MCQ Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 4

A stone is dropped from a height

$h$ . It hits the ground with a certain momentum

$P$ . If the same stone is dropped from a height

$100$ % more than the previous height, the momentum when it hits the ground will change by (approximately) :

0%
$41$ %
0%
$53$ %
0%
$60$ %
0%
$68$ %

Explanation

Velocity,

$v=\sqrt {2gh}$

Momentum,

$P=mv=m\sqrt {2gh}$

$P\propto \sqrt {h}$

$\cfrac{{P}_{2}}{{P}_{1}}=\sqrt {\cfrac{{h}_{2}}{{h}_{1}}}=\sqrt {\cfrac{2h}{h}}=\sqrt 2$

${P}_{2}=1.414{P}_{1}$

% change

$=\cfrac{{P}_{2}-{P}_{1}}{{P}_{1}}\times 100$ %

$\dfrac{1.414P_1-P_1}{P_1}\times 100$

% change

$=41.4$ %

The passengers in a boat are not allowed to stand because:

0%
This will raise the centre of gravity and the boat will be rocked
0%
This will lower centre of gravity and the boat will rocked
0%
The effective weight of system increases
0%
Of surface tension effects

Two particles of mass 1 kg and 0.5 kg are moving in the same direction with speed of 2 m/s and 6 m/s respectively on a smooth horizontal surface. The speed of centre of mass of the system is

0%
$\displaystyle\frac{10}{3}$ m/s
0%
$\displaystyle\frac{10}{7}$ m/s
0%
$\displaystyle\frac{11}{2}$ m/s
0%
$\displaystyle\frac{12}{3}$ m/s

Explanation

As we know that center of mass =

$\dfrac{{m}_{1}\times{v}_{1}+{m}_{2}\times{v}_{2}}{{m}_{1}+{m}_{2}}$

so C.O.M=

$\dfrac{{1}\times{2}+{0.5}\times{6}}{{1}+{0.5}}$ =

$\dfrac{10}{3}$ m/sec

A uniform disc of mass M and radius R is mounted on an axle supported in friction less bearings. A light cord is wrapped around the rim of the disc and a steady downward pull T is exerted on the cord. The angular acceleration of the disc is:

0%
$\displaystyle \frac{MR}{2 T}$
0%
$\displaystyle \frac{2T}{MR}$
0%
$\displaystyle \frac{T}{MR}$
0%
$\displaystyle \frac{MR}{T}$

Explanation

$\because \vec{\tau} = \vec R \times \vec F$

$\Rightarrow \tau = RF sin \theta$
Here,

$F= T$ (downward pull) and

$\theta = 90^o$

$\therefore \tau = RT$
Also

$\tau = 1 \alpha$

$\therefore I \alpha = TR$
For a disc.

$I = \displaystyle \frac{MR^2}{2}$

$\therefore \displaystyle TR = \frac{MR^2}{2}. \alpha$

$\Rightarrow \displaystyle \alpha = \frac{2T}{MR}$

A spinning top exhibits rectilinear motion. True or false.

0%
True
0%
False

The momentum of a body of mass

$0.5$ kg dropped from a certain height (h), when reaches the ground is

$10$ N s. The value h is ___ m. ( g

$\displaystyle {= 10\ ms }^{ -2 }$ )

0%
$20$
0%
$40$
0%
$10$
0%
$80$

Explanation

Initial velocity of body

$u = 0$

Velocity with which it reaches the ground

$V = \sqrt{2gh}$

Momentum of the body

$P = 10 \ N \ s$

Mass

$m = 0.5 \ kg$

Using

$P = mV$

$10 = 0.5\times \sqrt{2gh}$

$\implies \ h = 20 \ m$

A body of mass 100 g is moving with a velocity of 15 m/s. The momentum associated with the ball will be :

0%
0.5 kg m/s
0%
1.5 kg m/s
0%
2.5 kg m/s
0%
3.2 Ns

Explanation

momentum is multiple of mass and velocity so

mass of object in kg is

$0.1 kg$

momentum =

$1.5$ kg m/s

so the answer is B.

