MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 9

Q.1.

Swimming is possible on account of :-

15%
First law of motion
15%
Second law of motion
62%
Third law of motion
8%
Newton's law of motion

Explanation

Q.2.

The action and reaction force referred to in the third law

0%
must act on the same object
50%
must act on different object
50%
may act on different object
0%
act in the same direction

Explanation

Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of force acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.

Hence,

option

$B$ is correct answer.

Q.3.

An object of mass 10 kg moves at a constant speed of 10m/s. A constant force that acts for 4 sec on the object gives it a speed of 2m/s in opposite direction. The force acting on the object is:

0%
-3N
67%
-30N
17%
3N
17%
30N

Explanation

Force is equal to the rate change of momentum. i.e

$\vec{F}=m \dfrac{d \vec{v}}{dt}$
So,

$F= 10\times \dfrac{(-2)-10}{4} N = -30N$

Q.4.

A ball of mass

$0.2 \ kg$ is thrown vertically upwards by applying a force by hand. If the hand moves

$0.2 \ m$ while applying the force and the ball goes upto

$2 \ m$ height further, find the magnitude of force applied by the hand. Consider

$g =10$

${ ms }^{ -2 }$ .

11%
$20 \ N$
11%
$22 \ N$
56%
$4 \ N$
22%
$16 \ N$

Explanation

The height attained by the ball

$h=2 \ m$

So, the velocity of ball when leaving the hand

$=\sqrt{2gh}=\sqrt{40} \ m/s$

By using 3rd equation of motion for the duration while the force was applied,

$v^{2}-u^{2}=2as$

$40=2\times a\times 0.2$

$a=100m/s^{2}$

By Newton's second law of motion,

$F _{net}$ = Mass of ball

$\times$ acceleration

$=20$ N in upward direction

$F _{net} = F _{hand} - F _{gravity}$

$F_{hand}$ =

$20 + mg = 20+(0.2\times 10)$

$F_{hand}=22 \ N$

Q.5.

Assertion (A): According to Newton’s third law sum of action and reaction is not equal to zero

Reason (R) : The forces action and reaction acts on different bodies.

20%
A and R are correct and R is the correct explanation of A
60%
A and R are correct and R is not the correct explanation A
0%
A is true and R is false
20%
A is false and R is true

Explanation

Q.6.

A man is standing on a spring platform. The reading of spring balance is 60 kg wt. If the man jumps outside the platform, then the reading of the spring balance:

0%
remains the same
20%
decreases
20%
increases
60%
first increases and then decreases to zero

Explanation

Q.7.

When two identical balls are moving with equal speeds in opposite directions, which of the following is true for the system of two bodies?

0%
Momentum is zero, kinetic energy is zero
25%
Momentum is not zero, kinetic energy is zero
25%
Momentum is zero, kinetic energy not zero
50%
Momentum is not zero, kinetic energy is not zero

Explanation

Let mass of the balls be

$m_1$ and

$m_2$ and their velocities

$v_1$ and

$v_2$ respectively.

$m_1 = m_2 ,\ v_1 = -v_2$

Total momentum,

$\vec{p}_{total} = m_1 \vec{v_1} + m_2 \vec{u_2}$

$\vec{p}_{total} = m_1 \vec{v_1} - m_1 \vec{v_1} = 0$

Total kinetic energy,

$E = \dfrac{1}{2} m_1 v^2_1 + \dfrac{1}{2} m_2 v_2^2$

$E = \dfrac{1}{2} m_1 v^2_1 + \dfrac{1}{2} m_1 v^2_1 = m_1v^2_1$

So momentum is zero, kinetic energy is not zero.

Q.8.

A parachutist along with the parachute has a mass of 75 kg. He jumps from a large height and is coming down with a uniform velocity with the parachute fully opened. The magnitude of force due to the air resistance is: ( take g

$={10m/s}^{ 2 }$ ).

0%
$zero$
20%
$< 750 N$
60%
$> 750 N$
20%
$750N$

Explanation

The downward force on the parachute is

$mg=750N$ . Since he is coming down with a uniform velocity that means the downward force is balanced by the upward force due to air. So, the force due to the air resistance is equal to

$750N$ .

Q.9.

When you are holding a ball in your hand, the reaction force to the force of gravity on the ball is, the force exerted by the:

0%
ball on the earth.
25%
ball on the hand.
50%
hand on the ball.
25%
earth on the ball.

