1.

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(a) The distance covered by the object at any time interval is greater than any of the distances covered in previous time intervals. Therefore, the acceleration of the object is increasing.

(b) As per the second law of motion, force = mass × acceleration. Since the mass of the object remains constant, the increasing acceleration implies that the force acting on the object is increasing as well

2. 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)

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Given, mass of the car (m) = 1200 kg

When the third person starts pushing the car, the acceleration (a) is \( 0.2m/s^2 \) .

∴ The force applied by the third person (F) = ma

=> \( F = 1200 × 0.2 m/s^2 = 240 N \)

3. A hammer of mass 500 g, moving at 50 m/s, 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?

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Given, mass of the hammer (m) = 500 g = 0.5 kg

Initial velocity of the hammer (u) = 50 m/s

Terminal velocity of the hammer (v) = 0 {the hammer is stopped and reaches a position of rest} .

Time period (t) = 0.01 s

&$8756; the acceleration of hammer, \( a = \frac{v-u}{t} = \frac{0-50}{0.01} \ => \ a = -5000 \ m/s^2 \)

Therefore, the force exerted by the hammer on the nail (F) = ma

=> F = (0.5) &$215; (-5000) = -2500 N

4. 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.

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Given, mass of the car (m) = 1200 kg

Initial velocity (u) = 90 km/hour = 25 meters/sec

Terminal velocity (v) = 18 km/hour = 5 meters/sec

Time period (t) = 4 seconds

So, the acceleration of the car, \( a = \frac{v-u}{t} \ => \ a = \frac{5-25}{4} \ => \ a = -5 \ m/s^2 \)

Therefore, the acceleration of the car is \( -5 m/s^2 \) .

Initial momentum of the car = m × u = (1200kg) × (25m/s) = 30,000 kg m/s

Final momentum of the car = m × v = (1200) × (5) = 6,000 kg m/s

∴ change in momentum (final momentum – initial momentum) = (6,000 – 30,000) = -24,000 kg m/s

External force applied = mass of car × acceleration = (1200) × (-5) = -6000 N

A large truck and a car, both moving with a velocity of magnitude v, have a head-on collision and both of them come to a halt after that. If the collision lasts for 1 s:

(a) Which vehicle experiences the greater force of impact?

(b) Which vehicle experiences the greater change in momentum?

(c) Which vehicle experiences the greater acceleration?

(d) Why is the car likely to suffer more damage than the truck?

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(a) Both will experience same amount of force of impact by third law of motion.

(b) Mass of truck is greater then mass of car and initial ad final velocity of car ad truck are same i.e., v ad 0 respectively, therefore it will experience greater force of impact. Since, momentum is directly proportional to force.

(c) Car will have greater acceleration as for force of impact is same and mass of car is less than mass of truck.

(d) Car will suffer more as damage as its acceleration comes to 0 in shorter time interval.

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.

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Yes, it is possible for an object moving in some direction with constant velocity will continue in its state of motion as long as there are no external unbalanced forces acting on it. In order to change the motion of the object, some external unbalanced force must act upon it.

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

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When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.

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

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When some luggage is placed on the roof of a bus which is initially at rest, the acceleration of the bus in the forward direction will exert a force in the backward direction on the luggage. In a similar manner, when a bus which is initially in a state of motion suddenly comes to rest due to the application of brakes, a force in the forward direction is exerted on the luggage. This can cause luggage to fall from roof.

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.

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When the ball rolls on the flat surface of the ground, its motion is opposed by the force of friction. This frictional force will eventually cause ball to stop. Therefore, the correct answer is (c).

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 it’s mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)

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Given, distance covered by the truck (s) = 400 meters

Time taken to cover the distance (t) = 20 seconds

Initial velocity of the truck (u) = 0 {since it starts from rest}

We know that, \( s = ut + \frac{1}{2} at^2 \)

∴ \( 400 = 0 × 20 + \frac{1}{2} (a)(400) \ = \ 2 m/s^2 \)

Now, force = mass × acceleration

∴ force acting on truck = 7000 × 2 = 14000 N

6. A stone of 1 kg is thrown with a velocity of 20 ms-1 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?

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Given, mass of the stone (m) = 1 kg

Initial velocity (u) = 20 m/s

Terminal velocity (v) = 0 m/s {as stone reaches a position of rest}

Distance travelled by the stone (s) = 50 m

We know that, \( v^2 - u^2 = 2 a s \) => \( acceleration of stone = \frac{0 - 400}{100} = - 4 m/s^2 \)

Now, F = ma => force acting om stone = 1 × (-4) = -4N

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

(b) the acceleration of the train

(c) the force of wagon 1 on wagon 2.