A solid cylinder of radius r rolls down on an inclined plane without slipping. The speed of the centre of mass when it reaches the bottom, is:

0%
$\sqrt{2 gh}$
0%
$\sqrt{4 g}$
0%
$\sqrt{3 gh/4}$
0%
$\sqrt{4gh/3}$

If a body of mass M collides against a wall with velocity V and rebounds with the same speed, then its change of momentum will be

0%
Zero
0%
MV
0%
2MV
0%
-2MV

Explanation

Initial velocity of body before collision

$u = -V$

where minus sign indicates the body moves towards the wall.

Initial momentum of the body

$P_i = Mu = -MV$

Final velocity of body after collision

$v =V$

where positive sign indicates the body moves away from the wall.

Final momentum of the body

$P_f = Mv = MV$

Change in momentum

$\Delta P = P_f - P_i = MV - (-MV) = 2MV$

Moment of inertia of thin uniform rod of length

$\displaystyle l$ is equal to:

0%
$\displaystyle \frac { { ml }^{ 2 } }{ 6 }$
0%
$\displaystyle \frac { { ml }^{ 2 } }{ 12 }$
0%
$\displaystyle \frac { { ml }^{ 2 } }{ 4 }$
0%
$\displaystyle \frac { { ml }^{ 2 } }{ 24 }$

Explanation

Let a small element on the rod at a distance

$x$ from

$CM$ .

Mass of the small element of rod

$dm = \dfrac{m}{l}dx$

Moment of this element about

$CM$ ,

$dI = dm x^2$

$\therefore$ Total moment of inertia about

$CM$

$I = \int^{l/2}_{-l/2} x^2dm = \int^{l/2}_{-l/2} \dfrac{m}{l}x^2 dx$

$\therefore$

$I = \dfrac{m}{l} \times \dfrac{x^3}{3}\bigg|_{-l/2}^{l/2} = \dfrac{m}{l} \times \dfrac{2(l/2)^3}{3}$

$\implies$ Moment of inertia about

$CM$ ,

$I = \dfrac{ml^2}{12}$

A lead ball and rubber ball having same mass, strike normally on a smooth vertical wall with the same velocity. The lead ball falls down after striking but the rubber ball bounces back. Then

0%
momentum oflead ball is greater than that of rubber ball
0%
momentum of rubber ball is greater than that of lead ball
0%
the rubber ball suffers a greater change in momentum as compared to lead ball
0%
both the balls suffer an equal change in momentum

Explanation

Initial moment of both ball is same = mv

V = velocity and m in mass of both ball.

final momentum of lead =

$m\times 0=0$

final momentum of rubber ball =

$m\left( -v \right) =mv$

change in momentum of lead ball = mv - 0 = mv

change in momentum of rubber ball = mv - (-mv) = 2mv

So, option C - correct

A body of mass 5 kg undergoes a change in speed from 30 to 40 m/s. Its momentum would increase by :

0%
50 kg m/s
0%
75 kg m/s
0%
150 kg m/s
0%
350 kg m/s

Explanation

speed is increased by

$10$ m/s so the momentum will change by

$50$ kg m/s

so the answer is A.

A body of mass

${ m }_{ 1 }=4 \ kg$ moves at

$5i \ m/s$ and another body of mass

${ m }_{ 2 }=2 \ kg$ moves at

$10 i\ m/ s$ . The kinetic energy of centre of mass is

0%
$\dfrac{200 }{ 3} J$
0%
$\dfrac{500 }{ 3} J$
0%
$\dfrac{400 }{ 3} J$
0%
$\dfrac{800 }{ 3} J$

Explanation

$v_c=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}$

$v_c=\dfrac{40}{6}ms^{-1}$

Kinetic energy

$=\dfrac{1}{2}mv^2=\dfrac{1}{2}\times6\times(\dfrac{40}{6})^2$

$K_c=\dfrac{400}{3}J$

The radius of gyration of a body about an axis at a distance 6 cm from its centre of mass is 10 cm. Then, its radius of gyration about a parallel axis through its centre of mass will be :

0%
80 cm
0%
8 cm
0%
0.8 cm
0%
80 m

Explanation

Let radius of gyration for the axis not passing through center of mass be

$r$ and that for the axis passing through the center of mass be

$k$ and the distance between the two parallel axes be

$a$ .

