Force And Laws Of Motion

So students, welcome to today's science lesson. I am so happy to be here with you to learn about one of the most fundamental and interesting chapters in physics — Chapter 8: Force and Laws of Motion.

Now, before we begin, let me ask you something. Have you ever wondered why a ball that you kick on the ground eventually stops? Or why do you fall forward when a bus suddenly brakes? Or how is it that when you push a heavy box, it moves only when your push is strong enough? All these questions are related to something very important in science — the concept of force and motion. And today, we are going to understand all of this in detail.

So let's start from the very beginning. In the previous chapter, you learned about motion along a straight line — about position, velocity, and acceleration. You learned that motion can be uniform or non-uniform. But we never really understood what causes this motion. Why does the speed of an object change with time? Do all motions require a cause? If so, what is the nature of this cause? In this chapter, we are going to answer all these curious questions.

For many centuries, the problem of motion and its causes had puzzled scientists and philosophers. Think about this — when you place a ball on the ground and give it a small hit, it does not move forever. It eventually stops. Such observations led people to believe that rest is the "natural state" of an object. This was the common belief for a very long time, until two great scientists — Galileo Galilei and Isaac Newton — developed an entirely different approach to understand motion.

Now, let's think about our everyday experiences. In our daily life, we observe that some effort is required to put a stationary object into motion, or to stop a moving object. We experience this as a muscular effort, and we say that we must push, hit, or pull an object to change its state of motion. This push, hit, or pull is what we call a force.

So students, what is a force? Let me tell you — no one has ever seen, tasted, or felt a force directly. We cannot see force as we see a table or a chair. However, we always see or feel the effect of a force. It can only be explained by describing what happens when a force is applied to an object. When you push a shopping cart, it moves. When you hit a cricket ball with a bat, the ball flies away. When you pull a drawer, it opens. All these are examples of pushing, hitting, or pulling objects, and all these are ways of bringing objects into motion by applying a force.

Now, from your studies in earlier classes, you are also familiar with the fact that a force can be used to change the magnitude of velocity of an object. That means a force can make an object move faster or slower. A force can also change the direction of motion of an object. And we also know that a force can change the shape and size of objects. For example, when you press a spring, it compresses and changes its shape. When you squeeze a rubber ball, it becomes oblong. All these are effects of force.

Now, let us move on to a very important concept — Balanced and Unbalanced Forces.

Imagine a wooden block placed on a horizontal table. Two strings, X and Y, are tied to the two opposite faces of the block. If you apply a force by pulling string X, the block begins to move to the right. Similarly, if you pull string Y, the block moves to the left. But here is the interesting part — if the block is pulled from both sides with equal forces, the block will not move at all. Such forces are called balanced forces. Balanced forces do not change the state of rest or the state of motion of an object. They cancel each other out.

Now, consider a different situation. Suppose two opposite forces of different magnitudes pull the block. In this case, the block would begin to move in the direction of the greater force. The two forces are not balanced, and the unbalanced force acts in the direction the block moves. This suggests that an unbalanced force acting on an object brings it into motion.

Now, let me give you a very common example from everyday life. What happens when some children try to push a box on a rough floor? If they push the box with a small force, the box does not move. Why does this happen? It happens because of friction. Friction is a force that acts in a direction opposite to the push. This friction force arises between two surfaces in contact — in this case, between the bottom of the box and the floor. The friction force balances the pushing force, and therefore the box does not move.

Now, if the children push the box harder but still not hard enough, the box still does not move. This is because the friction force still balances the pushing force. But if the children push the box even harder, the pushing force becomes bigger than the friction force. Now there is an unbalanced force. So the box starts moving. This is exactly how things work in the real world.

Now, think about riding a bicycle. When you stop pedaling, the bicycle begins to slow down. This is again because of friction forces acting opposite to the direction of motion. In order to keep the bicycle moving, you have to start pedaling again.

So students, let me summarize what we have learned so far. An object moves with a uniform velocity when the forces acting on it are balanced and there is no net external force on it. If an unbalanced force is applied on the object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required. And the change in its speed or in the direction of motion would continue as long as this unbalanced force is applied. However, if this force is removed completely, the object would continue to move with the velocity it has acquired till then.

Now, let's move on to the First Law of Motion. This is a very important part of our chapter, so pay close attention.

Galileo Galilei, the famous Italian scientist, conducted many experiments to understand motion. He observed the motion of objects on inclined planes. He noticed that when a marble rolls down an inclined plane, its velocity increases. This is because the marble falls under the unbalanced force of gravity as it rolls down, and it attains a definite velocity by the time it reaches the bottom. When the marble climbs up an inclined plane, its velocity decreases.

