# Force and Laws of Motion : NCERT Exercise Questions

Q.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.
Sol.     When a net zero external unbalanced force is applied on the body, it is possible for the object to be travelling with a non-zero velocity. In fact, once an object comes into motion and there is a condition in which its motion is unopposed by any external force; the object will continue to remain in motion. It is necessary that the object moves at a constant velocity and in a particular direction.

Q.2     When a carpet is beaten with a stick, dust comes out of it. Explain.
Sol.     Beating of a carpet with a stick; makes the carpet come in motion suddenly, while dust particles trapped within the pores of carpet have tendency to remain in rest, and in order to maintain the position of rest they come out of carpet. This happens because of the application of Newton’s First Law of Motion which states that any object remains in its state unless any external force is applied over it.

Q.3     Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Sol.     Luggage kept on the roof of a bus has the tendency to maintain its state of rest when bus is in rest and to maintain the state of motion when bus is in motion according to Newton’s First Law of Motion. When bus will come in motion from its state of rest, in order to maintain the position of rest, luggage kept over its roof may fall down. Similarly, when a moving bus will come in the state of rest or there is any sudden change in velocity because of applying of brake, luggage may fall down because of its tendency to remain in the state of motion. This is the cause that it is advised to tie any luggage kept on the roof a bus with a rope so that luggage can be prevented from falling down.

Q.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.
Sol.     (c) There is a force on the ball opposing the motion. Explanation: When ball moves on the ground, the force of friction opposes its movement and after some time ball comes to the state of rest.

Q.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.)
Sol.     Given,  Initial velocity of truck (u) = 0 (Since, truck starts from rest) Distance travelled, s = 400 m
Time (t) = 20 Acceleration (a) = ?
We know that,
$s = ut + {1 \over 2}a{t^2}$
$\Rightarrow 400m = 0 \times 20s + {1 \over 2} \times a \times {\left( {20s} \right)^2}$
$\Rightarrow 400m = {1 \over 2} \times a \times 400{s^2}$
$\Rightarrow 400m = a \times 200{s^2}$
$\Rightarrow a = {{400m} \over {200{s^2}}} = 2m{s^{ - 2}}$
Force acting upon truck:
Given mass of truck = 7 ton = 7 X 1000 kg = 7000 kg
We know that, force, P = m x a
Therefore, P = 7000 kg x 2 m s - 2
Or, P = 14000 Newton
Thus, Acceleration = 2 m s - 2 and force acting upon truck in the given condition = 14000 N

Q.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?
Sol.     Given,
Mass of stone = 1 kg
Initial velocity, u = 20 m/s
Final velocity, v = 0 (as stone comes to rest)
Distance covered, s = 50 m
Force of friction = ?
We know that,
${v^2} = {u^2} + 2as$
$\Rightarrow {\left( 0 \right)^2} = {\left( {20m/s} \right)^2} + 2a \times 50m$
$\Rightarrow - 400{m^2}{s^{ - 2}} = 100ma$
$\Rightarrow a = {{ - 400{m^2}{s^{ - 2}}} \over {100m}} = - 4m{s^{ - 2}}$
Now, we know that, force, F = mass x acceleration
Therefore, F = 1 kg X -4ms-2
Or, F = -4ms-2
Thus, force of friction acting upon stone = -4ms-2. Here negative sign shows that force is being applied in the opposite direction of the movement of the stone.

Q.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;(b) the acceleration of the train; and(c) the force of wagon 1 on wagon 2.
Sol.     Given,
force of engine = 40000 NForce of friction = 5000 N
Mass of engine = 8000 kg
Total weight of wagons = 5 x 2000 kg = 10000 kg
(a) The net accelerating force
= Force exerted by engine – Force of fricition
= 40000 N – 5000 N = 35000 N
(b) The acceleration of the trainWe know that, F = mass x acceleration
Or, 35000 N = (mass of engine + mass )
Or $a = {{35000N} \over {18000kg}} = 1.94m{s^{ - 2}}$ of 5 wagons X a
Or, 35000 N = (8000 kg + 10000 kg) X a
Or, 35000N = 18000 kg X a
(c) The force of wagon 1 on wagon 2
Since, net accelerating force = 35000 N
Mass of all 5 wagons = 10000 kg
We know that, F = m x a
Therefore,
$a = {{35000N} \over {10000kg}} = 3.5m{s^{ - 2}}$
Therefore, acceleration of wagons = 3.5 $m{s^{ - 2}}$
Thus, force wagon 1 on 2 = mass of four wagons × acceleration
Or, F = 4 × 2000 kg × 3.5 $m{s^{ - 2}}$
Or, F = 8000 kg × 3.5 $m{s^{ - 2}}$
Or, F = 28000N
Thus, (a) The net accelerating force = 35000N
(b) The acceleration of train = 1.944 $m{s^{ - 2}}$
(c) The force of wagon 1 on 2 = 280000N 35000N = 10000 kg x a

