/g’ and the downward path ‘a = +g, speeding up, time = u/g’. Show the ground as a hatched horizontal line.
3.18 Mass And Weight
In everyday speech the two words are used as if they meant the same thing. In Physics they are two different quantities, with different units, measured by different instruments, and behaving differently when you carry a body from one place to another.
3.18.1 Mass
Mass is a scalar quantity, so it has no direction. It is constant for a given body at rest, and does not change by changing the place of the body. Carry a brick to the top of a mountain, or to the Moon, and the quantity of matter in it is untouched. Mass is an intrinsic property of the body.
Mass is measured by a physical balance, also called a beam balance. Such a balance compares the body in one pan with known masses in the other pan, and it works because the value of \(g\) is the same on both the pans. Whatever gravity does to one pan, it does equally to the other, so the comparison is unaffected by where the balance is being used.
There is one exception, and it never troubles us on Earth. The mass of a body increases with its velocity, but this change is perceptible only when the velocity of the body \(v\) becomes more than \(10^{6}\\ \\text{m s}^{-1}\), that is, when it reaches close to the speed of light \(c = 3 \\times 10^{8}\\ \\text{m s}^{-1}\). So for a body moving with a velocity less than \(10^{6}\\ \\text{m s}^{-1}\), its mass is taken to be constant. The relation itself was given in section 3.8.
3.18.2 Weight
Weight is a vector quantity. Its direction is downwards, towards the centre of the Earth.
The S.I. unit of weight is the newton (N) and the C.G.S. unit is the dyne, where \(1\\ \\text{N} = 10^{5}\\ \\text{dyne}\), exactly as for any other force — because weight is a force.
Relationship Between Weight And Mass
The weight of a body is related to its mass through the second law, since weight is the force of gravity and \(g\) is the acceleration it produces:
\[\\text{weight} = \\text{mass} \\times \\text{acceleration due to gravity} \\qquad \\text{or} \\qquad W = mg\]
where \(W\) is the weight in newton (N), \(m\) is the mass in kilogram (kg) and \(g\) is the acceleration due to gravity in \(\\text{m s}^{-2}\) at the place concerned.
From this relation, the S.I. unit of acceleration due to gravity \(g\) can also be written as newton per kilogram, \(\\text{N kg}^{-1}\), in place of metre per second squared, \(\\text{m s}^{-2}\). The two are the same thing, said from two points of view: how much force each kilogram feels, and how fast a free body speeds up.
Since the value of \(g\) varies from place to place, the weight of a given body also varies from place to place. The mass does not.
Gravitational Units Of Weight
Alongside the newton and the dyne, weight is also expressed in gravitational units. In the M.K.S. system the gravitational unit of weight is the kilogram force (kgf), and in the C.G.S. system it is the gram force (gf), where
\[1\\ \\text{kgf} = 9.8\\ \\text{N} \\qquad \\text{and} \\qquad 1\\ \\text{gf} = 980\\ \\text{dyne}\]
Obviously a body of mass \(m\) kilogram will weigh \(m\) kgf. These units are defined properly in the next section.
The weight of a body can be measured by a spring balance directly in newton, and also by a physical balance in kgf. Notice which instrument does which job: the spring balance responds to the pull of gravity itself, so it measures weight; the beam balance compares two pulls, so it measures mass.
3.18.3 Comparison Of Mass And Weight
3.19 Gravitational Units Of Force
The newton and the dyne are the absolute units of force, built from the kilogram, the metre and the second. But in shops, workshops and everyday speech, forces are often expressed in units built directly on the pull of gravity. These are called the gravitational units of force.
In the M.K.S. system, the gravitational unit of force is the kilogram force, whose symbol is kgf.