Galileo also studied a marble resting on an ideal frictionless plane that is inclined on both sides — what we call a double inclined plane. He argued that when the marble is released from the left side, it would roll down the slope and go up on the opposite side to the same height from which it was released. If the inclinations of the planes on both sides are equal, then the marble will climb the same distance that it covered while rolling down.

Now here is the interesting part. If the angle of inclination of the right-side plane is gradually decreased, then the marble would travel further distances till it reaches the original height. If the right-side plane is ultimately made horizontal — that is, the slope is reduced to zero — the marble would continue to travel forever, trying to reach the same height that it was released from. The unbalanced forces on the marble in this case are zero.

This suggests that an unbalanced external force is required to change the motion of the marble, but no net force is needed to sustain the uniform motion of the marble. In practical situations, it is difficult to achieve zero unbalanced force because of the presence of friction. So in practice, the marble stops after traveling some distance. The effect of friction can be minimized by using a smooth marble and a smooth plane and by providing a lubricant on top of the planes.

Now, Newton further studied Galileo's ideas on force and motion and presented three fundamental laws that govern the motion of objects. These three laws are known as Newton's laws of motion.

The first law of motion is stated as:

An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force.

In other words, all objects resist a change in their state of motion. This tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. This is why the first law of motion is also known as the law of inertia.

Now, let me explain this law with some real-life examples that you can relate to, especially from your experiences while traveling in a motorcar.

When you are sitting in a car and the driver applies brakes suddenly, the car slows down, but your body tends to continue in the same state of motion because of its inertia. This is why you feel a jerk forward. A sudden application of brakes may thus cause injury to you by impact with the panels in front. This is exactly why safety belts are worn. Safety belts exert a force on your body to make the forward motion slower, protecting you from getting hurt.

Now, the opposite experience happens when you are standing in a bus and the bus begins to move suddenly. You tend to fall backwards. This is because the sudden start of the bus brings motion to the bus as well as to your feet in contact with the floor of the bus. But the rest of your body opposes this motion because of its inertia.

Similarly, when a motorcar makes a sharp turn at a high speed, you tend to get thrown to one side. This can again be explained on the basis of the law of inertia. You tend to continue in your straight-line motion. When an unbalanced force is applied by the engine to change the direction of motion of the motorcar, you slip to one side of the seat due to the inertia of your body.

Now, let me tell you about some activities that demonstrate the law of inertia.

Activity 8.1: Make a pile of similar carom coins on a table. Now, attempt a sharp horizontal hit at the bottom of the pile using another carom coin or the striker. If the hit is strong enough, the bottom coin moves out quickly. Once the lowest coin is removed, the inertia of the other coins makes them fall vertically on the table. This happens because the coins above were at rest and wanted to remain at rest.

Activity 8.2: Set a five-rupee coin on a stiff card covering an empty glass tumbler standing on a table. Give the card a sharp horizontal flick with your finger. If you do it fast, the card shoots away, allowing the coin to fall vertically into the glass tumbler due to its inertia. The inertia of the coin tries to maintain its state of rest even when the card moves off.

Activity 8.3: Place a water-filled tumbler on a tray. Hold the tray and turn around as fast as you can. Observe that the water spills. Why does this happen? It happens because of inertia. The water tends to stay in its original position while the tray moves, causing the water to spill.

Now, let me tell you an interesting observation. Have you noticed that a saucer for a tea cup has a groove? This groove is provided for placing the tea cup. It prevents the cup from toppling over in case of sudden jerks, again because of inertia.

Now, let's move on to understand the relationship between inertia and mass.

All the examples and activities we have discussed illustrate that there is a resistance offered by an object to change its state of motion. If it is at rest, it tends to remain at rest. If it is moving, it tends to keep moving. This property of an object is called its inertia.

Now, the important question is — do all bodies have the same inertia? Let's think about this. Is it easier to push an empty box or a box full of books? Obviously, it is easier to push an empty box. Similarly, if you kick a football, it flies away. But if you kick a stone of the same size with equal force, it hardly moves. In fact, you might get an injury in your foot while doing so! Similarly, in Activity 8.2, if we use a one-rupee coin instead of a five-rupee coin, we find that a lesser force is required to perform the activity.

A force that is just enough to cause a small cart to pick up a large velocity will produce a negligible change in the motion of a train. This is because, in comparison to the cart, the train has a much lesser tendency to change its state of motion. We say that the train has more inertia than the cart.

Clearly, heavier or more massive objects offer larger inertia. Quantitatively, the inertia of an object is measured by its mass. We may thus relate inertia and mass as follows:

Inertia is the natural tendency of an object to resist a change in its state of motion or of rest. The mass of an object is a measure of its inertia.

Now, let's answer some questions based on what we have learned so far.

Question 1: Which of the following has more inertia: (a) a rubber ball and a stone of the same size? (b) a bicycle and a train? (c) a five-rupees coin and a one-rupee coin?

