9. Force and Laws of Motion

Class 9 Science Chapter 9. Force and Laws of Motion

Chapter 9. Force and Laws of Motion

Internal Questions :
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?

Answer: (a) The stone has more inertia than the rubber ball because it is denser and heavier.

(b) The train has more inertia than the bicycle because it is larger and has more mass.

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.

Answer: The velocity of the ball changes three times in the given example.

(i) The first change occurs when the first player kicks the ball to another player. The agent supplying the force is the first player's foot.

(ii) The second change happens when the second player kicks the ball towards the goal. The agent supplying the force is the second player's foot.

(iii) The third change takes place when the goalkeeper collects the ball and kicks it towards a player of his own team. The agent supplying the force is the goalkeeper's foot.

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

Answer: When we shake a tree branch vigorously, the force and movement can break the small stems or connections that hold the leaves. These connections are not very strong, so the leaves can easily detach and fall off the tree.
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?

Answer : When a moving bus suddenly brakes to a stop, you fall forward because your body tends to keep moving in the same direction due to inertia. When the bus accelerates from rest, you fall backward because your body resists the change in motion caused by the acceleration.

Example 9.1 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 to 7 m/s . 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: Given, m/s , m/s , s , kg

We know that,

If this force is applied for a duration of 5 s (t = 5 s), then the final velocity

m/s

Example 9.2 Which would require a greater force accelerating a 2 kg mass at 5 or a 4 kg mass at 2 ?

Solution: We know that,

Given, kg and m/s²

Again, kg and m/s²

So,

Thus, accelerating a 2 kg mass at 5 m/s² would require a greater force.

Example 9.3 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 km/h

m/s

And the final velocity of the motorcar m/s

The total mass of the motorcar along with its passengers kg and the time taken to stop the motorcar, s

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

Example 9.4 A force of 5 N gives a mass , an acceleration of 10 and a mass , an acceleration of 20 . What acceleration would it give if both the masses were tied together?

Solution: Given, , m/s² , m/s²

We know that,

Now, kg

And kg

If the two masses were tied together, the total mass

The acceleration of the combined mass by the 5 N force

m/s² .

Example 9.5 The velocity-time graph of a ball of mass 20 g moving along a straight line on a long table is given in Fig. 9.9.

How much force does the table exert on the ball to bring it to rest?

Solution: The initial velocity of the ball cm/s m/s

The final velocity of the ball m/s

The mass of the ball g kg and Time 10 s

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

The acceleration cm/s²

We know that,

The force exerted on the ball

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

Example 9.6 A bullet of mass 20 g is horizontally fired with a velocity 150 m/s from a pistol of mass 2 kg. What is the recoil velocity of the pistol ?

Solution: Given, the mass of bullet g kg

and the mass of the pistol kg

Initial velocities of the bullet and pistol m/s .

The final velocity of the bullet, m/s

Let be the recoil velocity of the pistol.

Total momenta of the pistol and bullet before the fire, when the gun is at rest

kgm/s

Total momenta of the pistol and bullet after it is fired =m1×v1+m2×v

kg m/s

According to the law of conservation of momentum,

Total momenta after the fire = Total momenta before the fire

m/s

Example 9.7 A girl of mass 40 kg jumps with a horizontal velocity of 5 m/s onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg . What is her velocity as the cart starts moving ? Assume that there is no external unbalanced force working in the horizontal direction.

Solution: Let be the velocity of the girl on the cart as the cart starts moving.

Given, kg , kg , m/s , m/s

The total momenta of the girl and cart before the interaction

kg m/s

Total momenta after the interaction

kg m/s

According to the law of conservation of momentum, the total momentum is conserved during the interaction.

So,

m/s

The girl on cart would move with a velocity of 4.65 m/s in the direction in which the girl jumped.

Example 9.8 Two hockey players of opposite teams, while trying to hit a hockey ball on the ground collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m/s while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m/s towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.

Solution: Given, kg , kg , m/s and m/s

The total momentum of the two players before the collision

kg m/s

let is the velocity of the two entangled players after the collision, the total momentum then

kg m/s

Equating the momenta of the system before and after collision, in accordance with the law of conservation of momentum, we get

m/s

Thus, the two entangled players would move with velocity 0.26 m/s from right to left, that is, in the direction the second player was moving before the collision.