The value follows from \(W = mg\):
\[1\\ \\text{kgf} = \\text{force due to gravity on a mass of }1\\ \\text{kg} = 1\\ \\text{kg} \\times g\\ \\text{m s}^{-2} = g\\ \\text{newton}\]
Since the average value of \(g\) is \(9.8\\ \\text{m s}^{-2}\),
\[1\\ \\text{kgf} = 9.8\\ \\text{N}\]
In the C.G.S. system, the gravitational unit of force is the gram force, whose symbol is gf.
By the same reasoning,
\[1\\ \\text{gf} = 1\\ \\text{g} \\times g\\ \\text{cm s}^{-2} = g\\ \\text{dyne}\]
Since the average value of \(g\) is \(980\\ \\text{cm s}^{-2}\),
\[1\\ \\text{gf} = 980\\ \\text{dyne}\]
Further, since \(1\\ \\text{kg} = 1000\\ \\text{g}\), it follows that \(1\\ \\text{kgf} = 1000\\ \\text{gf}\).
Getting A Feel For One Newton
To an approximation, \(1\\ \\text{kgf}\) is assumed to be nearly equal to \(10\\ \\text{N}\). On that approximation,
\[1\\ \\text{N} = 0.1\\ \\text{kgf} = 100\\ \\text{gf}\]
So one can actually feel a force of \(1\\ \\text{N}\) by holding a mass of \(100\\ \\text{g}\) on the palm, as shown in Fig. 3.19-1. That is roughly a small apple, or a bar of soap.
Obviously, if we say that \(1\\ \\text{kgf} = 9.8\\ \\text{N}\), we mean that we have to exert an upward force of \(9.8\\ \\text{N}\) to hold a mass of \(1\\ \\text{kg}\) at rest on our palm.
Diagram
Fig. 3.19-1 — A Force Of One Newton Exerted By The Palm An open hand, palm upwards, holding a small block resting on it. Label the block ‘100 g’. Draw a vertical arrow upwards from the palm through the block, labelled ‘1 N’. Add a short downward arrow from the block onto the palm labelled ‘weight of the 100 g mass’. Keep both arrows equal in length to show that the palm holds the mass at rest.
3.20 Summary
- A force applied on a body can produce two main effects: it can change the state of rest or of motion of the body, and it can change the size or the shape of the body.
- A force applied on a free rigid object changes only its motion, not its dimensions. A force applied on a non-rigid object changes its dimensions, and can also produce motion in it.
- Force is that physical cause which changes, or tends to change, either the size or the shape or the state of rest or of motion of a body. Force is a vector quantity, and its S.I. unit is the newton (N).
- Contact forces are applied on bodies by making a physical contact with them. Examples are the frictional force, the normal reaction force, the force of tension in a string, the force exerted by a spring, and the force exerted on two bodies during a collision.
- Friction acts along the surfaces in contact, in the direction opposite to the motion of the body, and comes into play as soon as the body tends to move. The normal reaction acts perpendicular to the surface in contact.
- For a body hanging at rest from a string, the tension equals the weight of the body, \(T = W\), where \(T\) and \(W\) are both in newton (N).
- A stretched or compressed spring exerts a restoring force \(F\) on the object at its free end, opposite to the displacement of that end, whose magnitude is directly proportional to the elongation or the compression.
- Non-contact forces are experienced by bodies even without being physically in touch. They are the gravitational force, the electrostatic force and the magnetic force.
- Non-contact forces act at a distance, may be attractive or repulsive, are long range, act even in vacuum, and vary inversely as the square of the separation: \[F \\propto \\frac{1}{r^{2}}\] where \(F\) is the force in newton (N) and \(r\) is the separation in metre (m). On doubling the separation, the force falls to one-fourth.
- Newton’s first law of motion: if a body is in a state of rest, it will remain at rest, and if it is in a state of motion, it will remain moving in the same direction with the same speed, unless an external force is applied on it.
- The first law gives the definition of inertia — the property of an object by virtue of which it tends to retain its state of rest or of motion — and the definition of force, as the external cause which tends to change the state of rest or the state of motion of an object.
