MCQ Questions for Class 9 Physics Force And Laws Of Motion Quiz 11

A stone is tied to the middle of a string and suspended from one end as shown in the given figure. Here S is the stone and O is the point of suspension. If we increase the pull at P gradually, the string will break

0%
below the stone
0%
at the point P itself
0%
above the stone
0%
nothing can be decided

Two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to 0.3 ms

$^{-1}$ . After collision the two balls come to rest then velocity of B is :

0%
-0.15 ms $^{-1}$
0%
0.15 ms $^{-1}$
0%
1.5 ms $^{-1}$
0%
none of these

Explanation

Here, two solid rubber balls A and B having masses 200 g and 400 g respectively are moving in opposite directions with velocity of A equal to

$0.3 ms^{-1}$ and after collision the two balls come to rest.

Final velocities

$v_A$ and

$v_B$ are zero because both come to rest.

Initial momentum=Final momentum

$m_A u_A + m_B u_B = m_A\times0 +m_B\times0$

$m_A u_A + m_B u_B = 0$

$m_B u_B = -m_A u_A$

$u_B = -\dfrac{m_A u_A}{m_B}$

$u_B = -\dfrac{200 \times 0.3}{400} = -0.15 ms^{-1}$

A book is at rest on a table. What is the "reaction" force according to Newton's third law to the gravitational force by the earth on the book?

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the normal force exerted by the table on the book
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the normal force exerted by the table on the ground
0%
the normal force exerted by the ground on the table
0%
the gravitational force exerted on the earth by the book.

Statement 1: When a man is standing on a stationary horizontal surface, the normal force exerted by the ground on him is equal and opposite to his weight.
Statement 2: According to Newton's

$3^{ rd }$ law , every action has an equal and opposite reaction.

0%
Statement 1 - is true, statement- 2 is true and statement-2 is correct explanation for statement-1
0%
Statement 1 - is true, statement- 2 is true and statement-2 is NOT the correct explanation for statement-1
0%
Statement 1 - is true, statement- 2 is false
0%
Statement 1 - is false, statement- 2 is true

A bullet of mass

$20g$ and moving with a velocity of

$200m/s$ strikes a heap of sand and comes to rest after penetrating

$3cm$ inside it. The force exerted by the sand on the bullet will be:

0%
$11.2\times { 10 }^{ 8 }dyne$
0%
$15.7\times { 10 }^{ 8 }dyne$
0%
$13.3\times { 10 }^{ 8 }dyne$
0%
$18.6\times{ 10 }^{ 8 }dyne$

Explanation

Given:

mass of the bullet (m) = 0.02 kg
initial velocity of the bullet (u) = 200 m/s
final velocity of the bullet (v) = 0 m/s
distance traversed by the bullet in the sand before coming to rest (S) = 0.03 m

From the equation of motion,

$v^2 – u^2 = 2 a S$

$\Rightarrow0 - 200^2 = 2 a S$

$\Rightarrow a = - \dfrac{2}{3}\times 10^6$ (Here a is the deceleration)

Hence the force exerted by the sand on the bullet is

$F = m (-a)$

$\Rightarrow F = 0.02 \times \dfrac{2}{3} \times 10^6$

$\Rightarrow F = 13.3 \times 10^3 N$

Note: $1 N = 10^5 dyne$

$\Rightarrow F = 13.3 \times 10^8 dyne$

Hence option C is correct.

A heavy uniform bar is being carried by two men on their shoulders. The weight of the bar is

$w$ . What is the work done by gravity?

0%
$w/4$
0%
$w/2$
0%
$w$
0%
$0$

$20 g$ bullet passes through a plate of mass

$1 kg$ and finally comes to rest inside another plate of mass

$2980 g$ . It makes the plates move from rest to same velocity. The percentage loss in velocity of bullet between the plate is:

0%
$0$
0%
$50$ %
0%
$75$ %
0%
$25$ %

Explanation

Using conservation of momentum

$0.02 u = 0.02 v + 1 V_1$ (i)

(where

$V_1$ be the velocity of plate of

$1 kg$ )

and

$0.02 v = (2.98 + 0.02) V_1$ (ii)

(

$\because$ plate of

$3 kg$ has also same velocity i.e.