Parallel axes theorem gives:

$m{ r }^{ 2 }=m({ k }^{ 2 }+{ a }^{ 2 })\Rightarrow { r }^{ 2 }={ k }^{ 2 }+{ a }^{ 2 }$ .

$\Rightarrow{k}=\sqrt{10^{2}-6^{2}}=8cm$ .

Thus, option B is the correct answer.

The moment of inertia of a solid sphere about an axis passing through centre of gravity is

$\dfrac {2}{5}MR^{2}$ , then its radius of gyration about a parallel axis at a distance

$2R$ from first axis is

0%
$5 R$
0%
$225\sqrt {R}$
0%
$\dfrac {5}{2}R$
0%
$\sqrt {\dfrac {12}{5}}R$

Explanation

The radius of gyration K of a body about a given axis of rotation is that radial distance from the axis, the square of which on being multiplied by the total mass of the body gives the moment of inertia of the body about that axis.
Thus,

$I=MK2=\Sigma mR2$
$$where M is the total mass of the body.
This means that

$K=\sqrt {\dfrac{I}{M} }$

According to theorem of parallel axis

$I=I_CG+M(2R)^2$
I=ICG+M(2R)2where,

$ICG$ ICG is moment of inertia about an axis through centre of gravity.

$\therefore I =25MR^2+4MR^2=225MR^2$ or

$MK^2=225MR^2$

$\therefore K=225\sqrt{R}$
MK2=225MR2^2∴K=225RThe total mass of a body may be supposed to be concentrated at a radial distance K from the axis of rotation, so far as the moment of inertia of the body about that axis is concerned.

1152631_460719_ans_4d915c430f864e24aa379b62cc376717.png

The ratio of angular speeds of minute hand and hour hand of a watch is.

0%
$1:12$
0%
$6:1$
0%
$12:1$
0%
$1:6$

Explanation

Hint: The rate of change of angular displacement is known as angular speed.
Explanation:
Step 1: Formula used:
We know that angular speed is given by:

$\omega = \dfrac{angular \ displacement}{time \ taken}$

Step 2: Find angular velocities in both cases:

Time taken by minute hand to rotate by an angle of

$2\pi \ radian$ ,

$t_m = 60\ minute$

$\therefore$ Angular speed of minute hand

$\omega_{min}=\displaystyle\frac{2\pi}{60}\frac{rad}{min}$

Time taken by hour hand to rotate by an angle of

$2\pi \ radian$ ,

$t_{hr} = 12 \ hr$

$= 12 \times 60 \ min$

$\therefore$ Angular speed of hour hand

$w_{hr} = \dfrac{2\pi}{12 \times 60}$

$rad/min$

Step 3: Conclusion:

$\omega_{min}=\displaystyle\frac{2\pi}{60}\frac{rad}{min}$
and

$\omega_{hr}=\displaystyle\frac{2\pi}{12\times 60}\frac{rad}{min}$

$\therefore \displaystyle\frac{\omega_{min}}{\omega_{hr}}=\frac{2\pi /60}{2\pi /12\times 60}$

$=\displaystyle\frac{12}{1}$

Option C is correct.

The angular speed of a body changes from

$\omega_{1}$ to

$\omega_{2}$ without applying a torque but due to changes in moment of inertia. The ratio of radii of gyration in two cases is

0%
$\sqrt {\omega_{2}} : \sqrt {\omega_{1}}$
0%
1
0%
$\sqrt {\omega_{2}^{2}} : \sqrt {\omega_{1}^{2}}$
0%
$\sqrt {\omega_{2}^{3}} : \sqrt {\omega_{1}^{3}}$

Explanation

$I_{1} \omega_{1} = I_{2} \omega_{2}$

$MK_{1}^{2} \omega_{1} = MK_{2}^{2} \omega_{2}$

$\left (\dfrac {K_{1}}{K_{2}}\right )^{2} = \dfrac {\omega_{2}}{\omega_{1}} \Rightarrow \dfrac {K_{1}}{K_{2}} = \sqrt {\omega_{2}} : \sqrt {\omega_{1}}$

The moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through one end is

$I$ . The same rod is bent into a ring and its moment of inertia about the diameter is

${I}_{1}$ . The ratio

$\dfrac{I}{{I}_{1}}$ is

0%
$\dfrac{4\pi}{3}$
0%
$\dfrac{8{\pi}^{2}}{3}$
0%
$\dfrac{5\pi}{3}$
0%
$\dfrac{8{\pi}^{2}}{5}$

Explanation

Moment of inertial of a thin uniform rod about perpendicular axis through it one end

$I = \dfrac{1}{3} M{l}^{2}$ ....(i)
Same rod is bent into a ring

$\therefore l = 2\pi r \Rightarrow \dfrac{l}{r} = 2\pi$ ....(ii)
Moment of inertia of ring about diameter

${I}_{1} = \dfrac{M{R}^{2}}{2}$ .....(iii)
Dividing equation (i) by (ii), we get

$\dfrac{I}{{I}_{1}} = \dfrac{2}{3} \dfrac{M{l}^{2}}{M{R}^{2}} = \dfrac{2}{3} \dfrac{{l}^{2}}{{R}^{2}}$

$\dfrac{I}{{I}_{1}} = \dfrac{2}{3} {\left(2\pi\right)}^{2} = \dfrac{8{\pi}^{2}}{3}$ [using equation (ii)]

What torque will increase angular velocity of a solid disc of mass

$16kg$ and diameter

$1m$ from zero to

$2$ rpm in

$8s$ ?

0%
$\cfrac { \pi }{ 4 } N-m$
0%
$\cfrac { \pi }{ 2 } N-m$
0%
$\cfrac { \pi }{ 3 } N-m$
0%
$\pi N-m$

Explanation

Torque

$\left( \tau \right) =I\alpha$

$\tau =\cfrac { 1 }{ 2 } \times M{ R }^{ 2 }\times \cfrac { 2\pi \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ t }$

$\therefore$

$\tau =16\times { \left( \cfrac { 1 }{ 2 } \right) }^{ 2 }\times \pi \cfrac { \left( 2-0 \right) }{ 8 } =\pi N-m$

A couple produces :

0%
no motion
0%
linear and rotational motion
0%
purely rotational motion
0%
purely linear motion

A particle is projected at an angle of

${30}^{0}$ with the horizontal with a momentum

$P$ . At the highest point its momentum is

0%
$\dfrac { \sqrt { 3 } }{ 2} P$
0%
$\dfrac { 2 }{ \sqrt { 3 } } P$
0%
$P$
0%
$\dfrac { 1 }{ 2 } P$

Explanation

At the point of projection,

$P=mu$

At the highest point, vertical component of velocity is 0.

Therefore,

$u_x=ucos30^0$

$P'=m\left( u\cos { \theta } \right)$

$=mu\dfrac { \sqrt { 3 } }{ 2 } \\$

$=\dfrac { P\sqrt { 3 } }{ 2 }$

Which one of the following is a scalar quantity?

0%
Displacement
0%
Velocity
0%
Acceleration
0%
Linear momentum
0%
Kinetic energy

Explanation

All quantities except kinetic energy have both direction and magnitude. So they are vectors but kinetic energy is a scalar because it is equal to

$\frac{1}{2}mv^2$ where

$v^2$ is a result of dot product of

$\vec v.\vec v$ and its a scalar.

A body of mass

$1\ kg$ is thrown with a velocity of

$\displaystyle 10{ ms }^{ -1 }$ at an angle of

$\displaystyle { 60 }^{ \circ }$ with the horizontal. Its momentum at the highest point is:

0%
$\displaystyle 2\ kg\ { ms }^{ -1 }$
0%
$\displaystyle 3\ kg\ { ms }^{- 1 }$
0%
$\displaystyle 4\ kg\ { ms }^{ -1 }$
0%
$\displaystyle 5\ kg\ { ms }^{ -1 }$

Explanation

At the highest point, its velocity in the upward direction is zero but its velocity in the horizontal direction remains constant at every instant. Hence , only the horizontal component of velocity is responsible for the momentum of the body at the highest point.