- Mass is a measure of inertia. The greater the mass of a body, the greater is its inertia.
- Inertia is of two kinds: inertia of rest, by which a body at rest remains at rest, and inertia of motion, by which a body in motion keeps moving with the same speed in the same straight line.
- The linear momentum of a body is the product of its mass and its velocity: \[p = mv\] where \(p\) is the momentum in \(\\text{kg m s}^{-1}\), \(m\) the mass in kilogram (kg) and \(v\) the velocity in \(\\text{m s}^{-1}\). It is a vector, directed along the velocity. Its C.G.S. unit is \(\\text{g cm s}^{-1}\).
- The change in momentum is \(\\Delta p = \\Delta(mv)\), which is always true. If the mass remains constant, this becomes \[\\Delta p = m,\\Delta v\] where \(\\Delta v\) is the change in velocity in \(\\text{m s}^{-1}\).
- For a constant mass, the rate of change of momentum equals the product of mass and acceleration: \[\\frac{\\Delta p}{\\Delta t} = ma\] where \(\\Delta t\) is the time in second (s) and \(a\) the acceleration in \(\\text{m s}^{-2}\).
- Newton found experimentally that \(a \\propto F\) for a constant mass, and \(F \\propto m\) for a constant acceleration. Combining these and choosing the unit of force suitably gives \[F = m \\times a\] where \(F\) is the force in newton (N). For a given force, \(a \\propto 1/m\).
- One newton is the force which, acting on a body of mass \(1\\ \\text{kg}\), produces an acceleration of \(1\\ \\text{m s}^{-2}\). One dyne is the force which, acting on a body of mass \(1\\ \\text{g}\), produces an acceleration of \(1\\ \\text{cm s}^{-2}\). They are related by \[1\\ \\text{N} = 10^{5}\\ \\text{dyne}\]
- Newton’s second law of motion: the rate of change of momentum of a body is directly proportional to the force applied on it, and the change in momentum takes place in the direction in which the force is applied. Its general form is \[F = \\frac{\\Delta p}{\\Delta t} = \\frac{\\Delta(mv)}{\\Delta t}\]
- This general form reduces to \(F = ma\) under two conditions: that the velocities are much smaller than the velocity of light, and that the mass remains constant.
- Putting \(F = 0\) in \(F = ma\) gives \(a = 0\), so a body not acted upon by any external force does not change its state of rest or of motion. Thus the first law is contained in the second law.
- Since \(\\Delta p = F \\times \\Delta t\), the same change in momentum can be produced by a large force acting for a short time or a small force acting for a long time. Increasing the stopping time therefore reduces the force — the reason a cricketer withdraws his hands while catching, an athlete lands on sand, and a glass vessel survives a fall on a carpet.
- Newton’s third law of motion: to every action there is always an equal and opposite reaction. For two bodies \(A\) and \(B\), \[\\vec{F}*{AB} = -,\\vec{F}*{BA}\] Action and reaction are equal in magnitude, opposite in direction, act simultaneously, and always act on two different bodies.
- Newton’s universal law of gravitation: the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, and acts along the line joining them: \[F = G,\\frac{m\_1 m\_2}{r^{2}}\] where \(F\) is in newton (N), \(m\_1\) and \(m\_2\) in kilogram (kg), \(r\) in metre (m), and \(G\) is the universal gravitational constant.
- \(G\) is numerically equal to the force of attraction between two masses of \(1\\ \\text{kg}\) each placed \(1\\ \\text{m}\) apart. Its value is \(G = 6.67 \\times 10^{-11}\\ \\text{N m}^{2},\\text{kg}^{-2}\), and it is the same everywhere, independent of the medium, the temperature and the nature of the bodies.
- The gravitational force is always attractive, obeys the inverse square law, and is significant between heavenly bodies but insignificant between ordinary bodies because of the small magnitude of \(G\).