Q.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?
Sol.     Given,
Mass of the vehicle, m = 1500 kg
Acceleration, a = - 1.7 m s -2
Force acting between the vehicle and road,F = ?
We know that, F = m x a
Therefore, F = 1500 kg X 1.7 m s-2
Or, F = - 2550 N
Thus, force between vehicle and road = - 2550 N. Negative sign shows that force is acting in the opposite direction of the vehicle.

Q.9     What is the momentum of an object of mass m, moving with a velocity v?(a) (mv)2 (b) mv2 ( c ) Ω mv2 (d) mv
Sol.     (d) mv
Explanation:
Given, mass = m, velocity = v,therefore, momentum =?
We know that, momentum, P = mass x velocity
Therefore, P = mv
Thus, option (d) mv is correct

Q.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?
Sol.       Since, a horizontal force of 200N is used to move a wooden cabinet, thus a friction force of 200N will be exerted on the cabinet. Because according to third law of motion, an equal magnitude of force will be applied in the opposite direction.

Q.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-1 before the collision during which they stick together. What will be the velocity of the combined object after collision?
Sol.       Since, two objects of equal mass are moving in opposite direction with equal velocity, therefore, the velocity of the objects after collision during which they stick together will be zero.
Explanation:
Given,
Mass of first object, m1 = 1.5 kg
Mass of second object, m2 = 1.5 kg
Initial velocity of one object, u1 = 2.5 m/s
Initial velocity of second object, u2 = -2.5 m/s (Since second object is moving in opposite direction)
Final velocity of both the objects, which stick after
We know that,
${m_1}{u_1} + {m_2}{u_2} + {m_1}{v_1} + {m_2}{v_2}$
$\Rightarrow 1.5kg \times 2.5m{s^{ - 1}} + 1.5kg \times \left( { - 2.5m{s^{ - 1}}} \right) = 1.5kg \times v + 1.5kg \times v$
$\Rightarrow 3.75kg\,m{s^{ - 1}} - 3.75kgm{s^{ - 1}} = v\left( {1.5kg + 1.5kg} \right)$
$\Rightarrow 0 = v \times 3.00kg$
$\Rightarrow v = {0 \over {3.00kg}} = 0$
collision, v = ?
Therefore, final velocity of both the objects after collision will be zero.

Q.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.
Sol.       Because of the huge mass of the truck, the force of static friction is very high. The force applied by the student is unable to overcome the static friction and hence he is unable to move the truck. In this case, the net unbalanced force in either direction is zero which is the reason of no motion happening here. The force applied by the student and the force because of static friction are cancelling out each other. Hence, the rationale given by the student is correct.

Q. 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 change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Sol.         Given,
Mass of hockey ball, m = 200 g = 200/1000 kg = 0.2 kg
Initial velocity of hockey ball, u = 10 m/s
Final velocity of hockey ball, v = - 5 m/s (because direction becomes opposite)
Change in momentum =?
We know  that,
Momentum = mass x velocity
Therefore, Momentum of ball before getting struck
= 0.2 kg x 10 m/s = 2 kg m/sMomentum of ball after getting struck = 0.2 kg x - 5m/s = - 1 kg m/s
Therefore, change in momentum = momentum before getting struck – momentum after getting struck= 2 kg m/s – (-1 kg m/s) = 2 kg m/s + 1 kg m/s = 3 kg m/s
Thus, change of momentum of ball after getting struck = 3 kg m/s

Q.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.
Sol.        Given,
Mass of bullet, m = 10 g = 10/1000 kg = 0.01 kg
Initial velocity of bullet, u = 150 m/s
Since bullet comes to rest, thus final velocity, v = 0
Time, t = 0.03 s
Distance of penetration, i.e. Distance, covered (s) = ?
Magnitude of force exerted by wooden block = ?
$v = u + at$
$\Rightarrow - 150m{s^{ - 1}} = a \times 0.03s$
$\Rightarrow a = - {{150m{s^{ - 1}}} \over {0.03s}} = - 500m{s^{ - 2}}$
We know that,
$s = ut + {1 \over 2}a{t^2}$
$\Rightarrow s = 150m{s^{ - 1}} \times 0.03s + {1 \over 2}\left( { - 5000m{s^{ - 2}}} \right) \times {\left( {0.03s} \right)^2}$
$\Rightarrow s = 4.5m - 2500m{s^{ - 2}} \times 0.0009{s^2}$
$\Rightarrow$ s = 4.5m – 2.25m
$\Rightarrow$ s = 2.25m
We know that,Magnitude of force exerted by wooden block
We know that, Force = mass x acceleration
Or, F = 0.01 kg x – 5000 m s-2 = - 50 N
Therefore,
Penetration of bullet in wooden block = 2.25 m
Force exerted by wooden block on bullet = - 50 N. Here negative sign shows that force is exerted in the opposite direction of bullet.

