AP 9th Class Physics 3rd Lesson Force and Laws of Motion Questions and Answers

AP 9th Class Physics 3rd Lesson Force and Laws of Motion Questions and Answers (Exercise)

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Question 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 not, provide a reason.

Answer:

Yes. Object can travel with non-zero velocity without experiencing net zero external unbalanced force in the space. However without unbalanced force the object can’t travel with non – zero velocity on the earth.

Question 2.
When a carpet is beaten with a stick, dust comes out of it. Explain.
Answer:
When a carpet is beaten with a stick it suddenly comes into motion.
But the dust in it continue to remain in rest due to inertia of rest.
Hence the dust comes out.

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

i) Backward – due to inertia of rest
ii) Forward – due to inertia of motion
iii) Sideways – due to inertia of direction
That’s why luggage should be tied.

Question 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.

Question 5.
A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20. Find its acceleration. Find the force acting on it if its mass is 7 tonnes
(Hint : 1 tonne = 1000 kg)
Answer:
Truck: Initial velocity u=0; Displacement s=400m
Time t=20sec ; Mass m=7 tonnes =7000 kg [∵ 1 tonne =1000kg]

Question 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?
Answer:

Mass m =1 kg
Initial velocity u=20ms-1
Final velocity
v=0 (comes to rest); Displacement
s=50 m
From Kinematics,


Note : Friction force is always opposite to the motion.

Question 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:
Mass of enging = 8000 kg;
Mass of each wagon =2000 kg
Force generated by Engine =40000 N; Force of Friction =5000 N
a) Net Accelerating Force =40000-5000=35000 N
b) Acceleration of the train F = ma (Newton’s 2nd law)

Question 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 ms-2?
Answer:
Automobile vehicle Visualisation :

Mass m =1500kg
Acceleration a =-1.7ms-2
F = ma × [By Newton’s 2nd law]
F=1500 × -1.7=-2550N
So, an opposite Force -2550 N should be applied to bring the vehicle to rest.
Question 9.
What is the momentum of an object of mass m, moving with a velocity v ?
a) (mv)2
b) mv2
c) 1/2 mv2
d) mv
Answer:
d) mv
Question 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:

Cabinet Visualisation:

To move the block with constant velocity across a floor the friction force exerted on the cabinet is -200 N.

Question 11.
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:
The boys reason is not correct. Heavier trucks have greater inertia, so it requires a greater unbalanced force to change its state of rest.

Question 12.
A hockey ball of mass 200 g travelling at 10 ms-1 is struck by a hockey stick so as to return it along its original path with a velocity at 5 ms-1. Calculate the magnitude of change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Answer:

Question 13.
A bullet of mass 10 g travelling horizontally with a velocity of 150 ms-1 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:

Question 14.
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, calcula inte the velocity of the combined object.
Answer:

Block – A
Mass m1= 1 kg
Initial velocity u1= 10 ms-1

Block – B
Mass m2 = 5 kg
Initial velocity u2= 0 ms-1

i) Block A is motion & Block B is at rest.
ii) Block A travelled in straight line & stuck with Block B.
iii) As they are moving together they will have a common velocity.
iv) Total momentum before collision = m1u1 + m2u2
= 1×10+5×0=10 kgms-1 or (N)

b) Momentum after Impact = 10 kg ms-1 (or) (N)
Reason : Momentum before collision is equal to momentum after collision.
c) Combined – velocity: m1u1 + m2u2 = m2v+ m2v
⇒ m1u1 + m2u2= v(m1+ m2)
⇒ 1 x 10 + 5 x 0 = v(1+ 5)

Question 15.
An object of mass 100 kg is accelerated uniformly from a velocity of 5 ms-1 to 8 ms-1 in 6s. Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object.
Answer:

Object:

Mass m = 100 kg Initial velocity u = 5 ms-1
Final velocity v = 8 ms-1,Time t, = 6 sec
Initial momentum P1 = mu = 100 x 5 = 500 kg ms-1 or (N)
Final momentum P2 = mv= 100 x 8 = 800 kg ms-1 (or) (N)

Question 16.
Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expresssway 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 to 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:

Kiran Akhtar’s reasons are not correct.
Rahul’s suggestion is correct. Force exerted by windshield and insect are equal opposite according to Newton’s 3rd law.
But the mass of the insect is lesser than the mass windshield the insect experiences a greater force due to this it dies.

Question 17.
How much momentum will a dumb – bell of mass 10 kg transfer to the floor if it falls from a height of 80cm ? Take its downward acceleration to be 10 ms-2.
Answer:

Dumb-Bell :

AP 9th Class Physics 3rd Lesson Force and Laws of Motion Questions and Answers