$V_1$ )

$V_1 = 0.02 \dfrac{v}{3}$

Substituting

$\displaystyle \therefore 0.02 u = 0.02v + 0.02 \dfrac{v}{3}$

$\displaystyle u = \dfrac{4v}{3}$ or

$\displaystyle v = \dfrac{3}{4}u$

percentage loss in velocity

$= \displaystyle |\dfrac{u-v}{u}| \times 100$

$= \displaystyle |\dfrac{u - \dfrac{3}{4} u}{u}| \times 100 = 25$ %

A rocket is moving at a constant speed in space by burning its fuel and ejecting out the burnt gases through a nozzle. Is there any force acting on the rocket? If yes, how much?

0%
Yes, equal to rate of change in momentum
0%
No
0%
Yes, equal to rate of change in velocity
0%
Yes, equal to change in mass

A force of

$5N$ acts horizontally on a body of weight

$9.8N$ . What is the acceleration produced (in

${ ms }^{ -2 }$ )?

0%
$0.51$
0%
$1.46$
0%
$49.00$
0%
$5.00$

Explanation

Suppose the mass of the body is

$m$ .

Weight of body

$= mg$

Here,

$mg=9.8$

$\Rightarrow m=1 kg$

So for acceleration,

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

$a=\dfrac{5}{1}$

$\Rightarrow a =5 ms^{-2}$

A body acted upon by a force

$F$ has an acceleration

$A$ . When it is acted upon by two forces each of magnitude

$F$ perpendicular to each other, its acceleration will be:

0%
$2A$
0%
$A$
0%
$A \sqrt 2$
0%
$\sqrt 2$

Explanation

As the acceleration of force

$F$ is

$A$ ,
so the resultant acceleration ,

$A_R=\sqrt{A^2+A^2}=A\sqrt 2$

Consider the following statements:
(a) No net force acts on a rain drop falling vertically with a constant speed.
(b) If net force acting on the body is zero, momentum of the body remains constant.
(c) Particle of different masses falls with different acceleration on earth.
(d) It is easier to start motion in a heavier body than a lighter body.
Which of the above statements are correct?

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(b) and (c)
0%
(a) and (b)
0%
(a), (b) and (d)
0%
All of the above

The speed-times relation of a car whose weight is

$1500kg$ as shown in the given graph. How much breaking force has been applied at the end of

$7$ sec, to stop the car in

$2$ sec?

0%
$2000N$
0%
$1200N$
0%
$4800N$
0%
$8400N$
0%
$9000N$

Explanation

Speed - time graph is given,

Slope of the speed-time graph gives an acceleration of the moving object.

In between 7th and 9th second,

Slope of line

$=\dfrac{0-12}{9-7}=-6m/s^{2}$

Slope of line is negative which show deacceleration of object. Hence When driver will apply break, car will deaccelerate with

$-6m/s^{2}$ .

Breaking force,

$F=ma=1500\times 6=9000\ N$

Two bodies of masses

$m$ and

$3m$ , moving with velocities

$3v$ and

$v$ respectively along same direction, collide with each other. After collision they stick together and move with a velocity

$V$ in the same direction. Then:

0%
$\displaystyle V= v$
0%
$\displaystyle V=\frac {3}{2}v$
0%
$\displaystyle V=2v$
0%
$\displaystyle V = \frac {4}{3}v$

Explanation

Here, the collision is inelastic. And law of conservation of momentum for inelastic collision is given by,

$m_{1}u_{1} + m_{2}u_{2} = \left ( m_{1} + m_{2} \right ) V$

$\Rightarrow V = \dfrac{m_{1}u_{1} + m_{2}u_{2}}{m_{1} + m_{2}}$

$\Rightarrow V = \dfrac{m \times 3v + 3m \times v}{m + 3m}$

$\Rightarrow V = \dfrac{3mv + 3mv}{4m}$

$\Rightarrow V = \dfrac{6mv}{4m} = \dfrac{3}{2}v$

When a net force acts on an object, the object will be accelerated in the direction of the force with an acceleration proportional to

0%
the force on the object
0%
the velocity of the object
0%
the mass of the object
0%
the inertia of the object

Explanation

According to Newton's second laws of motion the external force acting on a body is directly proportional to the rate of change of linear momentum.
i.e

$\displaystyle F \propto \frac{dp}{dt}$ or

$F = ma$

$\therefore a \propto F$

How can Newton's first law of motion be obtained from the second law of motion?