Q.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.
Sol.       Given, mass of moving object, m1 = 1 kg
Mass of the wooden block, m2 = 5kg
Initial velocity of object, u1 = 10 m/s
Initial velocity of wooden block, u2 = 0
Final velocity or moving object and wooden block, v = ?
Total momentum before collision and after collision = ?
We know that,
${m_1}{u_1} + {m_2}{u_2} = {m_1}{u_1} + {m_2}{u_2}$
$\Rightarrow 1kg \times 10m{s^{ - 1}} + 5kg \times 0 = 1kg \times v + 5kg \times v$
$\Rightarrow 10kgm{s^{ - 1}} = v\left( {1kg + 5kg} \right)$
$\Rightarrow 10kgm{s^{ - 1}} = v \times 6kg$
$\Rightarrow v = {{10kgm{s^{ - 1}}} \over {6kg}} = 1.66m/s...(i)$
Total momentum fo object and wooden block just before collision
$= {m_1}{u_1} + {m_2}{u_2}$
$= 1kg \times 10m{s^{ - 1}} + 5g \times 0 = 10kgm{s^{ - 1}}$
Total momentum just after collision
${m_1}{v_1} + {m_2}{v_2} = {m_1}{v_1} + {m_2}{v_2} = v\left( {{m_1} + {m_2}} \right)$ (Since both the objects move with same velocity 'v' after collision)
$= \left( {1kg + 5kg} \right) \times {{10} \over 6}m/s$ (From equation (i))
$= 6kg \times {{10} \over 6}m/s = 10kgm{s^{ - 1}}$
Thus,
Velocity of both the object after collision = 1.66 m/s
Total momentum before collision = 10 kg m/s
Total momentum after collision = 10 kg m/s

Q.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.
Sol.       Given,
Initial velocity, u = 5 m/s
Final velocity, v = 8 m/s
Mass of the given object, m = 100 kg
Time, t = 6 s
Initial momentum and Final momentum = ?
Magnitude of force exerted on the object = ?
We know that,
Momentum = mass x velocity
Therefore, initial momentum = mass x initial velocity
= 100 kg X 5 m/s = 500 kg m/s
Final momentum = mass x final velocity
= 100 kg x 8 m/s = 800 kg m/s
We know that,
$v = u + at$
$\Rightarrow 8m/s = 5m/s + a \times 6s$
$\Rightarrow 3m/s = a \times 6s$
$\Rightarrow a = {{3m/s} \over {6s}} = 0.5m{s^{ - 2}}$
Now, Force exerted on object = Mass x Acceleration
= 100 kg 0.5 m/s/s
= 50 N

Q.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.
Sol.       We know, that as per the Law of Conservation of Momentum; total momentum of a system before collision is equal to the total momentum of the system after collision. In this case, since the insect experiences a greater change in its velocity so it experiences a greater change in its momentum. From this angle, Kiran’s observation is correct. Motorcar is moving with a larger velocity and has a bigger mass; as compared to the insect. Moreover, the motorcar continues to move in the same direction even after the collision; which suggests that motorcar experiences minimal change in its momentum, while the insect experiences the maximum change in its momentum. Hence, Akhtar’s observation is also correct. Rahul’s observation is also correct; because the momentum gained by the insect is equal to the momentum lost by the motorcar. This also happens in accordance to the law of conservation of momentum.

Q.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.
Sol.       Given,
Mass of dumb-bell = 10 kg
Distance, s = 80 cm = 80/100 = 0.8 m
Acceleration, a = 10 m/s/s
Initial velocity of dumb-bell, u = 0
Momentum = ?
${v^2} = {u^2} + 2as$
$\Rightarrow {v^2} = 0 + 2 \times 10m{s^{ - 2}} \times 0.8m$
$\Rightarrow {v^2} = 16{m^2}{s^{ - 2}}$
$\Rightarrow v = \sqrt {16{m^2}{s^{ - 2}}} = 4m/s$
We know that,Now, we know that, momentum = mass x velocity
= 10 kg x 4 m/s = 40 kg m/s

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