- The force due to gravity on a body of mass \(m\) at the surface of the Earth of mass \(M\) and radius \(R\) is \[F = \\frac{GMm}{R^{2}}\] and it acts vertically downwards at the centre of gravity of the body. For \(m = 1\\ \\text{kg}\) it works out to \(9.8\\ \\text{N}\).
- All bodies falling freely in vacuum from the same height reach the ground together, with the same velocity, whatever their mass, size or shape. Their motion is uniformly accelerated, and this acceleration is called the acceleration due to gravity \(g\), measured in \(\\text{m s}^{-2}\) and directed vertically downwards.
- The mean value of \(g\) at the Earth’s surface is \(9.8\\ \\text{m s}^{-2}\). It is slightly less at the equator than at the poles, decreases with altitude and with depth, and is zero at the centre of the Earth. On the Moon’s surface it is nearly one-sixth of its value on the Earth.
- Equating \(mg\) with \(GMm/R^{2}\) gives \[g = \\frac{GM}{R^{2}}\] so the value of \(g\) on a planet or satellite depends only on its mass and its radius, and not on the falling body.
- For a body falling freely from rest, \(v = gt\), \(h = \\tfrac{1}{2}gt^{2}\) and \(v^{2} = 2gh\). If it starts with a downward velocity \(u\), then \(v = u + gt\), \(h = ut + \\tfrac{1}{2}gt^{2}\) and \(v^{2} = u^{2} + 2gh\), with \(h\) in metre (m) and \(t\) in second (s).
- For a body thrown vertically upwards with an initial velocity \(u\), the acceleration is \(-g\), so \(v = u - gt\), \(h = ut - \\tfrac{1}{2}gt^{2}\) and \(v^{2} = u^{2} - 2gh\). At the highest point \(v = 0\), giving \[h\_{\\text{max}} = \\frac{u^{2}}{2g}, \\qquad t = \\frac{u}{g}, \\qquad t' = \\frac{2u}{g}, \\qquad h' = \\frac{u^{2}}{g}\]
- Mass is the quantity of matter in a body. It is a scalar, measured in kilogram (kg) by a physical balance, and does not change with place. Weight is the force with which the Earth attracts a body. It is a vector directed downwards, measured in newton (N) by a spring balance, and it varies from place to place because \(g\) does.
- Weight and mass are related by \[W = mg\] where \(W\) is in newton (N), \(m\) in kilogram (kg) and \(g\) in \(\\text{m s}^{-2}\), which may also be written as \(\\text{N kg}^{-1}\).
- The gravitational units of force are the kilogram force and the gram force, where \(1\\ \\text{kgf} = 9.8\\ \\text{N}\), \(1\\ \\text{gf} = 980\\ \\text{dyne}\) and \(1\\ \\text{kgf} = 1000\\ \\text{gf}\). A body of mass \(m\) kilogram weighs \(m\) kgf. Approximately, \(1\\ \\text{kgf} = 10\\ \\text{N}\), so \(1\\ \\text{N} = 100\\ \\text{gf}\), which is the force needed to hold a \(100\\ \\text{g}\) mass on the palm.
Flowchart
FC 3.2 — From The Law Of Gravitation To The Weight Of A Body Universal law of gravitation, F = Gm₁m₂/r² → applied to the Earth and a body on its surface, F = GMm/R² → this force is the force due to gravity on the body → by the second law it equals mg → cancelling the mass m gives g = GM/R² → multiplying back by the mass gives the weight, W = mg
3.21 Key Terms
- Acceleration due to gravity — the rate at which the velocity of a freely falling body increases, produced by the gravitational attraction of the Earth; a vector directed vertically downwards. S.I. unit: m s⁻² (also written N kg⁻¹).
- Contact force — a force applied on a body by making physical contact with it. S.I. unit: newton (N).
