
01.Matter in Our Surroundings
11
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8

Lecture1.9

Lecture1.10

Lecture1.11


02.Is Matter Around Us Pure
11
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11


03.Atoms and Molecules
7
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7


04.Structure of The Atom
7
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6

Lecture4.7


05.Cell  Fundamental Unit of Life
11
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9

Lecture5.10

Lecture5.11


06.Tissues
8
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6

Lecture6.7

Lecture6.8


07.Diversity in Living Organisms
8
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8


08.Motion
11
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9

Lecture8.10

Lecture8.11


09.Force and Newtons Laws of Motion
12
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9

Lecture9.10

Lecture9.11

Lecture9.12


10.Gravitation
9
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8

Lecture10.9


11.Work and Energy
7
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6

Lecture11.7


12.Sound
8
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8


13.Why do We Fall Ill
7
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7


14.Natural Resources
11
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8

Lecture14.9

Lecture14.10

Lecture14.11


15.Improvements in Food Resources
7
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

NCERT Solutions – Force and Newtons Laws of Motion
Intext Questions
Q.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 onerupee coin?
Sol. (a) A stone.
(b) A train
(c) A five rupees coin.Explanation – Inertia is associated with mass. Objects having more mass have more inertia.
Q.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.
Sol. The velocity of football changes four times.First, when a football player kicks to another player, second when that player kicks the football to the goalkeeper. Third when the goalkeeper stops the football. Fourth when the goalkeeper kicks the football towards a player of his own team.Agent supplying the force –First case – First playerSecond case – Second playerThird case – GoalkeeperFourth case – Goalkeeper
Q.3 Explain why some of the leaves may get detached from a tree if we vigorously shake its branch.
Sol. The answer of this cause lies behind the Newton’s First Law of Motion. Initially, leaves and tree both are in rest. But when the tree is shaken vigorously, tree comes in motion while leaves have tendency to be in rest. Thus, because of remaining in the position of rest some of the leaves may get detached from a tree if we vigorously shake its branch.
Q.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?
Sol. In a moving bus, passengers are in motion along with bus. When brakes are applied to stop a moving bus, bus comes in the position of rest. But because of tendency to be in the motion a person falls in forward direction.Similarly, when a bus is accelerated from rest, the tendency to be in rest, a person in the bus falls backwards.
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Q.1 If action is always equal to the reaction, explain how a horse can pull a cart.
Sol. Horse pushes the ground in backward direction. In reaction to this, the horse moves forward and cart moves along with horse in forward direction.In this case when horse tries to pull the cart in forward direction, cart pulls the horse in backward direction, but because of the unbalanced force applied by the horse, it pulls the cart in forward direction.
Q.2 Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity.
Sol. When large amount of water is ejected from a hose at a high velocity, according to Newton’s Third Law of Motion, water pushes the hose in backward direction with the same force. Therefore, it is difficult for a fireman to hold a hose in which ejects large amount of water at a high velocity.
Q.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.
Sol. Given,Mass of rifle (m_{1}) = 4 kgInitial velocity of rifle (u_{1}) = 0Mass of bullet (m_{2}) = 50 g = 50/1000 kg = 0.05 kgInitial velocity of bullet (u_{2}) = 0Final velocity (v_{2}) of bullet = 35 m/sFinal velocity [Recoil velocity] of rifle (v_{1}) = ? We
know that,
m1u1+m2u2=m1v1+m2v2
⇒4kg×0+0.5kg×0=4kg×v1+0.005kg×35ms−1
⇒0=4kg×v1+1.75kgm/s
⇒−4kg×v1=1.75kgm/s
⇒vl=1.75kgm/s−4kg=−0.4375ms≈−0.44m/s
Here negative sign of velocity of rifle shows that rifle moves in the opposite direction of the movement of bullet. Therefore, recoil velocity of rifle is equal to 0.44 m/s.
Q.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.