Velocity of the body

$v = 10$

$ms^{-1}$

Velocity component of the body in the horizontal direction

$v_H = v cos 60^o = 10 \times 0.5 = 5$

$ms^{-1}$

$\therefore$ Momentum

$P = mv_H = 1 \times 5 =5$

$kg$

$ms^{-1}$

$2.5 kg$ ball is dropped onto a concrete floor. It strikes the floor with a momentum of

$20 kgm/s$ and bounces away from the floor with a momentum of

$16 kgm/s$ .What is the magnitude of change of momentum of the ball?

0%
$4 kg.m/s$
0%
$8 kgm/s$
0%
$32 kgm/s$
0%
$36 kgm/s$
0%
$40 kgm/s$

Explanation

Let the upward direction to be positive.

Initial momentum

$P_i = -20 \ kg \ m/s$

Final momentum

$P_f = 16 \ kg \ m/s$

Change in momentum

$\Delta P = P_f - P_i = 16-(-20) = 36 \ kg \ m/s$

A student performed an experiment and records the mass and speed of an object traveling at the constant velocity on a frictionless track. Find out which of above trial had the greatest momentum and the trial in which it had the greatest kinetic energy.

Greatest Momentum Greatest Kinetic Energy

0%
Trial 1 Trial 3
0%
Trial 2 Trial 2
0%
Trial 3 Trial 2
0%
Trial 3 Trial 3
0%
Trial 4 Trial 4

Explanation

Momentum ,

$p=mv$ and kinetic energy,

$K=\frac{1}{2}mv^2$

For Trial 1:

$p=0.5\times 4=2 kg.m/s$ and

$K=\frac{1}{2}\times 0.5\times 4^2=4 J$

For Trial 2:

$p=1\times 3=3 kg.m/s$ and

$K=\frac{1}{2}\times 1\times 3^2=4.5 J$

For Trial 3:

$p=2\times 2=4 kg.m/s$ and

$K=\frac{1}{2}\times 2\times 2^2=4 J$

For Trial 4:

$p=3\times 1=3 kg.m/s$ and

$K=\frac{1}{2}\times 3\times 1^2=1.5 J$

Thus, Trial 3 has greatest momentum and Trial 2 has greatest kinetic energy.

An object is moving with velocity V. The velocity of a moving object is gets doubled. Identify which of the following statements about the moving object is correct?

0%
The kinetic energy of the object increases by four
0%
The displacement of the object increases by four
0%
The momentum of the object increases by four
0%
The frictional force increases by four
0%
None of these quantities increase by four

Explanation

Kinetic energy,

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

If velocity v is doubled so kinetic energy will increase 4 times.

Since displacement and momentum is proportional to velocity so they will increase 2 times but frictional force does not related to velocity.

Two point objects of masses 1.5 g and 2.5 g respectively are at a distance of 16 cm apart, the centre of gravity is at a distance x from the object of mass 1.5 g where x is:

0%
10 cm
0%
6 cm
0%
13 cm
0%
3 cm

Explanation

Taking the moment of forces about centre of Gravity G is

$\displaystyle \left( 1.5 \right) gx=2.5g\left( 16-x \right)$

$\displaystyle \Rightarrow \quad 3x=80-5x$ or

$\displaystyle 8x=80$ or

$\displaystyle x=10cm$
1134121_470386_ans_25ca0dd4994c4b928f01bf870220e2d6.PNG

Find out the magnitude of the change in momentum of the object between

$t=0$ and

$t=0.4$ sec from the given graph between Total force acting on an object vs Time ?

0%
$2 kg-{m}/{sec}$
0%
$5 kg-{m}/{sec}$
0%
$10 kg-{m}/{sec}$
0%
$12 kg-{m}/{sec}$
0%
$15 kg-{m}/{sec}$

Explanation

We know that

$F=\dfrac{dp}{dt}$

$\implies \int dp=\int Fdt$

$\implies \Delta p$ =Change in momentum=

$\int Fdt$ = Area under the given curve for given interval of time.