Let me explain each part:

(a) Between a rubber ball and a stone of the same size, the stone has more mass. Since mass is a measure of inertia, the stone has more inertia.

(b) Between a bicycle and a train, the train has much more mass. So the train has more inertia.

(c) Between a five-rupees coin and a one-rupee coin, the five-rupee coin is heavier and has more mass. So the five-rupee coin has more inertia.

Question 2: In the following example, try to identify the number of times the velocity of the ball changes: "A football player kicks a football to another player of his team who kicks the football towards the goal. The goalkeeper of the opposite team collects the football and kicks it towards a player of his own team". Also identify the agent supplying the force in each case.

Let me analyze this step by step. The football is kicked by the first player — this is the first time the velocity changes. The second player kicks the football towards the goal — this is the second change in velocity. The goalkeeper collects the football and kicks it — this is the third change in velocity. So the velocity changes three times.

The agents supplying the force are: first player, second player, and goalkeeper.

Question 3: Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.

When we vigorously shake a branch of a tree, the branch moves suddenly. However, the leaves tend to remain in their state of rest due to their inertia. The force exerted by the branch on the leaves is not enough to make the leaves move along with it. As a result, some leaves get detached from the tree.

Question 4: Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?

When a moving bus brakes to a stop, the bus slows down, but our body tends to continue moving forward due to its inertia. So we fall in the forward direction.

When the bus accelerates from rest, the bus comes into motion, but our body tends to remain at rest due to its inertia. So we fall backwards.

Now, let's move on to the Second Law of Motion.

The first law of motion indicates that when an unbalanced external force acts on an object, its velocity changes — that is, the object gets an acceleration. Now we want to study how the acceleration of an object depends on the force applied to it and how we measure a force.

Let me recount some observations from our everyday life. During a game of table tennis, if the ball hits a player, it does not hurt him. On the other hand, when a fast-moving cricket ball hits a spectator, it may hurt him. A truck at rest does not require any attention when parked along a roadside. But a moving truck, even at speeds as low as 5 meters per second, may kill a person standing in its path. A small mass, such as a bullet, may kill a person when fired from a gun.

These observations suggest that the impact produced by the objects depends on their mass and velocity. Similarly, if an object is to be accelerated, we know that a greater force is required to give a greater velocity.

In other words, there appears to exist some quantity of importance that combines the object's mass and its velocity. One such property, called momentum, was introduced by Newton. The momentum, p of an object is defined as the product of its mass, m and velocity, v. That is,

p = mv

Momentum has both direction and magnitude. Its direction is the same as that of velocity. The SI unit of momentum is kilogram-metre per second, written as kg m s⁻¹.

Since the application of an unbalanced force brings a change in the velocity of the object, it is therefore clear that a force also produces a change of momentum.

Now, let me explain this with an example. Consider a situation in which a car with a dead battery is to be pushed along a straight road to give it a speed of 1 meter per second, which is sufficient to start its engine. If one or two persons give a sudden push, which is an unbalanced force, to it, it hardly starts. But a continuous push over some time results in a gradual acceleration of the car to this speed. It means that the change of momentum of the car is not only determined by the magnitude of the force but also by the time during which the force is exerted. It may then also be concluded that the force necessary to change the momentum of an object depends on the time rate at which the momentum is changed.

The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.

Now, let's derive the mathematical formulation of the second law of motion.

Suppose an object of mass m is moving along a straight line with an initial velocity u. It is uniformly accelerated to velocity v in time t by the application of a constant force F throughout the time t. The initial and final momentum of the object will be p₁ = mu and p₂ = mv respectively.

The change in momentum is p₂ – p₁, which equals mv – mu, which equals m × (v – u).

The rate of change of momentum is m × (v – u)/t.

Or, the applied force F is proportional to m × (v – u)/t.

Let me write this as F = kma, where a is the acceleration, which is the rate of change of velocity, equal to (v – u)/t. The quantity k is a constant of proportionality.

The SI units of mass and acceleration are kg and m s⁻² respectively. The unit of force is so chosen that the value of the constant k becomes one. For this, one unit of force is defined as the amount that produces an acceleration of 1 m s⁻² in an object of 1 kg mass. That is, 1 unit of force = k × (1 kg) × (1 m s⁻²).

Thus, the value of k becomes 1. From the equation F = kma, we get

F = ma

The unit of force is kg m s⁻² or newton, which has the symbol N. The second law of motion gives us a method to measure the force acting on an object as a product of its mass and acceleration.

