NCERT Solutions for Chapter 9 Force and Laws of Motion Class 9 Science
Book Solutions1
Answer
Inertia is the measure of the mass of the body. The greater is the mass of the body; the greater is its inertia and vice-versa.(a) Mass of a stone is more than the mass of a rubber ball for the same size. Hence, inertia of the stone is greater than that of a rubber ball.
(b) Mass of a train is more than the mass of a bicycle. Hence, inertia of the train is greater than that of the bicycle.
(c) Mass of a five rupee coin is more than that of a one-rupee coin. Hence, inertia of the five rupee coin is greater than that of the one-rupee coin.
2
"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 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 player
→ Second case – Second player
→ Third case – Goalkeeper
→ Fourth case – Goalkeeper
3
Answer
Some leaves of a tree get detached when we shake its branches vigorously because branches comes in motion while the leaves tend to remain at rest due to inertia of rest.4
Answer
In a moving bus, a passenger moves with the bus due to inertia of motion. As the driver applies brakes, the bus comes to rest. But, the passenger tries to maintain to inertia of motion. As a result, a forward force is exerted on him.
Similarly, the passenger tends to fall backwards when the bus accelerates from rest because when the bus accelerates, the inertia of rest of the passenger tends to oppose the forward motion of the bus. Hence, the passenger tends to fall backwards when the bus accelerates forward.
1
Answer
A horse pushes the ground in the backward direction. According to Newton's third law of motion, a reaction force is exerted by the Earth on the horse in the forward direction. As a result, the cart moves forward.2
Answer
When a fireman holds a hose, which is ejecting large amounts of water at a high velocity, then a reaction force is exerted on him by the ejecting water in the backward direction. This is because of Newton's third law of motion. As a result of the backward force, the stability of the firemandecreases. Hence, it is difficult for him to remain stable while holding the hose.
3
Answer
Mass of the rifle, m1= 4 kg
Mass of the bullet, m2= 50g= 0.05 kg
Recoil velocity of the rifle= v1
Bullet is fired with an initial velocity, v2= 35m/s
Initially, the rifle is at rest.
Thus, its initial velocity, v= 0
Total initial momentum of the rifle and bullet system= (m1+m2)v= 0
Total momentum of the rifle and bullet system after firing = m1v1 + m2v2
= 0.05 × 35
= 4v1 + 1.75
According to the law of conservation of momentum:
Total momentum after the firing = Total momentum before the firing 4v1 + 1.75= 0
v1 = -1.75/4 = -0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
4
Answer
Mass of one of the objects, m1 = 100 g = 0.1
kg
Mass of the other object, m2 = 200 g = 0.2
kg
Velocity of m1 before collision, v1=
2 m/s
Velocity of m2 before collision, v2=
1 m/s
Velocity of m1 after collision, v3=
1.67 m/s
Velocity of m2 after collision= v4
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
Therefore, m1v1 + m2v2 = m1v3
+ m2v4
⇒ 2(0.1) + 1(0.2) =
1.67(0.1) + v4(0.2)
⇒ 0.4 = 0.167 + 0.2v4
⇒ v4= 1.165
m/s
Hence, the velocity of the second object becomes 1.165 m/s after the collision.
1
Answer
Yes, an object may travel with a non-zero velocity even when the net external force on it is zero. A rain drop falls down with a constant velocity. The weight of the drop is balanced by the up thrust and the velocity of air. The net force on the drop is zero.2
Answer
When the carpet is beaten, it is suddenly set into motion. The dust particles tend to remain at rest due to inertia of rest, therefore the dust comes out of it.3
Answer
When a bust starts suddenly, the lower part of the luggage kept on the roof being in contact with the bus begins to move forward with the speed of bus, but the upper part tends to remain at rest due to inertia of rest. Therefore, the upper part is left behind and hence luggage falls backward. So, it is advised to tie any luggage kept on the roof of a bus with a rope.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 ball slows down and comes to rest due to opposing forces of air resistance and frictional force on the ball opposing its motion. Therefore the choice (c) there is a force on the ball opposing the motion is correct.5
Answer
Initial velocity, u = 0
Distance travelled, s = 400 m
Time taken, t = 20 s
We know, s = ut + ½ at2
Or, 400 = 0 + ½ a (20)2
Or, a = 2 ms–2
Now, m = 7 metric tonne = 7000 kg, a = 2 ms–2
Or, F = ma = 7000 × 2 = 14000 N Ans.