Thus

$\Delta p=\dfrac{1}{2}\times 0.2\times 100+(0.4-0.2)\times 25=15 kg-m/s$

The velocity vs Time graph of an object travelling in a straight line is given.
Select the correct option for the momentum vs time graph of the particle.

Explanation

We know that momentum

$p=mv$ where mass is independent of time t.

Thus, momentum vs time graph is same like velocity vs time. Hence the option A will be right.

The point through which the total weight of an object appears to act for any orientation of the object is ______.

0%
Centre of buoyancy
0%
Centre of gravity
0%
Centre of curvature
0%
Median of a triangle

A force acts on a mass and its variation with time is shown in the above graph.
Find out the change in momentum of the mass between 2s and 5s ?

0%
9 kg.m/s.
0%
18 kg.m/s.
0%
21 kg.m/s.
0%
30 kg.m/s.
0%
42 kg.m/s.

Explanation

Using

$\Delta P = Ft$

Area under the F-t graph gives the change in momentum of the object during that period.

$\therefore$ Change in momentum between 2 s and 5 s,

$\Delta P =$ Area

$=\dfrac{1}{2} \times 6 \times 3 = 9$ kg. m/s

How should the mass of a rotating body of radius r be distributed so as to maximize its angular velocity?

0%
The mass should be concentrated at the outer edge of the body
0%
The mass should be evenly distributed throughout the body
0%
The mass should be concentrated at the axis of rotation
0%
The mass should be concentrated at a point midway between the axis of rotation and the outer edge of the body
0%
Mass distribution has no impact on angular velocity

Explanation

The rotating body is an isolated system ,so its angular momentum is conserved (constant) according to the law of conservation of momentum .The angular momentum

$L$ of a body is given by ,

$L=I\omega$ ,

where

$I=$ moment of inertia of the body ,

$\omega=$ angular velocity of body ,

now

$I=\Sigma mr^{2}$ ,

where

$r=$ distance of mass from axis of rotation , (moment of inertia also depends upon the distribution of mass about axis of rotation) ,

now as

$L$ is constant ,it means product of moment of inertia

$I$ and angular velocity

$\omega$ will also be constant , so to maximize the

$\omega$ we have to reduce the

$I$ , for that we will reduce

$r$ , it means the mass should be concentrated at the axis of rotation .

Two bodies of unequal masses possess the same momentum. Which of the following can be concluded?

0%
Heavier mass possesses lesser K.E
0%
The masses of the two bodies are in the direct ratio of their velocities
0%
Lighter mass possesses less K.E,
0%
Both (A) and (B)

Explanation

Kinetic energy

$K = \dfrac{P^2}{2m}$

For same momentum,

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

Thus heavier mass possesses lesser kinetic energy.

In above shown figure, a ball of mass 4 kg slides over frictionless surface and strikes the post with velocity of 1 m/s and rebounds toward the north at the same speed. The change in the magnitude of the eastward component of the momentum of the disk is:

0%
-4kg.m/s
0%
-1kg.m/s
0%
0kg.m/s
0%
1kg.m/s
0%
4kg.m/s

Explanation

Initially the momentum of ball in the eastward direction=

$m\times v$

$4kg\times (+1m/s)=+4kg-m/s$

Finally the momentum of ball in eastward direction is zero since it moves vertically after striking the post.

Thus the change in momentum of the ball in eastward direction=Final momentum-Initial momentum=

$0-(+4kg-m/s)$

$=-4kg-m/s$

Refer to two 1 kg masses moving toward each other, one mass with velocity

$v_1$ = 10 m/s, the other with velocity

$v_2$ = 20 m/s. What is the velocity of the center of mass?