Internal Questions :

1. If action is always equal to the reaction, explain how a horse can pull a cart.

Answer: When a horse pulls a cart, the horse exerts a forward force on the cart. According to Newton's third law of motion, the cart also exerts an equal and opposite force on the horse. This force helps the horse move forward and pull the cart.
2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

Answer: It is difficult for a fireman to hold a hose that ejects large amounts of water at high velocity because of Newton's third law of motion. The water coming out of the hose exerts a strong backward force on the fireman, making it challenging to maintain a stable grip and control the hose.
3. From a rifle of mass 4 kg, a bullet of mass 50 g is fired with an initial velocity of 35 m/s . Calculate the initial recoil velocity of the rifle .

Solution: Given, The mass of the rifle kg

The mass of the bullet g kg

Initial velocity : m/s , m/s

The momentum of the bullet before the shot :

Initial momentum

kg m/s

Total momentum of the bullet after the shot :

Let be the final recoil velocity of the rifle .

Final momentum

A/Q,

m/s

Therefore, the initial recoil velocity of the rifle is 0.875 m/s.

4. Two objects A and B of masses 100 g and 200 g are moving along the same line in the same direction with velocities of 2 m/s and 1 m/s, respectively. They collide and after the collision, the object B moves at a velocity of 1.67 m/s . Determine the velocity of the object A.

Solution: Given, g kg , g kg , m/s , m/s

According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity.

Before the collision, the total momentum is the sum of the individual momenta of objects A and B:

Total momentum before

kg m/s

After the collision, the total momentum is the sum of the individual momenta of objects A and B:

Let be the velocity of the object A .

m/s

The velocity of object B after the collision:

Total momentum after

According to the conservation of momentum, the total momentum before and after the collision should be equal:

m/s

Therefore, the velocity of object A after the collision is 0.66 m/s.

Class 9 Force and Laws of Motion Exercises :

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 magnitude and direction of the velocity. If no, provide a reason.

Answer: No, it is not possible for an object to be traveling with a non-zero velocity if it experiences a net zero external unbalanced force.

According to Newton's first law of motion (the law of inertia), an object at rest or moving with a constant velocity will continue to do so unless acted upon by an external unbalanced force. If the net external force on an object is zero, it means there are no unbalanced forces acting on it, and therefore it will either remain at rest or continue to move with a constant velocity (zero velocity in this case) in a straight line.

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

Answer: When a carpet is beaten with a stick, the dust particles trapped within the carpet fibers exhibit inertia. The impact and vibrations from the stick transfer energy to the carpet, causing the carpet fibers and the dust particles to move. Due to their inertia, the dust particles resist the change in motion, leading to their detachment from the carpet and becoming airborne as dust.
3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?

Answer: It is advised to tie luggage on the roof of a bus with a rope to prevent it from moving or falling off due to its inertia. The rope provides a restraining force, counteracting the tendency of the luggage to maintain its state of motion, ensuring safety and stability during bus movements.
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.

Answer: The correct answer is (c) there is a force on the ball opposing the motion.

This causes the ball to slow down and eventually come to rest. Inertia, the tendency of an object to resist changes in its state of motion, is also involved as the ball resists the change in motion caused by the frictional force.

Option (a) is incorrect because the force of the hit by the batsman is not directly related to the ball slowing down.

Option (b) is incorrect because velocity is not solely determined by the force exerted on the ball, but also by other factors such as mass.

Option (d) is incorrect because the ball does experience an unbalanced force due to the friction acting against its motion.

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: Given, tonnes kg kg , m , s , m/s

We have,

m/s²

We know that ,

Therefore, the force acting on the truck is 14,000 Newtons.
6. A stone of 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?

Solution: Given, kg , m, m/s and m/s

We have,

We know that,

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

Therefore, the force of friction between the stone and the ice is 4 Newtons.

7. A 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: (a) The net accelerating force is the difference between the force exerted by the engine and the friction force:

Net accelerating force = Force exerted by the engine – Friction force

= 40000 N – 5000 N = 35000 N

Therefore, the net accelerating force is 35000 Newtons.

