MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 12

A body

$A$ of mass

$150 \ kg$ , travelling at

$20 \ m/s$ collides with a body

$B$ of mass

$100 \ kg$ travelling at

$10 \ m/s$ in the same direction. After collision the velocity of

$A$ becomes

$15 \ m/s$ . What is the magnitude of velocity of

$B$ after the collision?

0%
$22.5 \ m/s$
0%
$17.5 \ m/s$
0%
$5 \ m/s$
0%
$12.5 \ m/s$

Explanation

Given :

$m_A =150$ kg

$u_A = 20$ m/s

$m_B =100$ kg

$u_B = 10$ m/s

$v_A =15$ m/s

Let the speed of B after the collision be

$v_B$ .

Using conservation of linear momentum before and after the collision :

$P_i = P_f$

$\therefore$

$m_A u_A + m_B u_B = m_Av_A + m_B v_A$

$150 \times 20 + 100\times 10 = 150\times 15 + 100v_B$

$\implies v_B = 17.5$

$m/s$

What is the recoil velocity of the gun of mass

$8$ kg when a bullet of mass

$10$ g is fired from it with a velocity of

$400$ m/s ?

0%
$5$ m/s
0%
$2$ m/s
0%
$50$ m/s
0%
$0.5$ m/s

Explanation

Given :

mass of the gun

$m_g = 8$ kg

mass of the bullet

$m_b = 0.01$ kg

velocity of the bullet

$v_b = 400$ m/s

recoil velocity of the gun

$V_g = ?$

As no external force is acting on the system, the total linear momentum of the system is conserved.

Conservation of linear momentum :

$P_i = P_f$

Initial momentum

$P_i = 0$ as the gun and the bullet do not move.

Final momentum

$P_f = m_g V_g + m_b v_b$

$\therefore$

$0 = m_g V_g + m_b v_b$

$\Rightarrow 0 =8V_g + (0.01)(400)$

$\Rightarrow |V_g| =0.5$ m/s

Equal force F acts on isolated bodies X and Y as shown in figure. The mass of Y is three times that of X. The magnitude of the acceleration of X is ..........

0%
Three times that of Y
0%
1/3 that of Y
0%
Nine times that of Y
0%
1/9 that of Y

Explanation

$F=ma$

$a=\dfrac {F}{m}$
Let mass of X is

$m$ , so mass of Y will be

$3m$

The acceleration of X,

$A_X=\dfrac{F}{m}$ and acceleration of Y,

$A_Y=\dfrac{F}{3m}$ .

hence, The magnitude of the acceleration of X is Three times that of Y

Option A

The 35 kilogram girl is standing on a 20 kilogram wagon and jumps off, giving the wagon a kick that sends it off at 3.8 meters per second. How fast is the girl moving?

0%
1.2 m/s
0%
3 m/s
0%
4 m/s
0%
2.2 m/s

Explanation

Answer is D.

The total momentum is initially zero, and must remain zero. Therefore, the momentum acquired by the wagon in one direction equals the momentum acquired by the girl in the other case.
That is,

${ (mv) }_{ girl }={ (mv) }_{ wagon }$ .

${ 35\quad kg\quad \times \quad (v) }_{ girl }={ \quad 20\quad kg\quad \times \quad }3.8\quad m/s$ .
Therefore,

${ (v) }_{ girl }={ \quad 2.2\quad m/s }$ .
Hence, the girl is moving at a velocity of 2.2 m/s.

A boy sitting on the top most berth in the compartment of a train which is just going to stop on the railway station, drops an apple aiming at the open hand of his brother situated vertically below his hands at a distance of about

$2 \ m$ . The apple will fall

0%
In the hand of his brother
0%
Slightly away from the hand of his brother in the direction of the motion of the train.
0%
Slightly away from the hands of his brother in the direction opposite to the direction of the motion of the train.
0%
None of these

Explanation

Given, that the train is going to stop at a railway station, which means that the train is under retardation or is accelerating opposite to the direction of motion.

The boy sitting on the topmost berth drops an apple aiming at the open hand of his brother. Let at this instant, the velocity of the train be

$v$ . Since, the boy, his brother and the apple are in the same frame, they are also moving with velocity

$v$ . Hence, when the apple is released, its velocity will be

$v$ in the direction of motion of train.

Now, the apple falls downward only under the influence of gravitational force. Since, there are no forces acting in the horizontal direction, it will move with constant velocity

$v$ in the horizontal direction. Let it takes time

$t$ to come down a height of

$2 \ m$ . The horizontal distance it covered is given by ,

$d=vt$

Since, the train is under retardation, which means that the velocity of the train is decreasing. Hence, horizontal distance, the boy's hands will cover in time

$t$ will be less than

$vt$

Hence, the apple will fall slightly away from the hand of his brother in the direction of motion of the train.