Now, the second law of motion is often seen in action in our everyday life. Have you noticed that while catching a fast-moving cricket ball, a fielder in the ground gradually pulls his hands backwards with the moving ball? In doing so, the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased, and therefore the impact of catching the fast-moving ball is also reduced. If the ball is stopped suddenly, then its high velocity decreases to zero in a very short interval of time. Thus, the rate of change of momentum of the ball will be large. Therefore, a large force would have to be applied for holding the catch, which may hurt the palm of the fielder.

In a high jump athletic event, the athletes are made to fall either on a cushioned bed or on a sand bed. This is to increase the time of the athlete's fall to stop after making the jump. This decreases the rate of change of momentum and hence the force. This is why they don't get injured.

Now, here's an interesting point. The first law of motion can be mathematically stated from the mathematical expression for the second law of motion. From F = ma, we can write F = m(v – u)/t, or Ft = mv – mu.

That is, when F = 0, v = u for whatever time t is taken. This means that the object will continue moving with uniform velocity u throughout the time t. If u is zero, then v will also be zero. That is, the object will remain at rest. This is exactly what the first law of motion states.

Now, let's solve some numerical examples to understand the second law of motion better.

Example 8.1: A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m s⁻¹ to 7 m s⁻¹. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object?

Solution: We have been given that u = 3 m s⁻¹ and v = 7 m s⁻¹, t = 2 s and m = 5 kg.

From the equation F = m(v – u)/t, we have

F = 5 kg × (7 m s⁻¹ – 3 m s⁻¹) / 2 s = 5 kg × 4 m s⁻¹ / 2 s = 10 N.

So the magnitude of the applied force is 10 N.

Now, if this force is applied for a duration of 5 s (t = 5 s), then the final velocity can be calculated by rewriting the equation as v = u + Ft/m.

On substituting the values of u, F, m and t, we get

v = 3 m s⁻¹ + (10 N × 5 s) / 5 kg = 3 m s⁻¹ + 10 m s⁻¹ = 13 m s⁻¹.

So the final velocity would be 13 m s⁻¹.

Example 8.2: Which would require a greater force — accelerating a 2 kg mass at 5 m s⁻² or a 4 kg mass at 2 m s⁻²?

Solution: From the equation F = ma, we have

For the first case: m₁ = 2 kg, a₁ = 5 m s⁻², so F₁ = m₁a₁ = 2 kg × 5 m s⁻² = 10 N.

For the second case: m₂ = 4 kg, a₂ = 2 m s⁻², so F₂ = m₂a₂ = 4 kg × 2 m s⁻² = 8 N.

Since F₁ = 10 N and F₂ = 8 N, we can say that accelerating a 2 kg mass at 5 m s⁻² would require a greater force.

Example 8.3: A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Solution: The initial velocity of the motorcar u = 108 km/h.

We need to convert this to meters per second. We know that 1 km = 1000 m and 1 hour = 3600 s.

So u = 108 × 1000 m / (60 × 60 s) = 108 × 1000 / 3600 m s⁻¹ = 30 m s⁻¹.

The final velocity of the motorcar v = 0 m s⁻¹.

The total mass of the motorcar along with its passengers = 1000 kg, and the time taken to stop the motorcar, t = 4 s.

From the equation F = m(v – u)/t, we have

F = 1000 kg × (0 – 30) m s⁻¹ / 4 s = 1000 kg × (-30) m s⁻¹ / 4 s = -7500 kg m s⁻² or -7500 N.

The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar. This is the braking force.

Example 8.4: A force of 5 N gives a mass m₁ an acceleration of 10 m s⁻² and a mass m₂ an acceleration of 20 m s⁻². What acceleration would it give if both the masses were tied together?

Solution: From the equation F = ma, we have m₁ = F/a₁ and m₂ = F/a₂.

Given a₁ = 10 m s⁻², a₂ = 20 m s⁻² and F = 5 N.

Thus, m₁ = 5 N / 10 m s⁻² = 0.50 kg, and m₂ = 5 N / 20 m s⁻² = 0.25 kg.

If the two masses were tied together, the total mass m would be

m = 0.50 kg + 0.25 kg = 0.75 kg.

The acceleration a produced in the combined mass by the 5 N force would be

a = F/m = 5 N / 0.75 kg = 6.67 m s⁻².

Example 8.5: The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in a figure. How much force does the table exert on the ball to bring it to rest?

Solution: The initial velocity of the ball is 20 cm s⁻¹. Due to the frictional force exerted by the table, the velocity of the ball decreases down to zero in 10 s. Thus, u = 20 cm s⁻¹, v = 0 cm s⁻¹ and t = 10 s.

Since the velocity-time graph is a straight line, it is clear that the ball moves with a constant acceleration. The acceleration a is

a = (v – u)/t = (0 cm s⁻¹ – 20 cm s⁻¹) / 10 s = -2 cm s⁻² = -0.02 m s⁻².