(b) The total mass of the train is the sum of the mass of the engine and the mass of the wagons:

Total mass of the train = Mass of engine + (Number of wagons × Mass of each wagon)

= 8000 kg + (5 × 2000 kg)

= 18000 kg

We know that,

Therefore, the acceleration of the train is approximately 1.94 m/s².

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 ?

Solution: Given, kg , m/s²

We know that ,

The negative sign indicates that the force is in the opposite direction of the motion of the vehicle.

Therefore, the force between the vehicle and the road is 2550 Newtons .

9. What is the momentum of an object of mass m, moving with a velocity v ?
(a) (b) (c) (d)

Answer: An object of mass moving with a velocity , then the momentum . Correct option (d) .

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?

Answer: If the wooden cabinet is being moved across the floor at a constant velocity, it means that the net force acting on the cabinet is zero. In this case, the force of friction will be equal in magnitude and opposite in direction to the applied force.

Therefore, the friction force exerted on the cabinet will be 200 N, which is equal in magnitude to the applied force but acts in the opposite direction to prevent the cabinet from accelerating or decelerating.

11. Two objects, each of mass 1.5 kg, are moving in the same straight line but in opposite directions. The velocity of each object is 2.5 m/s before the collision during which they stick together. What will be the velocity of the combined object after collision?

Solution: Given, , , m/s , m/s

Total momentum before collision

kg m/s

Let, be the velocity of combined object .

Total momentum after collision

A/Q,

m/s

Therefore, the velocity of the combined object after the collision is 0 m/s.

12. 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.

Answer: The logic that the two opposite and equal forces cancel each other is incorrect in this context. While it is true that the forces are equal and opposite according to Newton's third law of motion, canceling out each other, the truck does not move because it has a significantly larger mass compared to the force exerted by the student, resulting in a negligible acceleration that is insufficient to overcome the truck's inertia.
13. A hockey ball of mass 200 g travelling at 10 m/s is struck by a hockey stick so as to return it along its original path with a velocity at 5 m/s . Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.

Solution: Given, kg kg , m/s and m/s

Initial Momentum kg m/s

Final Momentum kg m/s

Change in Momentum = Final Momentum - Initial Momentum

kg m/s kg m/s

The negative sign indicates that the direction of the momentum has changed (opposite to the initial direction).

Therefore, the magnitude of the change in momentum of the hockey ball, caused by the force applied by the hockey stick, is 1 kg m/s.

14. A bullet of mass 10 g travelling horizontally with a velocity of 150 m/s 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: Given, g kg , , m/s and m/s

We have,

Again,

We know that,

Therefore, the distance of penetration of the bullet into the block is 2.25 m, and the magnitude of the force exerted by the wooden block on the bullet is 50 Newtons.

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 m/s 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: Given, kg , kg , m/s and m/s

The total momentum just before the impact, we have

Total Momentum

kg m/s

Therefore, the total momentum just before the impact is 10 kg m/s .

Let be the velocity of the objects .

The total momentum just after the impact, we have

Total momentum

Total Momentum before impact = Total Momentum after impact

Therefore, the velocity of the combined object after the collision is approximately 1.67 m/s

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

Solution: Given, kg , m/s , m/s , s

We know that, Momentum

Initial momentum kg m/s

Final momentum kg m/s

Therefore, the initial momentum of the object is 500 kg m/s and the final momentum is 800 kg m/s.

Again, the equation of motion:

m/s²

Newton's second law of motion :

Therefore, the magnitude of the force exerted on the object is 50 Newtons .

17. 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.

Answer: Kiran's suggestion is incorrect. The change in momentum of the insect and the motorcar should be equal and opposite due to the conservation of momentum.

Akhtar's suggestion is also incorrect. The force experienced by an object is not solely determined by its velocity but also by its mass and the duration of the interaction.

Rahul's explanation is correct. Both the motorcar and the insect experience the same force and undergo a change in momentum, according to Newton's third law of motion.

18. 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 .

Solution: Given, Mass, kg , cm m , m/s and m/s²

The equation of motion, we have

m/s

We know that,

Momentum kg m/s

Therefore, the dumbbell will transfer a momentum of 40 kg m/s to the floor when it falls from a height of 80 cm .