Explanation

Q.10.

Inside a railway car a plumb bob (a heavy mass suspended by a thread) is suspended from the roof and a helium filled balloon is tied by a string to the floor of the car. When the railway car accelerates to the right, then

33%
Both the plumb bob and balloon move to the left
33%
Both the plumb bob and balloon move to the right
33%
Plumb bob moves to the left and the balloon moves to the right
0%
Plumb bob moves to the right and the balloon moves to the left

Explanation

Q.11.

A father and his seven year old son are facing each other on ice skates (assume no friction with the ground). With their hands, they push off against one another. Regarding the forces that act on them as a result of this and the accelerations they experience. Which of the following statements is correct?

0%
Father exerts more force on the son and experiences less acceleration.
33%
Son exerts less force on the father and experiences more acceleration.
33%
Father exerts as much force on the son as the son exerts on the father, but the father experiences less acceleration.
33%
Father exerts as much force on the son as the son exerts on the father, but the father experiences more acceleration.

Explanation

The normal forces between the father and son constitute an action-reaction pair. Therefore they are equal.
But since the mass of father is greater compared to son, the acceleration of father is lesser than the son.

Acceleration of son :

$a_{son} : \frac{F}{m}$
Acceleration of father :

$a_{father} : \frac{F}{M}$
As m<M hence

$a_{son}>a_{father}$

Q.12.

Two astronauts A and B off mass

$m_a$ and

$m_b$ are in space. A pushes B with some force. (

$m_a =m_b$ )

0%
acceleration of A is greater than that of B
33%
acceleration of A is less than that of B
0%
neither is accelerated
67%
their accelerations are equal in magnitude but opposite in direction

Explanation

A pushes B with some force, then by newton's third law same force will be applied by B on A, now since mass of both the astronaut are same,as

$F=ma$

As mass is same and just direction of force is opposote, so acceleration will be equal in magnitude but opposite in direction.

Q.13.

A body of mass

$2 \ kg$ moving on a horizontal surface with an initial velocity of

$4 \ ms^{-1}$ , comes to rest after

$2 \ s$ . If one wants to keep this body moving on the same surface with a velocity of

$4ms^{-1}$ , the force required is:

0%
zero
33%
$2 \ N$
67%
$4 \ N$
0%
$8 \ N$

Explanation

Given:

Mass

$m = 2 \ kg$

Initial velocity

$u=4 \ ms^{-1}$

Final velocity

$v=0$

Time

$t=2 \ s$

The body comes to rest due to a retarding acceleration provided by a force, say

$F_1$ .

Let us calculate this acceleration by first equation of motion.

$v=u+at$

$\Rightarrow 0=4+a(2)$

$\Rightarrow a= \dfrac{-4}{2}=-2 \ m/s^2$

By Newton's second law, the force providing this acceleration must be

$F_1=mass \times acceleration$

$\Rightarrow F_1=2 \times -2=-4N$

Negative sign indicates the force is opposite to the velocity.

Now, to keep the body moving with the same velocity, the net force on the body must be zero. The force

$F_2$ required to do so must be equal and opposite to

$F_1.$

$F_2=-(F_1)=-(-4 \ N)=+4 \ N$

Option (C) is correct.

Q.14.

A person weighing 60 kg in a small boat of mass 140 kg that is at rest, throws a 5 kg stone in the horizontal direction with a velocity of

$14ms^{-1}$ . The magnitude of velocity of the boat immediately after the throw is:

0%
$1.2 ms^{-1}$
0%
$0.5 ms^{-1}$
50%
$0.35 ms^{-1}$
50%
$0.65 ms^{-1}$

Explanation

As there is no external force applied, momentum of system will remain conserved.
Initial momentum

$=$ final momentum

$\Rightarrow$

$(5)(14) = (140+60)(V)$ , finally system (boat + person) will move with velocity

$V$ .
This gives

$V = 0.35 m/s.$

Q.15.

