NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion

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: Yes, if an object experiences net zero external unbalanced force, it is possible for the object to be travelling with non zero velocity. It happens when object moves with constant velocity in a particular direction. Therefore, no net unbalanced is acting on that body. Thus, the objects keep moving with non zero velocity. Inorder to change the state of motion of the body, there is need to apply a non zero external unbalanced force on the body.

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

Answer: When carpet is beaten using a stick, carpet is set into motion. But dust particles on the carpet try to resist their state of motion. By newton’s first law of motion, carpet moves and dust particles stay in their motion of rest. This makes dust particles to come out of the carpet.

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

Answer: When bus accelerates it acquires a state of motion. Luggage kept on the roof of the bus gains it inertia of motion and tries to be in state of rest. With the forward movement of the bus, luggage on roof top tries to be in its original position and thus falls from the roof. To avoid this situation, it is advised to tie luggage kept on the roof of the bus using the 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: (c) There is a force on the ball opposing the motion.

Ball stops after a while because of frictional force acting on it. Frictional force acts in opposite direction of 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.)

Answer: Given: Initial velocity u = 0 and distance travelled by truck is s = 400 m in time t = 20s.

Let’s consider second equation of motion, s = ut+ ½ at²,

∴ 400 = 0 + ½ a(20)²

∴ a = (400 x 2) / 400

∴ a = 2 m/s²

It is given that 1 metric tonne = 1000 kg

∴ 7 metric tonne = 7000kg

Let m be the mass of the truck

∴ m = 7000kg

According to newton’s second law, F = m x a

F = 7000 x 2

∴ F = 14000 N

Therefore, acceleration of the truck is 2 m/s² and the force acting on the truck is 14000 N.

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?

Answer: Given-

Initial velocity u = 20 m/s

Final velocity v = 0

Distance s covered by stone = 50 m

Considering newton’s third equation of motion.

v² = u² + 2as

0² = 20² + 2 a x 50

-400 / 100 = a

∴ a = -4 m/s²

Negative sign of acceleration indicates retardation.

Also, it is given that mass of the stone is 1 kg.

∴ m = 1 kg

Therefore, F = m x a ….. (From Newton’s second law of motion.)

F = 1 x (-4) = – 4 N

Thus, force of friction between stone and ice is -4N.

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.

Answer: (a) Given that:

Engine exerts force of 40000 N = F

Force of friction applied by the track is 5000 N = F_ft

Let, Net accelerating force = Fa

F_a = F – F_ft = 40000 – 5000 = 35000 N

∴ Net accelerating force is 35000 N

(b) Let a be the acceleration of the train

Force of 40000 N is exerted by engine on 5 wagons

Net accelerating force on wagons in 35000N

∴ Mass of wagon m = number of wagons x mass of wagons

∴ m = 5 x 2000 = 10000 kg

Mass of engine m’ = 8000 kg

∴ Total mass M_t= m + m’ = 10000 + 8000 = 18000 kg

Now consider Newton’s second law of motion,

F_a = M_t x a

∴ a = F_a / M_t = 35000 / 18000 = 1.944 m/s²

(c) Mass of all wagons except one = 4 x 2000 = 8000 kg

Wagons acceleration = 3.5 m/s²

Force exerted by wagons except wagon 1 = 8000 x 3.5 = 28000 N

This shows that force of one wagon on other four wagons is 28000N

Therefore, force exerted by wagon 1 on wagon 2 is 28000 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/s²?

Answer: Given –

Mass of automobile vehicle is 1500 kg

Final velocity v = 0 m/s

Acceleration a of automobile = – 1.7 m/s²

According to Newton’s second law of motion,

F = m x a = 1500 x (-1.7) = – 2550 N

Force between automobile and road is – 2550 N in the opposite direction to 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) mv

If m is mass of object and v is the velocity then

Momentum = mass x velocity = 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?

Answer: Newton’s third law states that every action has equal and opposite reaction. Here 200 N force is applied to move a wooden cabinet, therefore 200 N of friction force will be exerted on the cabinet.

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?

Answer: Let the mass of two objects m₁ and m₂ be 1.5 kg

Let v₁ and v₂ be the velocity of both the objects

∴ v₁= 2.5 m/s, v₂ = -2.5m/s

Also, we know that, after collision two objects stick to each other

Therefore total mass of combined objects = m₁ + m₂

And v be the combined velocity of both the objects.