The mass of the ball is 20 g, which is 20/1000 kg = 0.02 kg.

The force exerted on the ball F is

F = ma = 0.02 kg × (-0.02 m s⁻²) = -0.0004 N.

The negative sign implies that the frictional force exerted by the table is opposite to the direction of motion of the ball.

Now, let's move on to the Third Law of Motion.

The first two laws of motion tell us how an applied force changes the motion and provide us with a method of determining the force. The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object.

In the game of football, sometimes we, while looking at the football and trying to kick it with a greater force, collide with a player of the opposite team. Both feel hurt because each applies a force to the other. In other words, there is a pair of forces and not just one force. The two opposing forces are also known as action and reaction forces.

Let me explain this with an experiment. Consider two spring balances connected together. The fixed end of balance B is attached with a rigid support, like a wall. When a force is applied through the free end of spring balance A, it is observed that both the spring balances show the same readings on their scales. It means that the force exerted by spring balance A on balance B is equal but opposite in direction to the force exerted by balance B on balance A. Any of these two forces can be called as action and the other as reaction. This gives us an alternative statement of the third law of motion — to every action there is an equal and opposite reaction. However, it must be remembered that the action and reaction always act on two different objects, simultaneously.

Now, let me give you a very common example. Suppose you are standing at rest and intend to start walking on a road. You must accelerate, and this requires a force in accordance with the second law of motion. Which is this force? Is it the muscular effort you exert on the road? Is it in the direction we intend to move? No, you push the road below backwards. The road exerts an equal and opposite force on your feet to make you move forward. This is how we walk!

It is important to note that even though the action and reaction forces are always equal in magnitude, these forces may not produce accelerations of equal magnitudes. This is because each force acts on a different object that may have a different mass.

When a gun is fired, it exerts a forward force on the bullet. The bullet exerts an equal and opposite force on the gun. This results in the recoil of the gun. Since the gun has a much greater mass than the bullet, the acceleration of the gun is much less than the acceleration of the bullet.

The third law of motion can also be illustrated when a sailor jumps out of a rowing boat. As the sailor jumps forward, the force on the boat moves it backwards.

Now, let's do Activity 8.4. Request two children to stand on two separate carts. Give them a bag full of sand or some other heavy object. Ask them to play a game of catch with the bag. Does each of them experience an instantaneous force as a result of throwing the sand bag? Yes, they do! When one child throws the bag towards the other, the throwing child experiences a backward force due to the reaction of the bag. Similarly, when the other child catches the bag, they experience a force as well. This demonstrates Newton's third law of motion.

Now, let's solve the exercise questions at the end of the chapter.

Exercise 1: An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on the velocity.

Yes, it is possible. If the net external force on an object is zero, but the object is already in motion, it will continue to move with a constant velocity. This is in accordance with the first law of motion. The velocity must be constant in magnitude and direction — that is, uniform motion in a straight line.

Exercise 2: When a carpet is beaten with a stick, dust comes out of it. Explain.

When we beat a carpet with a stick, the carpet is set into sudden motion. However, the dust particles stuck in the carpet tend to remain in their state of rest due to their inertia. The carpet moves away from the dust, and the dust is left behind, appearing as if it comes out of the carpet.

Exercise 3: Why is it advised to tie any luggage kept on the roof of a bus with a rope?

When the bus is moving, the luggage on the roof tends to remain in its state of motion due to inertia. If the bus suddenly stops or changes its speed, the luggage may fall off due to its inertia. Tying it with a rope provides a force to keep it in place, preventing it from falling off.

Exercise 4: A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because

(a) the batsman did not hit the ball hard enough. (b) velocity is proportional to the force exerted on the ball. (c) there is a force on the ball opposing the motion. (d) there is no unbalanced force on the ball, so the ball would want to come to rest.

The correct answer is (c). There is a force on the ball opposing the motion — this is the force of friction between the ball and the ground. This friction force acts opposite to the direction of motion and gradually slows down the ball until it comes to rest.

Exercise 5: A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)

Solution: The truck starts from rest, so initial velocity u = 0. Distance traveled s = 400 m, time t = 20 s.

We use the equation s = ut + (1/2)at²

400 m = 0 × 20 + (1/2) × a × (20)²

400 = (1/2) × a × 400

400 = 200a

a = 400/200 = 2 m s⁻²

So the acceleration is 2 m s⁻².

Mass of the truck = 7 tonnes = 7 × 1000 kg = 7000 kg.

Force F = ma = 7000 kg × 2 m s⁻² = 14000 N.

Exercise 6: A stone of 1 kg is thrown with a velocity of 20 m s⁻¹ across the frozen surface of a lake and comes to rest after traveling a distance of 50 m. What is the force of friction between the stone and the ice?