0%
When net force is zero.
0%
When net force is positive.
0%
When moving with constant velocity.
0%
When velocity is zero.

A large truck and a car, both moving with a velocity of magnitude

$v$ , have a head-on collision. If the collision lasts for

$1\ s$ . Which vehicle experiences greater acceleration?

0%
Car
0%
Truck
0%
Both have equal acceleration
0%
Cannot say

Explanation

Let the mass of truck

$=M$ , the mass of car

$=m$

The velocity of truck

$=v$ ;

The velocity of car

$=-v$ , (Negative sign for the opposite direction of motion).

The time for which collision lasts,

$t=1 \ s$

Since acceleration

$a =\dfrac{force}{mass}$ ,

Since, force of impact is same,

Hence, acceleration

$a\propto \dfrac {1}{mass}$ .

As the mass of the car is smaller, therefore the acceleration produced in the car is greater than the acceleration of the truck.

A large truck and a car, both moving with a velocity of magnitude

$v$ , have a head-on collision. If the collision lasts for

$1\ s$ which vehicle experiences greater force of impact?

0%
Truck
0%
Car
0%
Both experience equal and opposite force
0%
Data insufficient

Explanation

Let, mass of truck

$=M$
Mass of car

$=m$
Velocity of truck

$=v$
Time for which collision lasts, t

$=1s$
Velocity of car

$= -v$
(Negative sign for opposite direction of motion).

According to third law of motion,on collision, both the vehicles experience the same force, as action and reaction are equal.

Suppose you push a spring with a force

$F_1$ as shown in figure, the spring also pushes up on your hand with a force

$F_2$ . What is the relationship between

$F_1$ and

$F_2$ ?

0%
$F_1 > F_2$
0%
$F_1 = 2F_2$
0%
$F_1 < F_2$
0%
$F_1 = F_2$

Explanation

According to third law of motion: For every action, there is an equal and opposite reaction. So, when you push a spring with a force

${F}_{1}$ , the spring also pushes up on your hand with a force

${F}_{2}$ of having same magnitude but in opposite direction hence,

${F}_{1}={F}_{2}$

A bullet of mass

$10\ g$ moving with the velocity of

$1.5\ ms^{-1}$ hits a thick wooden plank of mass

$90\ g$ . The plank is initially at rest, but when it gets hit by the bullet, the bullet remains in the plank, and both move with a certain speed. Calculate the speed with which the plank containing the bullet moves?

0%
$0.15\ ms^{-1}$
0%
$0.5\ ms^{-1}$
0%
$1.5\ ms^{-1}$
0%
$2\ ms^{-1}$

Explanation

Mass of the bullet,

$m = 10\ gm = 0.01\ kg$

Velocity of bullet ,

$v = 1.5\ ms^{-1}$

Mass of plank,

$m_{p} = 90\ gm = 0.09\ kg$

Initial velocity of plank

$= 0 m/s$

Momentum of the bullet,

$p = mv = 0.01 \times 1.5 = 0.015\ kg m/s$

Total mass of the plank

$= m + m_{p} = 0.01 +0.09 = 0.1\ kg$

The plank and the bullet both move with the same speed

The momentum of the plank and the embedded bullet is

$0.015\ kg m/s$

(

$\because$ the momentum is conserved in a collision.)

But the momentum of the plank and the embedded bullet = product of their mass and velocity.

$\therefore 0.015 = 0.1 \times v$

$\Rightarrow \therefore v = \dfrac{0.015}{0.1} = 0.15 m/s$

The speed of the plank containing the bullet is

$0.15 m/s$

Two small glass spheres of masses 10 g and 20 g are moving in a straight line in the same direction with velocities of

$3 ms^{-1}$ and

$2 ms^{-1}$ respectively. They collide with each other. After collision, glass sphere of mass 10 g moves with a velocity of

$2.5 ms^{-1}$ . Find the velocity of the second ball after collision.