By law of conservation of momentum,

Total momentum before collision of two bodies = total momentum after the collision of two bodies

∴ m₁v₁ + m₂ v₂ = (m₁ + m₂)v

∴ 1.5 (2.5) + 1.5 (-2.5) = (1.5 + 1.5)v

∴ 1.5(2.5 – 2.5) = 3 v

∴ 0 = 3v

∴ v = 0

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: Mass of the truck is very very large. Therefore, friction between road and the truck will be very high. In order to move the truck force should be higher than friction. Therefore, when a truck is pushed, it does not move. Thus, it is said that applied force is cancelled by the frictional force of same magnitude. So, student has correctly justified by answering that two opposite and equal forces cancel each other.

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.

Answer: Given that –

Let m be the mass of hockey ball.

∴ m = 200 g = 0.2 kg

Velocity of hockey ball = v₁ = 10m/s

Velocity of hockey ball in opposite direction = v₂ = – 5 m/s

Initial momentum = mv₁

Final momentum = mv₂

∴ Change in momentum = mv₁– mv₂ = m (v₁ – v₂)

∴ Change in momentum = 0.2 (10 – (-5)) = 0.2 (15) = 3 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.

Answer: Bullet is travelling with the speed 150m/s.

Therefore, velocity of bullet while entering the block is equal to initial velocity u = 150 m/s

∴ Final velocity of bullet v = 0 m/s

Bullet comes to rest in time t = 0.03 s.

Using first equation of motion,

v = u + at

Let a be the acceleration of bullet,

0 = 150 + a (0.03)

∴ a = – 5000 m/s²

Using third equation of motion,

v² = u² + 2as

0 = 150² + 2 (-5000)s

10000s = 22500

∴ s = 22500 / 10000

∴ s = 2.25 m

Using Newton’s second law,

F = m x a

Mass of bullet in kg = 10/1000 = 0.01 kg

∴ F = 0.01 x 5000

F = 50 N

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

Answer: Let m₁ be the mass of an object

∴ m₁ = 1 kg

Let v₁ be the velocity of an object before collision

∴ v₁ = 10 m/s

Let v₂ be the velocity of object after collision

∴ v₂ = 0 m/s

∴ total momentum before collision = m₁v₁ + m₂v₂

∴ total momentum before collision = 1(10) + 5(0) = 10 kg m/s

After collision, object and wooden block stick together

Therefore total mass of combined system = m₁ +m₂, and velocity of combined system = v

Therefore by law of conservation of momentum

m₁v₁ + m₂ v₂ = (m₁+m₂) v

1(10) + 5 (0) = (1+5)v

6v = 10

V = 10/ 6

V = 5/3 m/s

Therefore total momentum after collision = 10 kg m/s

Thus, momentum before the impact = 10 kg m/s

Therefore, momentum after the impact = (m₁ + m₂)v = (1+5)5/3

= 6 x 5 / 3 = 30/3 = 10 kg m/s

Therefore, velocity of combined object is 5/3 m/s.

16. An object of mass 100 kg is accelerated uniformly from a velocity of 5 m s–1 to 8 m s–1 in 6 s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.

Answer: Let u be the initial velocity of object .

∴ u = 5 m/s

Let v be the final velocity of an object

∴ v = 8 m/s

Also, given mass of the object is 100 kg

Object take 6 s to accelerate therefore time t = 6 s

Let initial momentum = mu = 100 x 5 = 500 kg m/s

∴ final momentum = mv = 100 x 8 = 800 kg m/s

Now force exerted on object = F

∴ F = (mv – mu)/ t

= m(v-u)/t

=100(8-5)/6= 300/6 = 50 N

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: Taking look over law of conservation of momentum,

Momentum of car and insect before collision = momentum of car and insect after collision.

Therefore, change in momentum of car and insect before collision is zero. Kiran’s suggestion is correct because – insect struck the windscreen shows that direction of insect is revered and car is moving with constant velocity, therefore, velocity of insect changes on great amount. Momentum of insect after collision is high as car is moving with high speed. Therefore, momentum gained by insect is momentum lost by the car. Moreover, mass of car is very large in comparison to mass of insect, therefore Akhtar’s conclusion is also correct. Rahul’s conclusion is also correct because both car and insect experienced equal opposite force on each other due to Newton’s action reaction law. However, statement made by Rahul is incorrect because, system suffers change in momentum because momentum before collision is equal to momentum after collision.

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/s².

Answer: Given that – 1) Mass of dumbbell = 10 kg

2) Distance covered by dumbell during the fall = 80 cm = 0.8 m

3) Acceleration of dumbbell a = 10 m/s²

Let u be initial velocity and v be the final velocity of the dumbbell

∴ u = 0

Using third equation of motion,

v² = u² + 2as

= 0 + 2 (10)(0.8)

= 16

v = 4m/s

∴ Momentum of dumbell = mv = 10 x 4 = 40 kg m/s

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