A bullet of mass

$10\ g$ moving with a horizontal velocity

$100\ m/s$ passes through a wooden block of mass

$100\ g$ . The block is resting on a smooth horizontal floor. After passing through the block the velocity of the bullet is

$10\ m/s$ . The velocity of the emerging bullet with respect to the block is:

100%
$10\ m/s$
0%
$9 \ m/s$
0%
$1 \ m/s$
0%
$5\ m/s$

Explanation

Using momentum conservation,

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

$m_1=10\ g$

$,m_2=100\ g$

$,u_1=100\ m/s$

$,u_2=0$

$,v_1=10\ m/s$

$\Rightarrow 10\times 100+100 \times 0= 10 \times 10+100\times v_2$

$\Rightarrow 1000 = 100 + 100v_2 \Rightarrow v_2=9\ m/s$

Therefore

$v_1$ with respect to

$v_2 =v_1-v_2=1\ m/s$

Q.16.

A man and a cart move towards each other. The man weighs 64 kg and the cart 32 kg. The velocity of the man is 5.4 km/hr and that of the cart is 1.8km/hr. When the man approaches the cart, he jumps on to it. The velocity of the cart carrying the man will be:

0%
30 km/hr
0%
3 km/hr
100%
1.8 km/ hr
0%
zero

Explanation

Using conservation of linear momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$
Thus we get

$64\times 1.5-32\times 0.5=96\times v$

$\Rightarrow v=\dfrac{80}{96}m/s$

$=3km/hr$ in the direction of man.

Q.17.

A

$60 \ kg$ boy lying on a surface of negligible friction throws a stone of mass

$1 \ kg$ horizontally with a speed of

$12 \ m/s$ away from him. With what kinetic energy does he move back due to result of this action?

0%
$2.4$ $J$
0%
$7.2$ $J$
50%
$1.2$ $J$
50%
$36$ $J$

Explanation

The given surface is frictionless so no external force is acting on the system. Hence linear momentum of the system will remain conserved.

$\Rightarrow$ Initial momentum

$=$ Final momentum

$\Rightarrow 0 = 60\times V + 1\times 12$

$\Rightarrow v = -\dfrac{1}{5}m/s$ (-ve sign show man moves in opposite direction of stone)

Kinetic energy of the boy after separation

$(K.E.) = \dfrac{1}{2} mv^{2}$

$K.E. = \dfrac{1}{2}\times 60\times \dfrac{1}{25}$

$K.E.= 1.2J$

Q.18.

When a gun of mass

$M$ fires a bullet of mass

$m$ , the total energy released in the explosion is found to be

$E$ . The kinetic energy of the bullet is:

100%
$\dfrac{EM}{(M+m)}$
0%
$\dfrac{Em}{(M+m)}$
0%
$\dfrac{E(M+m)}{M}$
0%
$\dfrac{E(M+m)}{m}$

Explanation

Using momentum conservation on bullet + gun as a system

$\Rightarrow$ initial momentum $=$ Final momentum

$0=-MV+mv$

$\Rightarrow mv=MV$ ..........(1)

Now total energy $=E$

$\Rightarrow \dfrac{1}{2}MV^{2}+\dfrac{1}{2}mv^{2}=E$

From (1)

$V=\dfrac{mv}{M}$

$\Rightarrow\dfrac{1}{2}M\dfrac{m^{2}}{M^2}v^{2}+\dfrac{1}{2}mv^{2}=E$

$\dfrac{1}{2}mv^{2}\left ( \dfrac{m}{M}+1 \right )=E$

$\dfrac{1}{2}mv^{2}=\dfrac{EM}{M+m}$

$\downarrow$

K.E bullet

Q.19.

If a net force

$F$ applied to an object of mass

$m$ will produce an acceleration of

$a$ , what is the mass of a second object which accelerates at

$5a$ when acted upon by a net force of

$2F$ ?

100%
$\dfrac{2}{5}m$
0%
$2m$
0%
$\dfrac{5}{2} m$
0%
$5m$

Explanation

According to Newton's second law

$F=ma$ .

For first object

$F=m\times a$ ...........(1)

For second object

and

$2F=m'\times5a$ ..........(2)

Solving these two equations, we get

$\Longrightarrow\ 2m \times a=m' \times 5a$

$\Longrightarrow m_2=\dfrac{2m}{5}$

Option A

Q.20.

A man stands on the surface of the earth. The number of action reaction pairs that can be seen here is/are

0%
1
100%
2
0%
4
0%
5

Explanation

Q.21.