0%
0 m/s
0%
5 m/s to the left
0%
10 m/s to the left
0%
15 m/s to the left
0%
20 m/s to the left

Explanation

Using law of conservation of momentum,

$Initial\quad momentum=Final\quad momentum\\ \quad \quad \quad \quad \quad \quad { { m }_{ 1 }{ v }_{ 1 } }+{ { m }_{ 2 }{ v }_{ 2 } }=({ { m }_{ 1 } }+{ { m }_{ 2 } }){ v }\\ \quad \quad \quad \quad \quad 10\times 1-20\times 1=(1+1)v\quad \quad \quad \quad \quad \quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad { v }=-\dfrac { 10 }{ 2 } m/s\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad v=5m/s\quad towards\quad left$

A large, massive man and a small boy of small mass are held far apart on an icy surface where friction is negligible. The man and the boy hold opposite ends of a bungee cord that is stretched.
When the boy and the man are released, they are each pulled by the bungee cord so that they approach one another.
When they get close to each other, how will their momenta, velocity, and kinetic energies compare?

0%
Magnitude of Momenta - man's greater then boy's, Magnitude of Velocity - boy's greater than man's, Kinetic Energy - man's greater than boy's
0%
Magnitude of Momenta - same for both, Magnitude of Velocity - boy's greater than man's, Kinetic Energy - same for both
0%
Magnitude of Momenta - same for both, Magnitude of Velocity - same for both, Kinetic Energy - same for both
0%
Magnitude of Momenta - boy's greater then man's, Magnitude of Velocity - boy's greater than man's, Kinetic Energy - same for both
0%
Magnitude of Momenta - same for both, Magnitude of Velocity - boy's greater than man's, Kinetic Energy - boy's greater than man's

Explanation

Initial momentum of the "man + boy" system is zero. Thus according to the law of conservation of momentum, the final momentum of the system must also be zero. Hence the magnitude of momenta for both is same.

Also

$P = mv =constant$

$\implies v \propto \dfrac{1}{m}$

As the mass of man is more, thus the boy will have greater velocity.

Kinetic energy

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

$\implies$

$K.E\propto \dfrac{1}{m}$ (

$\because P =constant$ )

Hence boy will have greater kinetic energy.

Option E is correct.

A filled shopping cart (30.0kg) is being pushed rapidly toward the back of a store so that its velocity is 2.0 m/s. What is the momentum of the cart. and what is the kinetic energy of the cart?

0%
$90 kg m/s$ and $60 J$
0%
$60 kg m/s$ and $90 J$
0%
$60 kg m/s$ and $60 J$
0%
$90 kg m/s$ and $90 J$

Explanation

Given :

$m = 30$ kg

$v = 2.0$ m/s

Momentum of the cart

$p = mv = 30 \times 2.0 = 60$ kg m/s

Kinetic energy of the cart

$K = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 30 \times (2.0)^2 = 60$ J

Find out the velocity of the centre of mass of two

$1 kg$ masses moving toward each other, one mass with velocity

${v}_{1} = 10 {m}/{s}$ , the other with velocity

${v}_{2} = 20 {m}/{s}$ .

0%
$0 {m}/{s}$
0%
$5 {m}/{s}$ to the left
0%
$10 {m}/{s}$ to the left
0%
$15 {m}/{s}$ to the left
0%
$20 {m}/{s}$ to the left

Explanation

Given :

$m_1 = m_2 = 1kg$

$v_1= 10 m/s$

$v_2 = -20 m/s$

Let the velocity of centre of mass be

$V_{cm}$ .

Using

$V_{cm} = \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$

$\therefore$

$V_{cm} = \dfrac{1 \times 10 + 1 \times (-20)}{1 + 1} = -5m/s$

Thus the centre of mass moves with speed

$5m/s$ towards left.

Rank the order of the magnitude of the momenta of the four objects pictured above, greatest first. Masses are shown inside the boundaries of the objects. Velocities are shown with the red arrows ant the corresponding values.

0%
blue and yellow tie, green and purple and orange tie
0%
purple and orange tie, blue, yellow, green
0%
blue, purple and orange tie, green yellow
0%
green and purple tie, or orange, blue and yellow tie
0%
green and purple and orange tie, blue and yellow tie

Explanation

Magnitude of momentum

$|p| = m |v|$

For Green :

$p_{g} = 5 \times 7 =35$ kg m/s

For Blue :

$p_{b} = 10 \times 3 =30$ kg m/s

For Violet :

$p_{v} = 5 \times 7 =35$ kg m/s

For Yellow :

$p_{y} = 10 \times 3 =30$ kg m/s

For Orange :

$p_{o} = 7 \times 5 =35$ kg m/s

Thus decreasing order of magnitude of momentum :

Green = Violet = Orange > Blue = Yellow

Which of the following quantities is zero, when a uniform object is being supported at its centre of gravity?