- Dyne — the C.G.S. unit of force; the force which, acting on a mass of 1 g, produces an acceleration of 1 cm s⁻².
- Electrostatic force — the non-contact force between electric charges, attractive between unlike charges and repulsive between like charges. S.I. unit: newton (N).
- Force — that physical cause which changes, or tends to change, the size, the shape, or the state of rest or of motion of a body; a vector quantity. S.I. unit: newton (N).
- Force of gravity — the force with which the Earth attracts a body towards its centre; the same as the weight of the body. S.I. unit: newton (N).
- Frictional force — the contact force acting along the surfaces in contact, opposite to the motion or the tendency to motion of a body. S.I. unit: newton (N).
- Gram force — the C.G.S. gravitational unit of force; the force due to gravity on a mass of 1 g, equal to 980 dyne.
- Gravitational constant — the universal constant G in Newton’s law of gravitation, numerically equal to the force between two masses of 1 kg each placed 1 m apart; its value is 6.67 × 10⁻¹¹ N m² kg⁻². S.I. unit: N m² kg⁻².
- Gravitational force — the always attractive non-contact force between any two masses in the universe. S.I. unit: newton (N).
- Inertia — the property of an object by virtue of which it tends to retain its state of rest or of motion; it has no unit, and mass is its measure.
- Inertia of motion — the property by which a body in motion continues to move with the same speed in the same straight line unless an external force acts on it.
- Inertia of rest — the property by which a body at rest remains at rest unless an external force acts on it.
- Kilogram force — the M.K.S. gravitational unit of force; the force due to gravity on a mass of 1 kg, equal to 9.8 N.
- Linear momentum — the product of the mass of a body and its velocity, a vector directed along the velocity. S.I. unit: kg m s⁻¹ (C.G.S. unit: g cm s⁻¹).
- Magnetic force — the non-contact force between magnetic poles, attractive between unlike poles and repulsive between like poles. S.I. unit: newton (N).
- Mass — the quantity of matter contained in a body; a scalar quantity that does not change with place. S.I. unit: kilogram (kg).
- Newton — the S.I. unit of force; the force which, acting on a mass of 1 kg, produces an acceleration of 1 m s⁻²; 1 N = 10⁵ dyne.
- Non-contact force — a force experienced by a body without any physical touch, also called a force at a distance. S.I. unit: newton (N).
- Normal reaction force — the contact force exerted by a surface on a body resting on it, acting perpendicular to the surface. S.I. unit: newton (N).
- Restoring force — the force exerted by a stretched or compressed spring, opposite to the displacement of its free end and proportional to the elongation or compression. S.I. unit: newton (N).
- Tension — the force developed in a stretched string, pulling the body attached to it along the string. S.I. unit: newton (N).
- Weight — the force with which the Earth attracts a body; a vector directed vertically downwards, given by W = mg. S.I. unit: newton (N).
3.22 Points To Ponder
Newton’s third law says the Earth pulls an apple down and the apple pulls the Earth up with an equal force. Why, then, does the apple visibly rush towards the Earth while the Earth shows no motion at all? Would your answer change if the apple were replaced by another planet?
A book lying on a table is pulled down by its weight and pushed up by the table, and these two forces are equal and opposite. Yet they are not an action–reaction pair in the sense of the third law. What exactly makes a pair of forces an action–reaction pair, and how would you decide the case for the book?
The relation \(g = GM/R^{2}\) contains no reference whatever to the falling body. Trace back through the derivation and identify the exact step at which the body’s mass disappeared. What would fall differently if that cancellation had not happened?
Friction is usually described as a nuisance that slows things down and wastes energy. But a person cannot walk, and a vehicle cannot start or stop, on a surface with no friction at all. Where would you place friction on the scale between an obstacle and a necessity, and what evidence from this chapter supports your view?
Both a bat hitting a ball and a cricketer catching a ball are explained by the same relation, \(\\Delta p = F \\times \\Delta t\). In one case we want the contact time short, in the other we want it long. What decides which of the two we should aim for in a given situation?