Solution: Mass of the stone m = 1 kg, initial velocity u = 20 m s⁻¹, final velocity v = 0 m s⁻¹, distance traveled s = 50 m.

We need to find the acceleration first. Using the equation v² = u² + 2as

0 = (20)² + 2 × a × 50

0 = 400 + 100a

100a = -400

a = -4 m s⁻²

The force of friction F = ma = 1 kg × (-4 m s⁻²) = -4 N.

The negative sign indicates that the force of friction acts opposite to the direction of motion.

Exercise 7: An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate:

(a) the net accelerating force and

(b) the acceleration of the train.

Solution: Mass of the engine = 8000 kg. Mass of each wagon = 2000 kg. Number of wagons = 5.

Total mass of the train = mass of engine + mass of wagons = 8000 kg + (5 × 2000 kg) = 8000 kg + 10000 kg = 18000 kg.

Force exerted by the engine = 40000 N. Friction force = 5000 N (opposing motion).

(a) Net accelerating force = Force by engine - Friction force = 40000 N - 5000 N = 35000 N.

(b) Acceleration a = F/m = 35000 N / 18000 kg = 1.94 m s⁻² (approximately).

Exercise 8: An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s⁻²?

Solution: Mass of the vehicle m = 1500 kg. Acceleration a = -1.7 m s⁻² (negative because it's stopping).

Force F = ma = 1500 kg × (-1.7 m s⁻²) = -2550 N.

The negative sign indicates that the force is opposite to the direction of motion, which is the braking force.

Exercise 9: What is the momentum of an object of mass m, moving with a velocity v?

(a) (mv)² (b) mv² (c) ½ mv² (d) mv

The correct answer is (d) mv. Momentum is defined as the product of mass and velocity.

Exercise 10: Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?

Solution: When an object moves with constant velocity, the net force on it is zero. This means the applied force is balanced by the friction force. Therefore, the friction force must be equal and opposite to the applied force. So the friction force is 200 N in the opposite direction.

Exercise 11: According to the third law of motion when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.

The student's logic is incorrect. The action and reaction forces always act on two different objects. When we push the truck, we apply a force on the truck, and the truck applies an equal and opposite force on us. These forces do not cancel each other because they act on different objects. The reason the truck does not move is that the force we apply is not enough to overcome the friction between the truck tires and the road. The truck has a large mass and therefore large inertia. To move the truck, we need to apply a force greater than the frictional force. The reaction force we feel is just our muscles feeling the resistance, but it does not prevent the truck from moving if the applied force is sufficient.

Exercise 12: A hockey ball of mass 200 g traveling at 10 m s⁻¹ is struck by a hockey stick so as to return it along its original path with a velocity of 5 m s⁻¹. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Solution: Mass of the hockey ball = 200 g = 0.2 kg. Initial velocity u = 10 m s⁻¹ (in one direction). Final velocity v = -5 m s⁻¹ (in the opposite direction).

Initial momentum = mu = 0.2 kg × 10 m s⁻¹ = 2 kg m s⁻¹.

Final momentum = mv = 0.2 kg × (-5 m s⁻¹) = -1 kg m s⁻¹.

Change in momentum = final momentum - initial momentum = (-1) - (2) = -3 kg m s⁻¹.

The magnitude of change of momentum is 3 kg m s⁻¹.

Exercise 13: A bullet of mass 10 g traveling horizontally with a velocity of 150 m s⁻¹ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Solution: Mass of the bullet = 10 g = 0.01 kg. Initial velocity u = 150 m s⁻¹. Final velocity v = 0 m s⁻¹. Time t = 0.03 s.

First, let's find the acceleration. Using v = u + at

0 = 150 + a × 0.03

a = -150 / 0.03 = -5000 m s⁻²

Now, distance of penetration s = ut + (1/2)at²

s = 150 × 0.03 + (1/2) × (-5000) × (0.03)²

s = 4.5 + (-0.5) × 5000 × 0.0009

s = 4.5 - 2.25 = 2.25 m

Wait, let me recalculate this more carefully.

s = ut + (1/2)at²

s = 150 × 0.03 + 0.5 × (-5000) × 0.0009

s = 4.5 - 2.25 = 2.25 m

Actually, this seems too large for a bullet penetration. Let me use another formula.

Using v² = u² + 2as

0 = (150)² + 2 × (-5000) × s

0 = 22500 - 10000s

10000s = 22500

s = 2.25 m

This is still 2.25 m, which seems reasonable for a bullet going through a wooden block, but let me check the calculation again.

Actually, wait. Let me recalculate the acceleration properly.

a = (v - u)/t = (0 - 150)/0.03 = -150/0.03 = -5000 m s⁻². This is correct.