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(a)

(a) Given, force exerted by the train (F) = 40,000 N

Friction exerts a force of = -5000 N {the negative sign indicates that the force is applied in the opposite direction}

∴ The net accelerating force = sum of all forces = 40,000 + (-5000 ) = 35,000 N

(b) Total mass of the train = mass of engine + mass of each wagon = 8000 + 5 × 2000 = 18000 kg

We know that, F = ma => : \( a = \fract{F}{m} \)

Therefore, acceleration of the train = (net accelerating force) / (total mass of the train)

=> \( \frac{35,000}{18,000} = 1.94 m/s^2 \)

(c) Force of wagon 1 on 2 = mass of 4 wagons × acceleration

=> \( F_{1,2} = 4 × 2000 × 1.94 \ => \ F = 1552 N \)

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 \( 1.7 m/s^2 \) ?

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Given, mass of the vehicle (m) = 1500 kg

Acceleration (a) = \( -1.7 m/s^2 \)

As per the second law of motion, F = ma

F = 1500 × (-1.7) = -2550 N

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

(a) (mv)^{2}

(b) mv^{2}

(c) ½ mv^{2}

(d) mv

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Since momentum is defined as the product of mass and velocity, the correct answer is (d), mv.

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?

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At constant velocity there is no unbalanced force. This implies that the magnitude of opposing frictional force is equal to the force exerted on the cabinet, which is 200 N. Therefore, the total friction force is -200 N.

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?

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Given, mass of the objects \( m_1 and m_2 \) = 1.5 kg

Initial velocity of the first object \( u_1 \) = 2.5 m/s

Initial velocity of the second object which is moving in the opposite direction \( u_2 \) = -2.5 m/s

When the two masses stick together, the resulting object has a mass of 3 kg i.e., \( m_1 + m_2 \)

Velocity of the resulting object (v) =?

By law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.

=> (1.5) (2.5) + (1.5) (-2.5) = 0

Therefore, total momentum after collision = \( m_1 + m_2 \) v = (3) v = 0

This implies that the object formed after the collision has a velocity of 0 meters per second.

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.

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Since the truck has a very high mass, the frictional force between the road and the truck is high. When we push the truck with a small force, the frictional force cancels out the applied force and the truck does not move. This implies that the two forces are equal in magnitude but opposite in direction. Therefore, the student’s logic is correct.

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.

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Given, mass of the ball (m) = 200 g

Initial velocity of the ball (u) = 10 m/s

Final velocity of the ball (v) = 5 m/s

Initial momentum of the ball = mu = 200 × 10 = 2000 gm/s

Final momentum of the ball = m v = 200 × 5 = 1000 gm/s

Therefore, the change in momentum (m v – m u) = 1000 – 2000 = -1000 gm/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.

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Given, mass of the bullet (m) = 10 g = 0.01 kg

Initial velocity of the bullet (u) = 150 m/s

Terminal velocity of the bullet (v) = 0 m/s

Time period (t) = 0.03 s

We know that, \( v = u +at \) => \( 0 = 150 + a(0.01) \) => \( a = - 5000 \ m/s^2 \)

Now, \( v^2 - u ^2 = 2 as \) => \( 0^2 - 150^2 = 2 × (-5000) s \) => \( s = 2.25 \ m \)

Now, force exerted by wooden block on bullet = mass of bullet × acceleration of bullet

=> \( F = 0.01 × (-5000) = -50 N \)

15. An object of mass 1 kg travelling in a straight line with a velocity of 10 ms–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.

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Given, mass of the object \( m_1 \) = 1 kg

Mass of the block \( m_2 \) = 5 kg

Initial velocity of the object \( u_1 \) = 10 m/s

Initial velocity of the block \( u_2 \) = 0

Mass of the resulting object \( = m_1 + m_2 = 6 kg \)

Total momentum before the collision = \( m_1 u_1 + m_2 u_2 = (1) × (10) + 0 = 10 kg m/s \)

By law of conservation of momentum, the total momentum before the collision is equal to the total momentum post the collision.

∴ The total momentum before the collision is also 10 kg m/s

Now, \( (m_1 + m_2) × v = 10 kg m/s \)

=> \( v = \frac{10}{6} = 1.66 m/s \)

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.

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Given, mass of the object (m) = 100 kg

Initial velocity (u) = 5 m/s

Final velocity (v) = 8 m/s

Time taken (t) = 6 s

We know that, \( momentum \ of \ object \ = \ mass \ of \ object \ × \ velocity \ of \ object \)

Now, initial momentum = m × u = 100 × 5 = 500 kg m/s

Final momentum = m × v = 100 × 8 = 800 kg m/s

Now, \( v = u + at \) => \( 8 = 5 + a(6) \ => \ a = 0.5 \ m/s^2 \)

Also, \( F = ma \ => \ F = 100 × 0.5 \ => \ F = 5 N \)

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.