Suppose the gravitational constant \(G\) were a hundred times larger than it actually is, everything else remaining the same. Which of the following would change noticeably, and which would not: the weight of a body on the Earth, the force between two students sitting a metre apart, the orbit of the Moon, and the reading of a beam balance?
3.23 Objective Questions
- A force acting on a body cannot: (1 mark)
- change its shape
- change its direction of motion
- change its mass
- bring it to rest
- Which of the following is a contact force? (1 mark)
- the force between two magnetic poles
- the restoring force of a spring
- the force between a comb and bits of paper
- the force of gravity on a falling coin
- If the separation between two masses is made three times larger, the gravitational force between them becomes: (1 mark)
- three times
- one-third
- one-sixth
- one-ninth
- The S.I. unit of linear momentum is: (1 mark)
- \(\\text{kg m s}^{-2}\)
- \(\\text{kg m s}^{-1}\)
- \(\\text{N s}^{-1}\)
- \(\\text{g cm s}^{-1}\)
- For a body of fixed mass, the graph of acceleration against the applied force is: (1 mark)
- a straight line through the origin
- a straight line not passing through the origin
- a hyperbola
- a horizontal straight line
- A body is taken from the Earth to the Moon. Which of these remains unchanged? (1 mark)
- its weight
- the force of gravity on it
- its mass
- the value of \(g\) acting on it
- Two bodies of masses \(m\) and \(3m\) are acted upon by the same force. The ratio of the accelerations produced in them is: (1 mark)
- \(1 : 3\)
- \(3 : 1\)
- \(1 : 1\)
- \(1 : 9\)
- One newton is equal to: (1 mark)
- \(10^{3}\) dyne
- \(10^{5}\) dyne
- \(10^{-5}\) dyne
- \(980\) dyne
Choose the answer based on the codes given below. a. Both A and R are true, and R is the correct explanation of A b. Both A and R are true, but R is not the correct explanation of A c. A is false but R is true d. A is true but R is false
Assertion (A): A body moving on a perfectly smooth horizontal surface needs no force to keep it moving. Reason (R): A body continues in its state of uniform motion unless an external force acts on it. (1 mark)
Assertion (A): The action and the reaction of a pair cancel each other, so no motion is ever produced. Reason (R): Action and reaction are equal in magnitude and opposite in direction. (1 mark)
Assertion (A): The value of \(g\) is the same everywhere on the surface of the Earth. Reason (R): The acceleration due to gravity does not depend on the mass of the falling body. (1 mark)
Assertion (A): A parachute reduces the force with which a jumper lands. Reason (R): It increases the time in which the jumper’s momentum is brought to zero. (1 mark)
Fill in the blanks.
- The property of a body by which it resists a change in its state of rest or of motion is called ……… , and ……… is its measure. (1 mark)
- The gravitational force between two bodies is ……… proportional to the product of their masses and ……… proportional to the square of the distance between them. (1 mark)
- For a body thrown vertically upwards, the acceleration due to gravity acts ……… to the direction of motion, so the body ……… down. (1 mark)
State whether the following statements are true or false.
- Action and reaction always act on the same body. (1 mark)
- The mass of a body measured by a beam balance would give the same reading on the Moon as on the Earth. (1 mark)
- The relation \(\\Delta p = m,\\Delta v\) holds for every moving body without exception. (1 mark)
Name the following.
- The force that comes into play as soon as a body tends to slide over a rough surface. (1 mark)
- The scientist whose experiments showed that a moving body needs no force to keep it moving. (1 mark)
- The kind of inertia shown when the dust in a carpet stays behind as the carpet is beaten. (1 mark)
Give one word or phrase for each of the following.
- The product of the mass of a body and its velocity. (1 mark)
- The force with which the Earth attracts a body towards its centre. (1 mark)
State the S.I. unit of the following.