A man fires a bullet of mass

$200\ g$ at a speed of

$5\ m/s$ . The gun is of mass

$1\ kg$ . The velocity with which the gun will recoil backward is:

0%
$3\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$1\ ms^{-1}$
0%
$0.01\ ms^{-1}$

Explanation

Given,

Mass of the bullet,

$m_b=200\ g=0.2\ kg$

Initial velocity of bullet,

$v_{b1}=0$

Muzzle velocity of bullet,

$v_{b2}=5\ m/s$

Mass of the gun,

$m_g=1\ kg$

Initial velocity of gun,

$v_{g1}=0$

Final velocity of gun,

$v_{g2}$

According to the law of conservation of momentum,

$Initial\ momentum=Final\ momentum$

$m_bv_{b1}+m_gv_{g1}=m_bv_{b2}+m_gv_{g2}$

$0=m_bv_{b2}+m_gv_{g2}$

$\implies v_{g2}=\dfrac{m_b}{m_g}v_{b2}$

$v_{g2}=\dfrac{0.2}{1}\times5$

$v_{g2}=-1 \ m/s$

The velocity with which the gun will recoil backward is

$1\ m/s$

Four forces are acting on a body. If the body doesn't change its position or shape, the forces :

0%
Must be of equal magnitude
0%
Must be parallel & opposite
0%
Must add up to zero when taken as vectors
0%
Must be in a single line

A boat at rest has two persons of unequal weights seated at either end. If they walk across the boat and interchange their positions,

0%
The boat will move in the direction of walking of the lighter person
0%
The boat will move in the direction of walking of the heavier person
0%
The boat will not move at all
0%
The boat will oscillate about a mean position and stop once the persons have moved to the new positions
0%
None of these

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

0%
While the bus is moving, luggage tends to remain in inertia of motion state
0%
When the bus stops, the luggage tends to resist the change and due to inertia of motion it move forward and may fall off
0%
Both a and b are true
0%
None of them is True

Consider two spring balances hooked as shown in the figure. We pull them in opposite directions. If the reading shown by A is 1.5 N, the reading shown by B will be :

0%
$1.5N$
0%
$2.5N$
0%
$3.0N$
0%
$zero$

Explanation

When we apply force F on spring balance on one end, it gives reaction of same force F at another end and it reads force as F. Now, generally another end is fixed to a support.

So, in practice, there are always two forces present at both ends, which is of same magnitude.

So,

in this case, both forces are

$1.5N$ on both ends reads

$1.5N$

According to Newton's Third Law of Motion, action and reaction are equal and opposite. The resultant force is not zero because:

0%
They act at different times
0%
They act on different bodies
0%
Their lines of action are different
0%
They act on the same body

A driver first accelerates his car at a rate of

$\displaystyle 1.8{ m }/{ { s }^{ 2 } }$ and then at a rate of

$\displaystyle 1.2{ m }/{ { s }^{ 2 } }$ . The ratio of the forces exerted by the engines respectively will be equal to

0%
2 : 3
0%
1 : 2
0%
2 : 1
0%
3 : 2

Explanation

$\underline{\text{First situation:}}$

Acceleration

$a_1 = 1.8\ m/s^2$

$F_1 = ma_1=m\times 1.8$

$\underline{\text{Second situation:}}$

Acceleration

$a_2 = 1.2\ m/s^2$

$F_2 = ma_2=m\times 1.2$

So,

$\dfrac{F_1}{F_2} = \dfrac{a_1}{a_2}= \dfrac{1.8}{1.2}= 3:2$

Every action has reaction. These action-reaction are

0%
Equal, opposite and acting on the same body
0%
Equal, in the same direction and acting on same body
0%
Equal, opposite and acting on different bodies
0%
Equal, in same direction and acting on different bodies.

A bullet of mass 0.01 kg is fired from a gun weighing 5.0 kg. If the initial speed of the bullet is 250 m/s, calculate the speed with which the gun recoils.

0%
-0.50 m/s
0%
- 0.25 m/s
0%
+ 0.05 m/s
0%
+0.25 m/s

Explanation

Let the recoil velocity of gun be

$V$ .

Mass of gun

$M_{gun} = 5.0 \ kg$

Mass of bullet

$M_{bullet} = 0.01 \ kg$

Velocity of bullet

$V_{bullet} = 250 \ m/s$

Initial momentum of the system

$P_i = 0$

Applying the conservation of linear momentum :

$P_f = P_i$

$M_{gun}V + M_{bullet}V_{bullet} = 0$

$5\times V+0.01\times 250 = 0$

$\implies \ V = -0.5 \ m/s$

Physical independence of force is a consequence of _____.