A horizontal force

$F$ produces an acceleration of

$6 \ m/ s^{2}$ on a block resting on a smooth horizontal surface. The same force produces an acceleration of

$3\ m/ s^{2}$ on a second block resting on a smooth horizontal surface. If the two blocks are tied together and the same force acts, the acceleration produced will be:

0%
$9\ m/ s^{2}$
100%
$2\ m/ s^{2}$
0%
$4\ m/ s^{2}$
0%
$\dfrac{1}{2}\ m/ s^{2}$

Explanation

Here,

$F=m_1a_1$ and

$F=m_2a_2$ .

Now if we tie both of them together then

$F=ma$ and

$m=m_1 +m_2$

$\displaystyle \frac{F}{a}=\frac{F}{a_{1}}+\frac{F}{a_{2}}$

If the same force acts, the acceleration produced will be,

$a=\dfrac{(a_1a_2)}{(a_1+a_2)}=2\ m/s^2$ .

Q.22.

A

$10\ N$ force applied on a body produces in it an acceleration of

$2\ m{ s }^{ -2 }$ . The mass of the body is

0%
$2\ kg$
100%
$5\ kg$
0%
$10\ kg$
0%
$20\ kg$

Explanation

Applying Newton’s second law of motion,

$F=ma$

$\Rightarrow a=\dfrac{F}{m}$ and

$m=\dfrac{F}{a}$

Here,

$F=10\ N$ and

$a=2\ m/{s}^{2}$

$\Rightarrow m=\dfrac{10}{2}=5\ kg$

Option

$-B$ is correct.

Q.23.

A trolley of mass

$60 kg$ moves on a smooth horizontal surface and has kinetic energy

$120 J$ .A mass of

$40 kg$ is lowered vertically on to the trolley. The total kinetic energy of the system after lowering the mass is:

0%
$60 J$
0%
$72 J$
100%
$120 J$
0%
$144 J$

Explanation

KE of trolley

$=120=0.5mV^2$

$\therefore 120=0.5\times 60\times V^2$

$V=2\ m/s$

Using momentum conservation:

$60\times 2=100\times v$

$1.2\ m/s=v$

$(K.E)total=\dfrac{1}{2}\times 100\times (1.2)^{2}$

$=72J$

Q.24.

A block weighing

$30 \ N$ is on a smooth horizontal surface and it is accelerated in the horizontal direction at a rate of

${4 \ m/s}^{ 2 }$ by a force. The net force in horizontal direction acting on the block is (g

$={ 10 m/s }^{ 2 }$ ):

100%
12 N
0%
120 N
0%
50 N
0%
10 N

Explanation

Weight of the block

$W=mg$

N is the normal reaction acting on the block due to the horizontal surface (mg is downward and N is upward)

So

$N=mg$ as there is no acceleration in vertical direction.

The net force experienced by the block will be in the horizontal direction and will be equal to the product of mass and acceleration.
Mass:

$m=\dfrac{W}{g}=\dfrac{30 \ N}{10 \ m/s^2} = 3 kg$

Hence, by Newton's second law of motion,

Force

$F=ma = 3 \ kg \times 4 \ m/s^2 = 12 \ N$

Q.25.

A boy weighing 50 kg throws a stone of mass 10 kg horizontally with a velocity of

$8 ms^{-1}$ . With what velocity does he move after throwing?

100%
$-1.6 ms^{-1}$
0%
$-16 ms^{-1}$
0%
$-160 ms^{-1}$
0%
$-1.9 ms^{-1}$

Explanation

Let mass of boy:

$m_1=50kg$

mass of stone:

$m_2=10 kg$

Velocity of boy after throwing stone:

$v_1$

Velocity of stone:

$v_2 = 8m/s$

Using conservation of momentum :

$m_1v_1+m_2v_2=0$ (initial momentum of the system = 0)

$50 \times v_1 + 10 \times 8 = 0$

$v_1 = \dfrac{-80}{50}$

$v_1 = -1.6m/s$

The boy moves with a speed of

$1.6 m/s$ in the opposite direction of the stone.

Q.26.