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Mass
0%
Weight
0%
Force
0%
Moment

State whether true or false.

A couple can never be replaced by a single force.

0%
True
0%
False

The kinetic energy of a body of mass

$4\ kg$ and momentum

$6\ Ns$ will be

0%
$2.5\ J$
0%
$3.5\ J$
0%
$4.5\ J$
0%
$5.5\ J$

Explanation

Given :

$m =4$ kg

$p =6$ Ns

Kinetic energy

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

$\therefore$

$K = \dfrac{6^2}{2\times 4} = 4.5$

$J$

A bucket is being lowered down into a wall through a rope passing over a fixed pulley f radius 10 cm. Assume that the rope does not slip on the pulley. Find the angular acceleration of the pulley at an instant when the bucket is going down at a speed of 20 cm/s and has a acceleration of

$4 ms^{-2}$ .

0%
$10 \ rad \ s^{-2}$
0%
$20 \ rad \ s^{-2}$
0%
$30 \ rad \ s^{-2}$
0%
$40 \ rad \ s^{-2}$

Explanation

Since there is no slipping, the linear acceleration of rope going down would be equal to the tangential acceleration of rotation of pulley at all points.

Hence,

$a=R\alpha$

$\implies \alpha=\dfrac{a}{R}=40rad/s$

State whether true or false.

A couple tends to produce motion in a straight line.

0%
True
0%
False

The total angular momentum of a rigid body can be written as

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$I \times \omega$ ; where I: moment of inertia of body, : angular velocity
0%
$I^2 \times \omega$ ; where I: moment of inertia of body, : angular velocity
0%
$I^2 \times {\omega}^2$ ; where I: moment of inertia of body, : angular velocity
0%
$I \times {\omega}^2$ ; where I: moment of inertia of body, : angular velocity

Explanation

Angular momentum is the rotational equivalent to linear momentum. Angular momentum of a rigid body is defined as the product of object's moment of inertia (

$I$ ) and its angular velocity (

$\omega$ ).
Hence,

$L=I\omega$

State whether true or false.

Only a couple can produce pure rotation in a body.

0%
True
0%
False

State whether true or false.

The centre of gravity of regular shaped objects is at their geometric centre.

0%
True
0%
False

What is the momentum of a

$6.0 kg$ bowling ball with a velocity of

$2.2 {m}/{s}$ ?

0%
11.2
0%
17.2
0%
15.2
0%
13.2

Explanation

Given,

Mass, $m=\,6\,kg$

Velocity, $v=2.2\,m{{s}^{-1}}$

Momentum, $P=mv=6\times 2.2\,=13.2\,Ns$

Hence, momentum is $13.2\,Ns$

An object will not undergo rotational motion when:

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the forces are acting on it at different positions
0%
every forces is creating different turning effects
0%
every moment has the same amplitude
0%
all the forces are acting at its centre of gravity

If v, P and E denote the velocity, momentum and kinetic energy of the particle, then:

0%
P = dE/dv
0%
P = dE/dt
0%
P = dv/dt
0%
none of these.

Explanation

Dimensions of momentum

$[P] = [MLT^{-1}]$

Dimensions of velocity

$[v] = [LT^{-1}]$

Dimensions of energy

$[E] = [ML^2T^{-2}]$

Option A :

Dimensions of LHS

$= [ML T^{-1}]$

Dimensions of RHS

$= \dfrac{ML^2 T^{-2}}{LT^{-1}} = MLT^{-1}$

LHS = RHS

Thus option A is correct.

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Systems Of Particles And Rotational Motion Quiz 4 - MCQExams.com

0:0:3

Practice Class 11 Engineering Physics Quiz Questions and Answers

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