0%
first law of motion
0%
second law of motion
0%
third law of motion
0%
all ofthese laws

The mass of balloon and its contents is M. It is descending with an acceleration a. By how much the mass should be decreased, keeping the volume constant, so that the balloon starts ascending with the same acceleration?

0%
$\displaystyle \frac { a }{ a+g } M$
0%
$\displaystyle \frac { g }{ a+g } M$
0%
$\displaystyle \frac { 2a }{ a+g } M$
0%
$\displaystyle \frac { 2g }{ a+g } M$

Explanation

The forces acting on the balloon when the balloon is coming downwards with an acceleration a are
(i) weight, mg acting downwards and
(ii) the force of upthrust, F acting upwards. Since the accelerated movement is in the downward direction, the force Mg must be greater than F. So for the downwards accelerated movement of the block,

$\displaystyle Mg-F=Ma$ ..........(i)

Let m be the mass removed from the balloon, the weight of the balloon now becomes

$\displaystyle (M-m)g$ . Now for the upward accelerated movement with the same acceleration, we have

$\displaystyle F-(M-m)g=(M-m)a$ .......(ii)

Solving equations, (i) and (ii)

$(Mg−Mg)−Mg+mg=Ma−ma$

$m(g+a)=2Ma$

$m=\dfrac{2Ma}{g+a} .$

What is the momentum of a 100000 kg truck whose velocity is 2 m/s?

0%
$2 \times 10^5 kg m/s$
0%
$1 \times 10^5 kg m/s$
0%
4 kg m/s
0%
none of these

Explanation

Given:

Mass of the truck

$m = 100000\ kg = 10^{5}\ kg$

Velocity of the truck

$v=2\ m/s$

We know that,

$Momentum(P)= Mass(m) \times Velocity(v)$

$\therefore P= 2 \times 10^5\ kg m/s$

A ball A of mass 2.5kg moving at 5 m/s collides with an identical ball B at rest. A stops and B starts moving. What is the magnitude of velocity of B after the collision?

0%
5 m/s
0%
12.5 m/s
0%
2.5 m/s
0%
10 m/s

Explanation

Given :

$m_A=m_B = 2.5$ kg

$u_A = 5$ m/s

$u_B = 0$ m/s

$v_A =0$

Conservation of momentum:

Intial momentum

$=$ Final momentum

$m_Au_A + m_Bu_B = m_Av_A + m_Bv_B$

$\therefore$

$2.5(5) + 0 = 0 + 2.5v_B$

$\implies v_B = 5$ m/s

The rocket engine lifts a rocket from the earth, because hot gases:

0%
push it against the air with very high velocity
0%
push it against the earth with very high velocity
0%
heat up the air which lifts the rocket with very high velocity
0%
react against rocket and push it up with very high velocity

Two bodies of equal masses (M) moving with equal velocities (V) in opposite direction collide. Resultant velocity of the combination:

0%
$V$
0%
$Va$
0%
$-V$
0%
Zero

Explanation

Given :

$m_1=m_2 = M$

$v_1 = V$

$v_2 = -V$

Momentum of

$m_1$ is

$P_1=MV$

Momentum of

$m_2$ is

$P_2=-MV$

Their sum is

$P_R=P_1+P_2=MV-MV=0$ . . . . (i)

Let us consider the resultant velocity of the system as

$V_R$

Resultant mass of the system be

$M_R=m_1+m_2=2M$

The Resultant momentum of the system

$P_R=M_RV_R=2MV_R$ . . . . (ii)

Comparing (i) and (ii),

$2MV_R=0$

$\Rightarrow V_R=0$

Mark the incorrect option:

0%
Balanced forces cannot set any stationary body into motion
0%
Balanced forces cannot change the speed of a moving body
0%
Balanced forces change the shape and size of a soft body
0%
None

If the momentum of a body is doubled, the kinetic energy is

0%
Halved
0%
Unchanged
0%
Doubled
0%
Becomes 4 times

Explanation

The correct answer is option (D).

Hint: Relate the formula of momentum and kinetic energy of any body.

Step 1: Find the relation between momentum and kinetic energy.

By formula,

$p=mv$ (where,

$p$ is the momentum of the body)

$v=\dfrac{p}{m}$

Also,

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

Substituting the value of

$v$ in the above equation, we get,

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

$K.E.=\dfrac{1}{2}m\dfrac{p^2}{m^2}$

$K.E.=\dfrac{p^2}{2m}$

Hence,

$K.E.\propto p^2$

Step 2: Calculate the final kinetic energy.

From the above relation, we can write,

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{p_2^2}$

According to the question,

$p_2=2p_1$

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{(2p_2)^2}$

$\dfrac{K.E._1}{K.E._2}=\dfrac{p_1^2}{4p_1^2}$

$K.E._2=4K.E._1$

Hence, the kinetic energy becomes four times.

The correct answer is option (D).