0%
$2.25 ms^{-1}$
0%
$5.5 ms^{-1}$
0%
$2.75 ms^{-1}$
0%
$7.5 ms^{-1}$

Explanation

Here,

$m_1=10g=\dfrac {10}{1000}=0.01 kg$

$m_2=20 g=0.02 kg$

$u_1=3 ms^{-1}; u_2=2 ms^{-1}$

$v_1=2.5 ms^{-1}; v_2=?$

Total momentum of both the spheres before collision

$=m_1u_1+m_2u_2$

$=0.01\times 3+0.02\times 2=0.07 kg ms^{-1}$

Total momentum of both the spheres after collision

$=m_1v_1+m_2v_2$

$=0.01\times 2.5+0.02v_2=0.025+0.02v_2$

Now, according to the law of conservation of momentum,

Total momentum after collision

$=$ Total momentum before collision

$\therefore 0.025+0.02v_2=0.07$

$0.02v_2=0.07-0.025=0.045$

$v_2=\dfrac {0.045}{0.02}=2.25 ms^{-1}$

The earth and the moon are attracted to each other by gravitational force. Is the force with which the Earth attracts the Moon greater, smaller or the same as the force with which the Moon attracts the Earth?

0%
Same and opposite
0%
Greater
0%
Smaller
0%
Data insufficient

Explanation

Gravitation force between two bodies is equal in magnitude but opposite in direction

$F= \dfrac{Gm_e m_m}{r^2}$

$F_{em}=-F_{me}$

$F_{em}\rightarrow Force \ on \ earth \ due \ to \ moon$

$F_{me}\rightarrow Force \ on \ moon \ due \ to \ earth$
Hence, Earth and Moon will be attracted to each other with a force equal in magnitude but opposite in direction.

Student A and student B sit in identical office chairs facing each other, as shown in figure. Student A is heavier than student B. Student A suddenly pushes with his feet. Which of the following occurs?

0%
Neither student applies a force on each other
0%
A exerts a force that is applied to B, but A experiences no force
0%
Each student applies a force to the other, but A exerts the larger force
0%
The students exert the same amount of force on each other

Explanation

Newtons third law states that all forces exist in pairs: if one object A exerts a force

${F}_{A}$ on a second object B, then B simultaneously exerts a force ${F}_{B}$ on A, and the two forces are equal and opposite:

${F}_{A}$ $=$ - ${F}_{B}$
So,the students exert same amount of force on each other in opposite direction.

Student A and student B sit in identical office chairs facing each other, as shown in figure. Student A is heavier than student B. Student A suddenly pushes student B with his feet. Suppose

$v_A$ and

$v_B$ are the velocities of student A and student B respectively, then

0%
$v_A > v_B, v_A$ is leftward and $v_B$ is rightward
0%
$v_A < v_B, v_A$ is leftward and $v_B$ is rightward
0%
$v_A = v_B, v_A$ is leftward and $v_B$ is rightward
0%
$v_A > v_B, v_A$ and $v_B$ both are rightward

Two solid rubber balls,

$A$ and

$B$ having masses

$200 \ g$ and

$400 \ g$ respectively are moving in opposite directions with velocity of

$A$ equal to

$0.3\:m/s$ . After collision the two balls come to rest, then the velocity of

$B$ is-

0%
$0.15\:m/s$
0%
$1.5\:m/s$
0%
$-0.15\:m/s$
0%
None of the above

Explanation

Given, mass of ball A

$m_A = 200 \ g = 0.2 \ kg$

mass of ball B

$m_B = 400 \ g = 0.4 \ kg$

initial velocity of ball A

$u_A = 0.3 \ m/s$

initial velocity of ball B

$u_B = ?$

final velocity of ball A

$v_A = 0 \ m/s$

final velocity of ball B

$v_B = 0\ m/s$

By law of conservation of momentum

$m_Au_A+m_Bu_B = m_Av_A+m_Bv_B$

$0.2\times 0.3+0.4\:u_B=0$ or

$u_B=-0.15\:m/s$

A space-craft of mass

$M$ moves with a velocity

$V$ and suddenly explodes into two pieces. If a part of mass

$m$ becomes stationary, then the velocity of another part will be :

0%
$\dfrac{MV}{(M - m)}$
0%
$\dfrac{MV}{(M + m)}$
0%
$\dfrac{mV}{M - m)}$
0%
$\dfrac{mV}{M + m)}$

Explanation

Given:

Mass of the spacecraft =

$M$

Velocity of the spacecraft =

$V$

Mass of one part =

$m$ , velocity of this part =

$0$

Mass of the other part =

$M-m$ , velocity of this part =

$v=?$

No external force is acting on the spacecraft in the explosion. Therefore, the total linear momentum must be conserved.