A gun fires a bullet of mass 50 g with a velocity of mass

$130$ ms

$^{-1}$ . Because of this, the gun is pushed back with a velocity of

$1$ ms

$^{-1}$ . The mass of the gun is :

0%
15 kg
0%
30 kg
100%
6.5 kg
0%
20 kg

Explanation

Mass of the gun =

$m_1=?$

Mass of the bullet

$m_2=50\ g= 0.050\ kg$

Final velocity of the gun

$v_1=-1\ m/s$ (The minus sign indicates that the gun is pushed back)

Final velocity of the bullet

$v_2=130\ m/s$

Let the system consists of the gun and the bullet. Now the initial momentum is zero.
By using momentum conservation, we get,

$m_1v_1 + m_2 v_2=0$
But

$v_1=-1$ ms

$^{-1},v_2=130$ ms

$^{-1}, m_2=\dfrac{50}{1000}$ kg

$\Rightarrow m_1(-1)+\dfrac{50}{1000}\times 130= 0$

$\Rightarrow m_1 = 6.5$ kg

Q.27.

The net force on a rocket with a weight of

$1.5\times 10^{4}N$ is

$2.4 \times 10^{4}N$ . About how much time is needed to increase the rocket's speed from

$12 m/s$ to

$36 m/s$ near the surface of the Earth at take off? (Take

$g=10m/s^2$ )

0%
$15 s$
100%
$0.78 s$
0%
$1.5 s$
0%
$3.8 s$

Explanation

Weight of the rocket

$=1.5\times10^4 N$

Weight = mass

$\times$ gravitational acceleration

$\therefore$ Mass of the rocket

$= \dfrac{Weight}{gravitational \ acceleration}=\dfrac{1.5\times10^4}{10}=1.5\times10^3 \ kg$

According to Newton's second law of motion,

$F_{net}=mass \times acceleration$
Acceleration of the rocket

$=\dfrac{Net \ force}{mass}=\dfrac{2.4\times10^4}{1.5\times10^3}=16\ ms^{-2}$

Given:

$v=36m/s$

$u=12m/s$

$a=16 m/s^2$

Using the equation of motion:

$v=u+at$

$t=\dfrac{36-12}{16}=1.5\ sec$

Q.28.

A bullet of mass 20 gm is fired from a rifle of mass 20 kg with a muzzle velocity of

$200\ ms^{-1}$ . Find the velocity of recoil of the rifle.

0%
$0.4ms^{-1}$
0%
$0.5ms^{-1}$
100%
$0.2ms^{-1}$
0%
$0.3ms^{-1}$

Explanation

By the law of conservation of momentum,

Initial momentum

$=$ find momentum

$\therefore P_{1} = P_{2}$

$\therefore m_{1}v_{1} = m_{2}v_{2}$

here,

$m_{1} = 20\,gm. v_{1} = 200\,m\,s^{-1},m_{2} = 20kg = 20,000\,gm$

$\therefore v_{2} = \dfrac{m_{1}v_{1}}{m_{2}}$

$= \dfrac{20\times 200}{20,000}$

$\therefore \boxed{v_{2} = 0.2}m/s$

$\therefore$ option (c) is correct.

Q.29.

A bullet is fired from a gun with a velocity 600 m/s. The recoil velocity of the gun is 3 m/s. What is the ratio of the mass of the gun and bullet ?

0%
100:1
0%
400:1
100%
200:1
0%
300:2

Explanation

Given,

Initially both Gun and bullet will be in rest ie, initial velocity of Gun and bullet

$=0$

After fired,

$V_b=600\ m/s\\V_r=3\ m/s$

Let mass of gun is

$M$ and mass of bullet is

$m$ .

Using momentum conservation:

Initial momentum (Before fired)

$=$ Final momentum (After fired)

$M\times 0+m\times 0= (-3 \times M) + 600 \times m$

$0=-3M+600m$

$\Rightarrow \dfrac{M}{m} = \dfrac{600}{3}=\dfrac{200}1$

Ration of mass of Gun to bullet is

$\dfrac{200}1$ Option C.

Q.30.

A bullet of mass 125 gm leaves a rifle with a velocity of

$500ms^{-1}$ . The rifle recoils with a velocity of

$5ms^{-1}$ . Find the mass of the rifle.

0%
100 kg
100%
12.5 kg
0%
1.25 kg
0%
125 kg

Explanation

Using linear momentum conservation,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$u_1=u_2=0$ and

$v_1=-5$

$v_2=500$

Let mass of rifle be

$m$

$0 = m \times -5 + 125 \times 500$

$m = 12500 gm$
Mass of rifle

$m=12.5 kg$

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