6
Answer
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0
Distance covered by the stone, s = 50 m
Since, v2 - u2 = 2as,
Or, a = –4 ms-2
Force of friction, F = ma = – 4N
7
(a) the net accelerating force;
(b) the acceleration of the train; and
(c)the force of wagon 1 on wagon 2.
Answer
(a) Force exerted by the engine, F = 40000 NFrictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F − Ff = 40000 − 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon x Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass, M = m = 10000 kg
From Newton’s second law of motion:
Fa= Ma
a = Fa/M = 35000 10000 = 3.5 ms-2
Hence, the acceleration of the wagons and the train is 3.5 m/s2.
(c) Mass of all the wagons except wagon 1 is 4 × 2000 = 8000 kg
Acceleration of the wagons = 3.5 m/s2
Thus, force exerted on all the wagons except wagon 1
= 8000 × 3.5 = 28000 N
Therefore, the force exerted by wagon 1 on the remaining four wagons is 28000 N.
Hence, the force exerted by wagon 1 on wagon 2 is 28000 N.
8
Answer
Mass of the automobile vehicle, m= 1500 kgFinal velocity, v= 0 (finally the automobile stops)
Acceleration of the automobile, a= −1.7 ms−2
From Newton’s second law of motion:
Force = Mass x Acceleration = 1500 x (−1.7) = −2550 N
Hence, the force between the automobile and the road is −2550 N, in the direction opposite to the motion of the automobile.
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
Answer
(d) mvMass of the object = m
Velocity = v
Momentum = Mass x Velocity
Momentum = mv
10
Answer
The cabinet will move with constant velocity only when the net force on it is zero.Therefore, force of friction on the cabinet = 200 N, in a direction opposite to the direction of motion of the cabinet.
11
Answer
Mass of one of the objects, m1 = 1.5 kg Mass of the other object, m2 = 1.5 kg
Velocity of m1 before collision, u1 = 2.5 m/s
Velocity of m2, moving in opposite direction before collision, u2 = −2.5 m/s
Let v be the velocity of the combined object after collision. By the law of conservation of momentum,
Total momentum after collision = Total momentum before collision,
Or, (m1 + m2) v = m1u1 + m2u2
Or, (1.5 + 1.5) v = 1.5 × 2.5 +1.5 × (–2.5) [negative sign as moving in opposite direction]
Or, v = 0 ms–1
12
Answer
The logic is that Action and Reaction always act on different bodies, so they can not cancel each other. When we push a massive truck, the force of friction between its tyres and the road is very large and so the truck does not move.13
Answer
Mass of the hockey ball, m =
200 g = 0.2 kg Hockey ball travels with velocity, v1 = 10 m/s
Initial momentum = mv1
Hockey ball travels in the opposite direction with
velocity, v2 = −5 m/s
Final momentum = mv2
Change in momentum = mv1 − mv2 = 0.2 [10 − (−5)] = 0.2 (15) = 3 kg ms−1
Hence, the change in momentum of the hockey ball is 3
kg ms−1.
14
Answer
Initial velocity, u = 150 m/s
Final velocity, v = 0 (since the bullet finally comes to rest)
Time taken to come to rest, t = 0.03 s
According to the first equation of motion, v = u + at
Acceleration of the bullet, a
0 = 150 + (a × 0.03 s)a = -150/0.03 = -5000 m/s2
(Negative sign indicates that the velocity of the bullet is decreasing.)
According to the third equation of motion:v2= u2+ 2as
0 = (150)2+ 2 (-5000)
= 22500 / 10000
= 2.25 m
Hence, the distance of penetration of the bullet into the block is 2.25 m.
From Newton’s second law of motion:
Force, F = Mass × Acceleration
Mass of the bullet, m = 10 g = 0.01 kg
Acceleration of the bullet, a = 5000 m/s2
F = ma = 0.01 × 5000 = 50 N
Hence, the magnitude of force exerted by the wooden block on the bullet is 50 N.
15
Answer
Mass of the object, m1 = 1 kgVelocity of the object before collision, v1 = 10 m/s
Mass of the stationary wooden block, m2 = 5 kg
Velocity of the wooden block before collision, v2 = 0 m/s
∴ Total momentum before collision = m1 v1 + m2 v2
= 1 (10) + 5 (0) = 10 kg ms−1
It is given that after collision, the object and the wooden block stick together.
Total mass of the combined system = m1 + m2
Velocity of the combined object = v
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
m1 v1 + m2 v2 = (m1 + m2)v
⇒ 1 (10) + 5 (0) = (1 + 5)v
⇒ v = 10/6
= 5/3
The total momentum after collision is also 10 kg m/s.