- Linear momentum. (1 mark)
- The universal gravitational constant. (1 mark)
- Acceleration due to gravity, in two equivalent forms. (1 mark)
Pick the odd one out and give a reason for your choice.
- Frictional force, tension, magnetic force, normal reaction force. (1 mark)
- Newton, dyne, kilogram force, kilogram. (1 mark)
Match each item in Column A with the correct item in Column B. (1 mark)
| Column A | Column B |
|---|---|
| a. \(G\) | i. \(\\text{m s}^{-2}\) |
| b. \(g\) | ii. \(\\text{kg m s}^{-1}\) |
| c. \(p\) | iii. newton |
| d. \(W\) | iv. \(\\text{N m}^{2},\\text{kg}^{-2}\) |
3.24 Check Your Concepts
Define a non-contact force and give one example of it. (2 marks)
State the S.I. and C.G.S. units of force, and write the relation between them. (2 marks)
Give a reason: a spring balance measures weight, but a beam balance measures mass. (2 marks)
What is meant by the acceleration due to gravity? State its mean value at the Earth’s surface, with the unit. (2 marks)
Why does a body weigh less on the Moon than on the Earth, although the quantity of matter in it is unchanged? (2 marks)
Differentiate between mass and weight, stating the basis of difference in each case. (3 marks)
State Newton’s first law of motion. How does this law give both the definition of inertia and the definition of force? (3 marks)
Give a reason: a person stepping out of a moving bus is advised to run for a short distance in the direction the bus was moving. (3 marks)
State the universal law of gravitation, with the condition on direction, and write its mathematical form naming every symbol with its unit. (3 marks)
What would happen to the value of \(g\) at the surface of a planet if its mass were doubled while its radius stayed the same? Justify your answer from a relation. (3 marks)
Show that the rate of change of momentum of a body is equal to the product of its mass and its acceleration. Under what condition does this relation hold? (4 marks)
Explain, with the help of \(\\Delta p = F \\times \\Delta t\), why a glass tumbler is far more likely to break on a stone floor than on a thick carpet. (4 marks)
State Newton’s third law of motion. Explain, with one example, why the action and the reaction never cancel each other even though they are equal and opposite. (4 marks)
Derive Newton’s first law of motion from the second law, and explain what the derivation tells us about the relation between the two laws. (4 marks)
Starting from the force of gravity on a body at the Earth’s surface, derive the relation \(g = GM/R^{2}\). Explain why the mass of the falling body does not appear in the final relation, and what this means for two bodies of different masses dropped together in vacuum. (5 marks)
3.25 Diagrams, Graphs & Constructions
Draw a neat labelled diagram to show the forces acting on a block resting on a horizontal table top. Name each force and mark its direction and point of application. (3 marks)
Fig. 3.2-3 shows a body hanging at rest from a string tied to a rigid support. Redraw the diagram, label the two forces acting on the body, and state the relation between them, giving the condition under which it holds. (3 marks)
Fig. 3.10-1 shows the graph of acceleration against applied force for a body of fixed mass. State what quantity the gradient of this line represents, and explain why the line must pass through the origin. (3 marks)
Sketch the shape of the graph of acceleration against mass when the same force is applied to bodies of different masses. Label both axes with their units and state the relation the shape represents. (3 marks)
Fig. 3.13-4 shows a rocket rising as gases are expelled from its nozzle. Redraw the outline and mark the action and the reaction, stating clearly which body each force acts on. (3 marks)