Sol. Given,Mass of first object (m_{1}) = 100 g = 100/1000 kg = 0.1 kgInitial velocity of first object (u_{1}) = 2 m/sFinal velocity of first object after collision (v_{1}) = 1.67 m/sMass of second object (m_{2}) = 200 g = 200/1000 kg = 0.2 kgInitial velocity of second object (u_{2) = 1 m/s}Final velocity of second object after collision (v_{2}) = ? We know that,
m1u1+m2u2=m1v1+m2v2
⇒0.1kg×2ms−1+0.2kg×1ms−1=0.1kg×1.67ms−1+0.2kg×v2
⇒0.2kgms−1+0.2kgms−1=0.167kgms−1+0.2kg×v2
⇒0.4kgms−1−0.167ms−1=0.2kg×v2
⇒v2=0.233kgms−10.2kg=1.165ms−1
Thus, velocity of the second object after collision = 1.165 ms−1
Exercise
Q.1 An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a nonzero 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 nonzero 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+12at2
⇒400m=0×20s+12×a×(20s)2
⇒400m=12×a×400s2
⇒400m=a×200s2
⇒a=400m200s2=2ms−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,
v2=u2+2as
⇒(0)2=(20m/s)2+2a×50m
⇒−400m2s−2=100ma
⇒a=−400m2s−2100m=−4ms−2
Now, we know that, force, F = mass x acceleration
Therefore, F = 1 kg X 4ms2
Or, F = 4ms2
Thus, force of friction acting upon stone = 4ms2. 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=35000N18000kg=1.94ms−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=35000N10000kg=3.5ms−2
Therefore, acceleration of wagons = 3.5 ms−2
Thus, force wagon 1 on 2 = mass of four wagons × acceleration
Or, F = 4 × 2000 kg × 3.5 ms−2
Or, F = 8000 kg × 3.5 ms−2
Or, F = 28000N
Thus, (a) The net accelerating force = 35000N
(b) The acceleration of train = 1.944 ms−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 s2?
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 s2
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 s1 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,
m1u1+m2u2+m1v1+m2v2
⇒1.5kg×2.5ms−1+1.5kg×(−2.5ms−1)=1.5kg×v+1.5kg×v
⇒3.75kgms−1−3.75kgms−1=v(1.5kg+1.5kg)
⇒0=v×3.00kg
⇒v=03.00kg=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
⇒−150ms−1=a×0.03s
⇒a=−150ms−10.03s=−500ms−2
We know that,
s=ut+12at2
⇒s=150ms−1×0.03s+12(−5000ms−2)×(0.03s)2
⇒s=4.5m−2500ms−2×0.0009s2
⇒ s = 4.5m – 2.25m
⇒ 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 s2 = – 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,
m1u1+m2u2=m1u1+m2u2
⇒1kg×10ms−1+5kg×0=1kg×v+5kg×v
⇒10kgms−1=v(1kg+5kg)
⇒10kgms−1=v×6kg
⇒v=10kgms−16kg=1.66m/s...(i)
Total momentum fo object and wooden block just before collision
=m1u1+m2u2
=1kg×10ms−1+5g×0=10kgms−1
Total momentum just after collision
m1v1+m2v2=m1v1+m2v2=v(m1+m2) (Since both the objects move with same velocity ‘v’ after collision)
=(1kg+5kg)×106m/s (From equation (i))
=6kg×106m/s=10kgms−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
⇒8m/s=5m/s+a×6s
⇒3m/s=a×6s
⇒a=3m/s6s=0.5ms−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 dumbbell 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/s2.
Sol. Given,
Mass of dumbbell = 10 kg
Distance, s = 80 cm = 80/100 = 0.8 m
Acceleration, a = 10 m/s/s
Initial velocity of dumbbell, u = 0
Momentum = ?
v2=u2+2as
⇒v2=0+2×10ms−2×0.8m
⇒v2=16m2s−2
⇒v=16m2s−2−−−−−−−√=4m/s
We know that,Now, we know that, momentum = mass x velocity
= 10 kg x 4 m/s = 40 kg m/s
Exemplar
Multiple Choice Questions :
Q.1 Which of the following statement is not correct for an object moving along a straight path in an accelerated motion?
(a) Its speed keeps changing
(b) Its velocity always changes
(c) It always goes away from the earth
(d) A force is always acting on it
Sol. (c)
Q.2 According to the third law of motion, action and reaction
(a) Always act on the same body
(b) Always act on different bodies in opposite directions
(c) Have same magnitude and directions
(d) Act on either body at normal to each other
Sol. (b)
Q.3 A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goal keeper to
(a) Exert larger force on the ball
(b) Reduce the force exerted by the ball on hands
(c) Increase the rate of change of momentum
(d) Decrease the rate of change of momentum
Sol. (b)
Q.4 The inertia of an object tends to cause the object
(a) To increase its speed
(b) To decrease its speed
(c) To resist any change in its state of motion
(d) To decelerate due to friction
Sol. (c)
Q.5 A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is
(a) Accelerated
(b) Uniform
(c) Retarded
(d) Along circular tracks
Sol. (a)
Q.6 An object of mass 2 kg is sliding with a constant velocity of 4 m s–1 on a friction less horizontal table. The force required to keep the object moving with the same velocity is
(a) 32 N
(b) 0 N
(c) 2 N
(d) 8 N
Sol. (b)
Q.7 Rocket works on the principle of conservation of
(a) Mass
(b) Energy
(c) Momentum
(d) Velocity
Sol. (c)
Q.8 A water tanker filled up to 2 3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would
(a) Move backward
(b) Move forward
(c) Be unaffected
(d) Rise upwards
Sol. (b)