Now, s = ut + (1/2)at² = 150 × 0.03 + 0.5 × (-5000) × (0.03)²

= 4.5 + (-2500) × 0.0009

= 4.5 - 2.25

= 2.25 m

Yes, 2.25 m is correct.

Now, force exerted by the wooden block on the bullet:

F = ma = 0.01 kg × (-5000 m s⁻²) = -50 N.

The magnitude of the force is 50 N.

Exercise 14: An object of mass 1 kg traveling in a straight line with a velocity of 10 m s⁻¹ collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.

Solution: Mass of the object m₁ = 1 kg, velocity u₁ = 10 m s⁻¹. Mass of the wooden block m₂ = 5 kg, velocity u₂ = 0 m s⁻¹.

Total momentum before impact = m₁u₁ + m₂u₂ = (1 × 10) + (5 × 0) = 10 kg m s⁻¹.

When they stick together, the total mass becomes m₁ + m₂ = 1 + 5 = 6 kg.

By the law of conservation of momentum, total momentum after impact = total momentum before impact = 10 kg m s⁻¹.

Velocity of the combined object v = total momentum / total mass = 10 / 6 = 1.67 m s⁻¹ (approximately).

So the total momentum just before the impact is 10 kg m s⁻¹, just after the impact is also 10 kg m s⁻¹, and the velocity of the combined object is 1.67 m s⁻¹.

Exercise 15: An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s⁻¹ to 8 m s⁻¹ in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Solution: Mass m = 100 kg. Initial velocity u = 5 m s⁻¹. Final velocity v = 8 m s⁻¹. Time t = 6 s.

Initial momentum p₁ = mu = 100 kg × 5 m s⁻¹ = 500 kg m s⁻¹.

Final momentum p₂ = mv = 100 kg × 8 m s⁻¹ = 800 kg m s⁻¹.

Change in momentum = p₂ - p₁ = 800 - 500 = 300 kg m s⁻¹.

Force F = change in momentum / time = 300 / 6 = 50 N.

Alternatively, we can find acceleration first: a = (v - u)/t = (8 - 5)/6 = 3/6 = 0.5 m s⁻².

Then F = ma = 100 × 0.5 = 50 N.

Exercise 16: Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.

Let me analyze this situation carefully. According to Newton's third law, the force exerted by the motorcar on the insect is equal and opposite to the force exerted by the insect on the motorcar. So Rahul is correct in saying that both experience the same force.

However, the change in momentum depends on both the force and the time during which the force acts. The insect comes to rest very quickly (in a very short time), so the change in its momentum is large and happens in a very short time. The motorcar's velocity changes very slightly (almost negligible), so its change in momentum is very small.

Kiran is partially correct in saying that the insect suffered a greater change in momentum, but the reason is not just the change in velocity but also the very short time interval.

Akhtar's reasoning is not entirely correct because the force is the same on both (according to Newton's third law), not larger on the insect.

So the best explanation is Rahul's — both the motorcar and the insect experience the same force (action and reaction), but because of the huge difference in their masses, the effects are very different. The insect has very small mass, so even a small force produces a large acceleration and change in velocity. The motorcar has very large mass, so the same force produces negligible acceleration and change in velocity.

Exercise 17: How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 m s⁻².

Solution: Mass of the dumb-bell m = 10 kg. Height h = 80 cm = 0.8 m. Acceleration g = 10 m s⁻².

When the dumb-bell falls from a height, it gains velocity. We need to find the velocity just before it hits the floor.

Using the equation v² = u² + 2gh, where u = 0 (starts from rest)

v² = 0 + 2 × 10 × 0.8 = 16

v = 4 m s⁻¹

Momentum just before impact p = mv = 10 kg × 4 m s⁻¹ = 40 kg m s⁻¹.

When the dumb-bell hits the floor and comes to rest, it transfers its momentum to the floor. So the momentum transferred is 40 kg m s⁻¹.

Now, let's solve the Additional Exercises.

Additional Exercise A1: The following is the distance-time table of an object in motion:

Time in seconds: 0, 1, 2, 3, 4, 5, 6, 7 Distance in metres: 0, 1, 8, 27, 64, 125, 216, 343

(a) What conclusion can you draw about the acceleration? Is it constant, increasing, decreasing, or zero?

(b) What do you infer about the forces acting on the object?

Solution: Let's look at the pattern in the distance-time data.

At t = 1 s, distance = 1 m = 1³ At t = 2 s, distance = 8 m = 2³ At t = 3 s, distance = 27 m = 3³ At t = 4 s, distance = 64 m = 4³ At t = 5 s, distance = 125 m = 5³ At t = 6 s, distance = 216 m = 6³ At t = 7 s, distance = 343 m = 7³

So the distance is proportional to t³, which means s ∝ t² (since s = k × t², where k is constant).