$\underline{\text{Applying conservation of momentum:}}$

$\text{Total initial linear momentum of the spacecraft}=\text{Total final momentum of the two parts}$

$\Rightarrow MV=m\times0+(M-m)v$

$\Rightarrow v=\dfrac{MV}{M-m}$

Hence, option A is correct.

If a constant external force starts acting on a moving particle, then:

0%
The line of motion of the particle will keep changing
0%
The speed of the particle will keep changing
0%
The particle will keep moving
0%
All of the above are correct.

A bullet of mass m = 50 gm strikes a bag of mass M = 5 kg hanging from a fixed point, with a horizontal velocity

$\bar{V}_p$ . If bullet sticks to the sand bag then just after collision the ratio of final & initial kinetic energy of the bullet is approximately :

0%
$10^{-2}$
0%
$10^{-3}$
0%
$10^{-6}$
0%
$10^{-4}$

Explanation

By momentum conservation we have

$mV_p=(m+M)V$ ,m=0.05 kg

$0.05\times V_p=(5+0.05)V\Rightarrow V_p=101V$

initial KE of bullet=

$\dfrac{mV_p^2}{2}$

Final KE=

$\dfrac{mV^2}{2}$

So ratio =

$\dfrac{mV^2}{mV_p^2}=10^{-4}$

In which of the following cases the net force is not equal to zero?

0%
A kite skillfully held stationary in the sky
0%
A ball falling freely from a height
0%
A helicopter hovering above the ground
0%
A cork floating on the surface of water

Explanation

According to newton's 1st law of motion, A body maintains its states ie Rest or motion with uniform velocity until and unless there is no external force apply on that.

So, A kite skillfully held stationary in the sky, A helicopter hovering above the ground, A cork floating on the surface of water, All these examples are maintaining state of rest so resultant or net force on it will be zero.

A freely falling ball have acceleration that will equal to

$g$ So, Net force on it will not be equal to zero.

A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if the shoots 40 bullets a second at the speed of 500 m/s. If the mass of a bullet is 49 gm, what is the mass of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8

$m/s^2$ .

0%
60 kg
0%
75 kg
0%
100 kg
0%
125 kg

Explanation

We know

$F=\dfrac{\Delta P}{\Delta t}$

$\Rightarrow \Delta P=F. \Delta t$ . . . (i)

where

$\Delta P$ is change in momentum and

$\Delta t$ is the time interval

Let the mass of man with gun be

$M$ .

The force acting on the man alongwith the gun is

$F=Mg$ downwards

Now this force is balanced by the change in momentum in a given time say

$\Delta t= 1 \ sec$

The mass of one bullet

$m=49 \ gm=0.049 \ kg$

Velocity of each bullet

$v= 500 \ m/s$

The momentum change due to one bullet

$\Delta p=p_f-p_i=mv-0=mv$

$n$ bullets are fired per sec,

then the total momentum change

$\Delta P= P_f-P_i= (n\times mv) -(n \times 0)=nmv$

Given

$n=40$

We consider a time interval of

$\Delta t=1 \ sec$

So using (i),

$\Delta P=F. \Delta t$

$nmv=Mg \times (1)$

$M \times 9.8 = 40 \times 0.049 \times 500$

$M= \dfrac{40\times49\times 500}{9.8\times 1000}$

$M = 100 \ kg$
So the mass of the man along with the gun is 100 kg.

A massive ball moving with speed v collides head-on with a tiny ball at rest having a very small mass as compared to the first ball. If the collision is elastic, then immediately after the impact, the second ball will move with a speed approximately equal to

0%
v
0%
2v
0%
v/2
0%
$\infty$

Explanation

$m_1>>m_2$ ; velocity df massive ball

$=$ v
In a head-on elastic collision where the projectile is much more massive than the target, the velocity of the target particle after the collision will be about twice that of the projectile and the projectile velocity will be essentially unchanged. So, the second ball will move with a speed approximately equal to 2v.

CBSE Questions for Class 9 Physics Force And Laws Of Motion Quiz 11 - 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 11 - MCQExams.com

0:0:1

Practice Class 9 Physics Quiz Questions and Answers

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