Total momentum just before the impact = 10 kg ms−1
Total momentum just after the impact = (m1 + m2)v = 6 × 5/3 = 10 kg ms-1
Hence, velocity of the combined object after collision = 5/3 ms-1
16
Answer
Initial velocity of the object, u = 5 m/sFinal velocity of the object, v = 8 m/s
Mass of the object, m = 100 kg
Time take by the object to accelerate, t = 6 s
Initial momentum = mu = 100 × 5 = 500 kg ms−1
Final momentum = mv = 100 × 8 = 800 kg ms−1
Force exerted on the object, F = (mv - mu)/ t
= m (v-u)/t
= 800 - 500
= 300/6
= 50 N
Initial momentum of the object is 500 kg ms−1.
Final momentum of the object is 800 kg ms−1.
Force exerted on the object is 50 N.
17
Answer
The suggestion made by Kiran that the insect suffered a greater change in momentum as compared to the change in momentum of the motor car is wrong.
The suggestion made by Akhtar that the motor car exerted a larger force on the insect because of large velocity of motor car is also wrong. The explanation put forward by Rahul is correct. On collision of insect with motor car, both experience the same force as action and reaction are always equal and opposite. Further, changes in their momenta are also the same. Only the signs of changes in momenta are opposite, i.e., change in momenta of the two occur in opposite directions, though magnitude of change in momentum of each is the same.
18
Answer
Mass of the dumbbell, m = 10 kgDistance covered by the dumbbell, s = 80 cm = 0.8 m
Acceleration in the downward direction, a = 10 m/s2
Initial velocity of the dumbbell, u = 0
Final velocity of the dumbbell (when it was about to hit the floor) = v
According to the third equation of motion:
v2 = u2 + 2as
⇒ v2 = 0 + 2 (10) 0.8
⇒ v = 4 m/s
Hence, the momentum with which the dumbbell hits the floor is
= mv = 10 × 4 = 40 kgms−1
A1
Time in seconds |
Distance in metres |
0 |
0 |
1 |
1 |
2 |
8 |
3 |
27 |
4 |
64 |
5 |
125 |
6 |
216 |
7 |
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 ?
Answer
(a) There is an unequal change of distance in an equal interval of time.Thus, the given object is having a non-uniform motion. Since the velocity of the object increases with time, the acceleration is increasing.
(b) The object is in accelerated condition. According to Newton's second law of motion, the force acting on an object is directly proportional to the acceleration produced in the object. So, we can say unbalanced force is acting on the object.
A2
(Assume that all persons push the motorcar with the same muscular effort
Answer
Mass of the motor car = 1200 kgOnly two persons manage to push the car. Hence, the acceleration acquired by the car is given by the third person alone.
Acceleration produced by the car, when it is pushed by the third person,
a = 0.2 m/s2
Let the force applied by the third person be F.
From Newton’s second law of motion:
Force = Mass × Acceleration
F = 1200 × 0.2 = 240 N
Thus, the third person applies a force of magnitude 240 N.
Hence, each person applies a force of 240 N to push the motor car.
A3
Answer
Mass of the hammer, m = 500 g = 0.5 kgInitial velocity of the hammer, u = 50 m/s
Time taken by the nail to the stop the hammer, t = 0.01 s
Velocity of the hammer, v = 0 (since the hammer finally comes to rest)
From Newton’s second law of motion:
Force, f = m(v-u)/t
= 0.5(0-50)/0.01
= -2500 N
The hammer strikes the nail with a force of −2500 N. Hence, from Newton’s third law of motion, the force of the nail on the hammer is equal and opposite, i.e., +2500 N.
A4
Answer
Mass of the motor car, m = 1200 kgInitial velocity of the motor car, u = 90 km/h = 25 m/s
Final velocity of the motor car, v = 18 km/h = 5 m/s
Time taken, t = 4 s
According to the first equation of motion:
v = u + at
⇒ 5 = 25 + a (4)
⇒ a = − 5 m/s2
Negative sign indicates that its a retarding motion i.e. velocity is decreasing.
Change in momentum = mv − mu = m (v−u)
= 1200 (5 − 25) = −24000 kg m s−1
∵ Force = Mass × Acceleration
= 1200 × −5 = −6000 N
Acceleration of the motor car = −5 m/s2
Change in momentum of the motor car = −24000 kg m s−1
Hence, the force required to decrease the velocity is 6000 N.
(Negative sign indicates retardation, decrease in momentum and retarding force)