This is the equation of motion for constant acceleration: s = (1/2)at²

Comparing, we have (1/2)a = 1, so a = 2 m s⁻².

Since the acceleration is constant (2 m s⁻²), the acceleration is constant, not increasing, decreasing, or zero.

(b) Since the acceleration is constant, according to Newton's second law of motion, the net force acting on the object must be constant. This is because F = ma, and if m is constant and a is constant, then F is constant.

Additional Exercise A2: Two persons manage to push a motorcar of mass 1200 kg at a uniform velocity along a level road. The same motorcar can be pushed by three persons to produce an acceleration of 0.2 m s⁻². With what force does each person push the motorcar? (Assume that all persons push the motorcar with the same muscular effort.)

Solution: When two persons push the motorcar at a uniform velocity, the net force is zero. This means the pushing force by the two persons is balanced by the frictional force. So, force by two persons = frictional force.

When three persons push the motorcar, they produce an acceleration of 0.2 m s⁻². The net force is the pushing force by three persons minus the frictional force.

Let the force exerted by each person be F. Then the force by two persons is 2F, and by three persons is 3F.

When two persons push: 2F = frictional force (since velocity is uniform, net force = 0)

When three persons push: net force = 3F - frictional force = ma = 1200 kg × 0.2 m s⁻² = 240 N.

But we know frictional force = 2F from the first case.

So, 3F - 2F = 240 N

F = 240 N

So each person pushes with a force of 240 N.

Additional Exercise A3: A hammer of mass 500 g, moving at 50 m s⁻¹, strikes a nail. The nail stops the hammer in a very short time of 0.01 s. What is the force of the nail on the hammer?

Solution: Mass of the hammer = 500 g = 0.5 kg. Initial velocity u = 50 m s⁻¹. Final velocity v = 0 m s⁻¹. Time t = 0.01 s.

The acceleration of the hammer is a = (v - u)/t = (0 - 50)/0.01 = -5000 m s⁻².

Force F = ma = 0.5 kg × (-5000 m s⁻²) = -2500 N.

The negative sign indicates that the force exerted by the nail on the hammer is opposite to the direction of motion of the hammer. The magnitude of the force is 2500 N.

Additional Exercise A4: A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Solution: Mass of the motorcar m = 1200 kg.

Initial velocity u = 90 km/h. Convert to m s⁻¹: 90 × 1000 / 3600 = 25 m s⁻¹.

Final velocity v = 18 km/h. Convert to m s⁻¹: 18 × 1000 / 3600 = 5 m s⁻¹.

Time t = 4 s.

Acceleration a = (v - u)/t = (5 - 25)/4 = -20/4 = -5 m s⁻².

The negative sign indicates deceleration (slowing down).

Change in momentum = m(v - u) = 1200 × (5 - 25) = 1200 × (-20) = -24000 kg m s⁻¹.

The magnitude of the force required F = ma = 1200 × 5 = 6000 N (taking magnitude only).

Alternatively, F = change in momentum / time = -24000 / 4 = -6000 N. The magnitude is 6000 N.

Now, students, we have covered the entire chapter. Let me give you a complete summary of everything we have learned today.

Summary:

In this chapter on Force and Laws of Motion, we learned about:

1. Force: Force is a push, hit, or pull that can change the state of motion, speed, direction, shape, or size of an object.

2. Balanced and Unbalanced Forces: When equal forces act on an object in opposite directions, they balance each other and the object does not move. When forces of different magnitudes act in opposite directions, the object moves in the direction of the greater force. This is called an unbalanced force.

3. First Law of Motion: An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. This is also known as the law of inertia.

4. Inertia: The natural tendency of an object to resist a change in its state of motion or rest is called inertia. Heavier or more massive objects have more inertia.

5. Mass: The mass of an object is a measure of its inertia. Its SI unit is kilogram (kg).

6. Second Law of Motion: The rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, where F is force, m is mass, and a is acceleration. The SI unit of force is newton (N), which is kg m s⁻².

7. Momentum: The momentum of an object is the product of its mass and velocity. It has both magnitude and direction. Its SI unit is kg m s⁻¹.

8. Third Law of Motion: To every action, there is an equal and opposite reaction. Action and reaction always act on two different objects.

We also learned about various real-life applications of these laws, such as why we fall forward when a bus brakes, why safety belts are important, how a fielder catches a ball, how we walk, and how a gun recoils when fired.

We solved numerical problems based on calculating force, acceleration, momentum, and velocity using the equations of motion and Newton's laws.

Students, I hope this lesson has helped you understand the concepts of force and laws of motion thoroughly. Remember, physics is all around us, and these laws explain many phenomena we observe in our daily life. Keep practicing the numerical problems, and you will master this chapter.

Thank you for listening attentively. See you in the next lesson!