Draw a labelled diagram of the arrangement of two spring balances used to demonstrate Newton’s third law, and state the observation that is made. (4 marks)
Draw a sketch of the path of a ball thrown vertically upwards from the ground. Mark on it the point where the velocity is zero, the maximum height, and the direction of the acceleration on the way up and on the way down. (4 marks)
3.26 Practice With Numericals
A body of mass \(4\\ \\text{kg}\) is moving with a velocity of \(3\\ \\text{m s}^{-1}\). Calculate its linear momentum. (2 marks)
A force of \(20\\ \\text{N}\) acts on a body of mass \(4\\ \\text{kg}\). Find the acceleration produced in it. (2 marks)
The linear momentum of a ball of mass \(200\\ \\text{g}\) is \(1.2\\ \\text{kg m s}^{-1}\). Find its velocity. (2 marks)
Express a force of \(3.5\\ \\text{N}\) in dyne. (2 marks)
Calculate the weight of a body of mass \(8\\ \\text{kg}\) in (a) newton and (b) kilogram force. Take \(g = 9.8\\ \\text{m s}^{-2}\). (2 marks)
Two bodies attract each other with a force of \(36\\ \\text{N}\) at a certain separation. Find the force of attraction between them when the separation is made three times as large, the masses remaining the same. (2 marks)
A body of mass \(500\\ \\text{g}\), initially at rest, is acted upon by a constant force of \(2\\ \\text{N}\) for \(4\\ \\text{s}\). Calculate (i) the acceleration produced, (ii) the velocity acquired, and (iii) the change in momentum of the body. (3 marks)
A cricket ball of mass \(200\\ \\text{g}\) moving at \(15\\ \\text{m s}^{-1}\) is brought to rest by a fielder in \(0.04\\ \\text{s}\). Find the average force applied by the fielder, and state its direction. (3 marks)
Calculate the gravitational force of attraction between two bodies of masses \(50\\ \\text{kg}\) and \(70\\ \\text{kg}\) placed \(5\\ \\text{m}\) apart. Take \(G = 6.67 \\times 10^{-11}\\ \\text{N m}^{2},\\text{kg}^{-2}\). (3 marks)
A stone is dropped from rest from the top of a tower and reaches the ground in \(5\\ \\text{s}\). Taking \(g = 10\\ \\text{m s}^{-2}\), find the height of the tower and the velocity with which the stone strikes the ground. (3 marks)
The weight of a body on the Earth is \(147\\ \\text{N}\), where \(g = 9.8\\ \\text{m s}^{-2}\). Find (a) its mass, and (b) its weight on the Moon, where \(g = 1.6\\ \\text{m s}^{-2}\). (3 marks)
A ball is thrown vertically upwards with a velocity of \(30\\ \\text{m s}^{-1}\). Taking \(g = 10\\ \\text{m s}^{-2}\), calculate (i) the maximum height reached, (ii) the time taken to reach that height, and (iii) the total time of journey. (3 marks)
A car of mass \(1000\\ \\text{kg}\) moving at \(72\\ \\text{km h}^{-1}\) is brought to rest in \(8\\ \\text{s}\) by applying the brakes. Calculate the retardation produced and the braking force applied. (4 marks)
A planet has a mass of \(3.0 \\times 10^{23}\\ \\text{kg}\) and a radius of \(2.0 \\times 10^{6}\\ \\text{m}\). Taking \(G = 6.67 \\times 10^{-11}\\ \\text{N m}^{2},\\text{kg}^{-2}\), calculate the acceleration due to gravity at its surface, and find the weight there of a body of mass \(6\\ \\text{kg}\). (4 marks)
A bullet of mass \(20\\ \\text{g}\) moving at \(200\\ \\text{m s}^{-1}\) strikes a wooden block and is brought to rest in \(0.02\\ \\text{s}\). Calculate (i) the initial momentum of the bullet, (ii) the change in its momentum, and (iii) the average resistive force exerted by the block. (4 marks)
A body of mass \(2\\ \\text{kg}\) is released from rest and falls freely for \(3\\ \\text{s}\). Taking \(g = 9.8\\ \\text{m s}^{-2}\), calculate the velocity it gains, the momentum it gains, and the distance it falls through in that time. (5 marks)