Force And Laws Of Motion(9th Grade > Physics ) Questions and answers for exam Preparation

9th Grade > Physics

FORCE AND LAWS OF MOTION MCQs

Total Questions : 39 | Page 2 of 4 pages

Question 11. Jaikishan and Arun are pushing each other as shown in the given figure. Jaikishan’s push is weaker than that of Arun. As a result, they move in the direction shown in the figure.

Jaikishan and Arun move in the given direction because of:

Tension force
Unbalanced force
Gravitational force
Balanced force

Discuss Question

Answer: Option B. -> Unbalanced force
:
B
It is clear from the given information that Arun is pushing harder, i.e., their pushes are unequal. Hence, the net force between them will be unbalanced and the direction of force will be as shown in the given figure.

Question 12. A cannon ball of mass

2 k g

is fired at a target

500 m

away. The

400 k g

cannon recoils with a velocity of

2 m s^{- 1}

. Assuming cannon ball moves with constant velocity, when will it hit the target?

1.25 s
2 s
1.5 s
0.2 s

Discuss Question

Answer: Option A. -> 1.25 s
:
A
Let mass of cannon be

m_{1}

400 k g

Let mass of cannon ball be

m_{2}

2 k g

Let initial velocities of the cannon and cannon ball be

u_{1}

and

u_{2}

respectively.
Let final velocity of cannon and cannon ball be

v_{1}

and

v_{2}

respectively.
From thelaw of conservation of momentum,
total momentum before firing = total momentum after firing.

∴

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

As the cannon and cannon ball is at rest initially, the initial velocity will be zero, i.e

0 = m_{1} v_{1} + m_{2} v_{2}

Substituting the values:

0 = 400 \times (- 2) + 2 \times v

\Rightarrow v = \frac{400 \times 2}{2} = 400 m s^{- 1}

Hence, the velocity of the cannonball will be

400 m s^{- 1}

Since we know,

t i m e = \frac{d i s p l a c e m e n t}{v e l o c i t y}

Thus, time taken to hit the target=

\frac{500}{400}

1.25 s

Question 13. When a bullet is fired from a pistol, in which direction the pistol moves (ignore the force exerted by hand)?

Opposite to the bullet's direction.
Perpendicular to the bullet's direction.
It will be in the same position as before.
Along the bullet's direction

Discuss Question

Answer: Option A. -> Opposite to the bullet's direction.
:
A
The total momentum of the bullet-pistol system was initially zero as both of them were at rest with respect to ground. During the shooting of the bullet, net external force on the gun and bullet system is zero. Hence, according to the law of conservation of momentum, the total momentum of the system will be zero even after firing the shot. Since the momentum of the bullet will be in forward direction, in order to balance it, the momentum of the pistol should be in backward direction or the pistol should move backwards.

Question 14. A bullet of mass

20 g

moving with a velocity of

100 {m s}^{- 1}

strikes a wooden block of mass

80 g

and gets embedded into it. Calculate velocity of the combined system.

100 ms−1
30 ms−1
20 ms−1
5 ms−1

Discuss Question

Answer: Option C. -> 20 ms−1
:
C
Let mass of bullet be

m_{1} = 20 g

Let mass of the wooden block be

m_{2} = 80 g

Let initial velocity of the bullet be

u_{1}

100 m s^{- 1}

and final velocity of the bullet be

v_{1}

.
Let the initial velocity of wooden block be

u_{2}

, since it wasat rest, the initial velocity is zero, i.e

u_{2} = 0 m s^{- 1}

Let final velocity of the wooden block be

v_{2}

.
By applying law of conservation of momentum,
intialmomentum of the system = final momentum of the system
i.e,

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

After the bullet is fired, the bullet gets embedded on the block.
Therefore, it acts as a system
Hence , the final velocity of the bullet and wooden block after collision will be equal as it works as a system
i.e

v_{1} = v_{2} = v

where v is the velocity of the system after collision.
Substituting onthe equation, we get

(0.020 \times 100) + (0.080 \times 0) = v \times (0.020 + 0.080)

\Rightarrow

2 = 0.1 \times v

\Rightarrow

v = 20 {m s}^{- 1}

∴

Velocity of system after collision =

20 {m s}^{- 1}

Question 15.
It is easier to stop a

60 g

tennis ball as compared to a

150 g

cricket ball, both having the same velocity.

True
False
20 ms−1
5 ms−1

Discuss Question

Answer: Option A. -> True
:
A
As we know,

Force = mass \times acceleration

. Hence, force is directly proportional to mass. Here, cricket ball has more mass than the tennis ball. So, even if the velocity is the same, the mass of the cricketball and the associated momentum(which is the product of mass and velocity) is greaterthan that of the tennis ball and hence, it is easier to stop a tennis ball as compared to a cricket ball having the same velocity.

Question 16. A man is standing at an edge of a boat. He starts moving towards the centre of the boat. As no external force is acting on the system of boat and man, (neglect friction due to water) the boat will move in opposite direction.

True
False

Discuss Question

Answer: Option A. -> True
:
A
Initially, the total momentum was zero and after the man starts moving, the total momentum would still be zero as no external force is acting on the system of the man and the boat (law of conservation of momentum). So,the boat will move in the direction opposite to the direction of motion of the man.

Question 17. Newton's second law of motion gives us the mathematical expression of force.

True
False
20 ms−1
5 ms−1

Discuss Question

Answer: Option A. -> True
:
A
Newton's second law of motion gives us the quantitative analysis of force. It gives us the mathematical expression of force as it states thatthe rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

Question 18. The rate of change of momentum of a body is directly proportional to the force applied and change in momentum takes place in the same direction as the applied force.

True
False
20 ms−1
5 ms−1

Discuss Question

Answer: Option A. -> True
:
A
According to newton's second law of motion-"the rate of change of momentum is directly proportional to force applied and the change in momentum is in the direction of force applied." Newton's second law of motion gives mathematical expression of force.
Force

\propto

\frac{d p}{d t}

Question 19. An object of mass

2 k g

is sliding with a constant velocity of

4 m s^{- 1}

on a frictionless horizontal table. The force required to keep the object moving with the same velocity is:

32 N
0
2 N
8 N

Discuss Question

Answer: Option B. -> 0
:
B
Ideally, a frictionless horizontal table has no resistance to motion.
Therefore, the object will continue to slide at

4 m s^{- 1}

without any application of force.

∴

The force required to keep the object moving =

0

Question 20.

A car of mass $1, 000 k g$ is moving at a speed of $20 m s^{- 1}$ .
It is brought to rest at a distance of $50 m$ . Calculate the net force acting on the car.

$4, 000 N$
$3, 000 N$
$2, 000 N$
$1, 000 N$

Discuss Question

Answer: Option A. ->

4, 000 N

:
A

For calculating force, we need to calculate deceleration first. From the third equation of motion,
$v^{2} - u^{2} = 2 a S$ where, v- final velocity, u- initial velocity, a- acceleration and S- distance travelled.
Given, u= $20 m s^{- 1}$ , S $= 50 m$ , mass $m = 1000 k g$ and v $= 0$ .
Therefore, $0 - 20^{2} = 2 \times a \times 50$
$\Rightarrow a = - 4 m s^{- 2}$
Let the force acting on the car be $F$ .
From Newton's second law,
$F = m a$
$\Rightarrow F = 1000 \times (- 4) = - 4000 N$
Hence, the net force is $- 4, 000 N$ . The negative sign indicates that the force is acting opposite to the direction of initial velocity and causes deceleration. This net force is actually the frictional force acting between the ground and the tyre of the car.