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Kiran’s suggestion is correct. The mass of the insect is very small when compared to the mass of the car. As per the law of conservation of momentum, the total momentum before the collision between the insect and the car is equal to the total momentum after the collision. Therefore, the change in the momentum of the insect is much greater than the change in momentum of the car. Akhtar’s suggestion is also correct, since the mass of the car is very high, the force exerted on the insect during the collision is also very high. Rahul’s suggestion is also correct, as by the third law of motion, the force exerted by the insect on the car is equal and opposite to the force exerted by the car on the insect. However, Rahul’s suggestion that the change in the momentum is the same contradicts the law of conservation of momentum.

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 \( 10 m/s^2\).

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Given, mass of the dumb-bell (m) = 10 kg

Distance covered (s) = 80 cm = 0.8 m

Initial velocity (u) = 0

Acceleration (a) = \( 10 m/s^2 \)

Momentum of the dumb-bell when it hits the ground = m v

We know that, \( v^2 - u^2 = 2 a s \ => \ v^2 - 0 = 2(10)(0.8) \ => \ v = 4 \)

The momentum transferred by the dumb-bell to the floor = (10) × (4) = 40 kg m/s

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?

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The following objects hold greater inertia because of their mass.

(a) Stone (b) Train (c) Five-Rupee coin
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.

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The velocity of the ball changes 4 times that are:

1. When first player kicks the football velocity changes from 0 to u. So, the agent is first player.

2. When second player kicks football towards goal, velocity again changes. So, the agent is second player.

3. When the goalkeeper collects football, velocity becomes 0. So, the agent is goalkeeper.

4. When goalkeeper kicks football towards a player from his team, velocity changes from 0 to v. So, the agent is again goalkeeper.

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

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When the branch of the tree is shaken, the branch moves in motion. However, the inertia of the leaves resists the motion of the branch. Therefore, the leaves that are weakly attached to the branch fall off due to inertia.

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?

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We fall in the forward direction when on a moving bus brakes are applied because of inertia. As, our body continues to move while the bus gets stop.Same goes when a bus is accelerated from rest out body tends to be in rest while bus starts moving.

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

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When the horse walks in the forward direction (with the cart attached to it), it exerts a force in the backward direction on the Earth. An equal force in the opposite direction (forward direction) is applied on the horse by the Earth. This force moves the horse and the cart forward.

2. Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.

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For the hose to eject water at high velocities, a force must be applied on the water which is usually done with the help of a pump or a motor. Now, the water applies an equal and opposite force on the hose. For the fireman to hold this hose, he must apply a force on it to overcome the force applied on the hose by the water. The higher the quantity and velocity of the water coming out of the hose, the greater the force that must be applied by the fireman to hold it steady.

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.

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Given, mass of bullet( \( m_1 \) ) = 50 g

Mass of rifle( \( m_2 \) ) = 4000 g

Initial velocity of the fired bullet( \( v_1 \) ) = 35 m/s

Let the recoil velocity be \( v_2 \).

Since the rifle was initially at rest, the initial momentum of the rifle = 0

Total momentum of the rifle and bullet after firing = \( m_1 v_1 + m_2 v_2 \)

Now we know that, the total momentum of the rifle and the bullet after firing = 0 { By law of conservation of momentum}

∴ \( m_1 v_1 + m_2 v_2 \) = 0 => \( v_2 = - \frac{m_1 v_1}{m_2} \)

= - \frac{50 × 35}{4000} = -0.4375 m/s

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

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Assuming that the first object is object A and the second one is object B, it is given that:

Mass of A (m_{1}) = 100g

Mass of B (m_{2}) = 200g

Initial velocity of A (u_{1}) = 2 m/s

Initial velocity of B (u_{2}) = 1 m/s

Final velocity of A (v_{1}) = 1.67 m/s

Final velocity of B (v_{2}) =?

Total initial momentum = Initial momentum of A + initial momentum of B

= m_{1}u_{1} + m_{2}u_{2}

= (100) × (2) + (200) × (1) = 400 gm/sec

As per the law of conservation of momentum, the total momentum before collision must be equal to the total momentum post collision.

∴ \( m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 = 400 \)

=> \( (100)(1.67) + (200)v_2 = 400 \)

∴ \( v_2 = \frac{400-167}{200} = 1.165 m/s \)