$80\;\text{kg}$ person is parachuting and is experiencing a downward acceleration of

$2.8\;\text{m/s}^2$ . The mass of the parachute is

$5\;\text{kg}$ . The net force on the system (parachute+man) is (Take

$\text{g=9.8 m/s}^2$ ):

0%
$238\;N$
0%
$675\;N$
0%
$456\;N$
0%
$925\;N$

Explanation

The net downward acceleration is $2.8 m/s^2$

According to Newton second law

$F = ma$

$m$ = $80+5=85$ $kg$

Hence $F = 85 \times 2.8$ N = $238 N$

The force applied on a body depends on the area over which it acts

0%
True
0%
False

Explanation

The force applied to a body does not depend on the area on which it acts.

Pressure due to a force depends on the area because we know that

$\text{Pressure}=\dfrac{\text{Force}}{\text{Area}}$

Mark the correct statement.

0%
Acceleration produced in a body $\propto$ force applied on the body
0%
Acceleration produced in a body $\propto \dfrac {1}{\text {Mass of the body}}$
0%
Both
0%
None

Explanation

From Newton's second law of motion

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

$\implies$

$a \propto F$ and

$a \propto \dfrac{1}{m}$

Hence option C is correct.

A body moves with uniform speed along a straight line. So, it is acted upon by:

0%
Unbalanced force
0%
Balanced force
0%
No net external force
0%
None of these

Explanation

Under the action of a balanced force or in absence of any net external force, a body moves with uniform speed along a straight line.

According to Newton's second law of motion.

$F = ma$

$F$ : Net external forces

$m$ : mass of the object

$a$ : acceleration produced by forces

In this case, the net force is zero so acceleration will be zero.

From a rifle of mass

$40\ kg$ , a bullet of mass

$50\ g$ is fired with an initial velocity of

$35\ m/s$ . Calculate the initial recoil velocity of the rifle:

0%
$0.04375\ m/s$
0%
$0.4375\ m/s$
0%
$0.2235\ m/s$
0%
$0.00223\ m/s$

Explanation

$\textbf{Step 1 - Initial momentum of (rifle + bullet) system}$

mass of rifle

$(m_{1}) = 40\ kg$

mass of bullet

$(m_{2}) = 0.05\ kg$

Initial velocity of rifle

$(u_{1}) = 0\ m/s$

Initial velocity of bullet

$(u_{2}) = 0\ m/s$

Initial momentum,

$P_{1} = m_{1}u_{1} = m_{2}u_{2}$

$P_{1} = 0 + 0 = 0\ m/s$

$....(1)$

$\textbf{Step 2 - Final momentum of (rifle + bullet) system Refer Figure}$

final velocity of bullet

$= V_{2} = 35\ m/s$

final velocity of rifle

$= V_{1} (let)$

final momentum of system

$P_{2} = m_{1}v_{1} + m_{2}v_{2}$

$P_{2} = 40v_{1} + 0.05 \times 35$

$....(2)$

$\textbf{Step 3 - Apply momentum conservation}$

As no external forces are acting on

$(bullet + rifle)$ system.

So momentum of system will be conserved.

$\text {Initial momentum} = \text{final momentum}$

$P_{1} = P_{2}$

$0 = 40v_{1} + 0.05 \times 35$

$\Rightarrow v_{1} = -0.04375$

This negative sign indicates that the rifle recoils

in a direction opposite to bullet i.e. backward

So recoil speed

$= 0.04375$

Hence option (A) correct.

Two tugboats are moving a barge. Tugboat A exerts a force of

$3000$ newtons on the barge. Tugboat B exerts a force of

$5000$ newtons in the same direction. What is the combined force on the barge?

0%
$2000\ N$
0%
$8000\ N$
0%
$3000\ N$
0%
$5000\ N$

Explanation

forces in the same direction combine by addition.

thus,

$3000N+5000N$

$=8000$

1131544_448831_ans_d27c1978d2c748869be16b8dfe2b45b0.PNG

X and Y are two objects with masses

$100\ kg$ and

$75\ kg$ respectively, then :

0%
Both will have the same inertia
0%
Y will have more inertia
0%
X will have more inertia
0%
Can not be compared

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 a stop because

0%
The batsman did not hit the ball hard enough
0%
Velocity is proportional to the force exerted on the ball
0%
There is a force on the ball opposing the motion
0%
There is no unbalanced force on the ball, so the ball would want to come to rest

Explanation

Hint:

According to Newton's laws of motion, force is required to change the velocity of an object. The force is the product of mass and acceleration.

Step 1: Explaining the change in the velocity of the ball.

$\bullet$ According to Newton's first law of motion, if a body is in a rest position, then it will remain in rest unless an external force is applied to it.

$\bullet$ Therefore a force must be applied against its direction of motion to the ball to stop it.

Therefore option (C) is the correct answer.

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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