Center Of Mass(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

CENTER OF MASS MCQs

Total Questions : 30 | Page 1 of 3 pages

Two 20 kg cannon balls are chained together and fired horizontally with a velocity of 200 m/s from the top of a 30 m wall. The chain breaks during the flight of the cannon balls and one of them strikes the ground at t = 2 s, at a distance of 250 m from the foot of the wall, and 5 m to the right of the line of fire. Determine the position of the other cannon ball at that instant. Neglect the resistance of air.

(550, 20, -5)
(250, 20, 5)
(550, 10, -5)
(550, 20, 0)

Discuss Question

Answer: Option A. -> (550, 20, -5)
:
A
As no external force acts in z direction, hence z-coordinate of the centre of mas of the ball should be zero. To make z-coordinate zero other ball should fall symmetrically with respect to z-axis.
Hence, z-coordinate of other ball = –5 m.
The balls do not have any external force in x-direction. Hence in x-direction the centre of mass should move with constant velocity. X-coordinate of centre of mass at

t = 2 s = 200 \times 2 = 400 m

Hence,

x_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}}

400 = \frac{20 \times 250 + 20 x_{2}}{20 + 20}

x_{2} = 800 - 250 = 550 m

Height fallen by centre of mass at t = 2s,

h = \frac{1}{2} \times 10 \times (2)^{2} = 20 m

Hence, y-coordinate of centre of mass

= 30 - 20 = 10 m

Hence,

y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}}

10 = \frac{20 \times 0 + 20 \times y_{2}}{20 + 20}

\Rightarrow y_{2} = 20 m

Question 2.

Figure below shows the system is at rest initially with x = 0. A man and a woman both are initially at the extreme ends of the platform. The man and the woman start to move towards each other. Obtain an expression for the displacement 's' of the platform when the two meet in terms of the displacement ' $x_{1}$ ' of the man relative to the platform.

$\frac{m_{1} x_{1} + m_{2} l}{m_{1} + m_{2} + m_{0}}$
$\frac{m_{1} x_{1} - m_{2} l}{m_{1} + m_{2} + m_{0}}$
$\frac{m_{1} x_{1} + m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}$
$\frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}$

Discuss Question

Answer: Option D. ->

\frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

:
D
As no external force is acting on the system, the displacement of the centre of mass should be zero. Let the displacement of the platform toward right be s,

Δ {\to x}_{1} = Δ {\to x}_{1, p} + Δ {\to x}_{p} = (- x_{1} + s)

Δ {\to x}_{2} = Δ {\to x}_{2, p} + Δ {\to x}_{p} = (l - x_{1} + s)

Hence,

0 = \frac{m_{1} (- x_{1} + s) + m_{2} (l - x_{1} + s) + m_{0} s}{m_{1} + m_{2} + m_{0}}

s (m_{1} + m_{2} + m_{0}) - m_{1} x_{1} + m_{2} (l - x_{1}) = 0

s = \frac{m_{1} x_{1} - m_{2} (l - x_{1})}{m_{1} + m_{2} + m_{0}}

Question 3.

A block was released on the convex surface of hemispherical wedge as shown in the figure below. Determine the displacement of the wedge when the block reaches the angular position $θ$ .

$\frac{m R c o s θ}{M + m}$ to the right
$\frac{m R s i n θ}{M + m}$ to the right
$\frac{m R c o s θ}{M + m}$ to the left
$\frac{m R s i n θ}{M + m}$ to the left

Discuss Question

Answer: Option D. ->

\frac{m R s i n θ}{M + m}

to the left
:
D

m_{1} Δ {\to x}_{1} + m_{2} Δ {\to x}_{2} = 0

m (R s i n θ - x) - M x = 0

x = \frac{m R s i n θ}{m + M}

Question 4.

A block of mass 'm' is initially lying on a wedge of mass 'M' with an angle of inclination $θ$ , as shown in the figure. Calculate the displacement of the wedge when the block reaches bottom of the wedge.

$\frac{m h}{M + m}$
$\frac{m h c o t θ}{M + m}$
$\frac{m h c o s θ}{M + m}$
$\frac{m h s i n θ}{M + m}$

Discuss Question

Answer: Option B. ->

\frac{m h c o t θ}{M + m}

:
B

A Block Of Mass 'm' Is Initially Lying On A Wedge Of Mass 'M...

l = h c o s θ

Let displacement of wedge be x towards right:

Δ x_{C M} = \frac{m (l + x) + M x}{M + m}

0 = m l + (M + m) x

x = \frac{- m l}{M + m} = - \frac{m h c o t θ}{M + m}

Question 5.

A cart of mass 'M' is at rest on a frictionless horizontal surface and a pendulum bob of mass 'm' hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes several collisions on the floor and finally lands up in a small slot made on the floor. The horizontal distance between the string and the slot is 'L'. Find the displacement of the cart during this process.

$\frac{m l}{M + m}$ to the right
$\frac{m l}{M + m}$ to the left
$\frac{M l}{M + m}$ to the right
$\frac{M l}{M + m}$ to the left

Discuss Question

Answer: Option B. ->

\frac{m l}{M + m}

to the left
:
B
Let displacement of the cart be

x_{c} = - x

A Cart Of Mass 'M' Is At Rest On A Frictionless Horizontal S...

Displacement of the bob is

x_{b} = (L - x)

(x_{C M})_{s y s t e m} = 0

M_{c a r t} x_{c a r t} + M_{b o b} x_{b o b} = 0

x = \frac{m L}{m + M}

Question 6.

Two blocks of masses $m_{1}$ and $m_{2}$ connected by a massless spring of spring constant 'k' rest on a smooth horizontal plane as shown in figure. Block 2 is shifted a small distance 'x' to the left and the released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

$\frac{\sqrt{m_{2} K}}{m_{1} + m_{2}}$ x
$\frac{\sqrt{m_{1} K}}{m_{1} + m_{2}}$ x
$\frac{\sqrt{(m_{1} + m_{2}) K}}{m_{1} + m_{2}} x$
zero

Discuss Question

Answer: Option A. ->

\frac{\sqrt{m_{2} K}}{m_{1} + m_{2}}

x
:
A
If

m_{2}

is shifted by a distance x and released, the mass

m_{1}

will break off from the wall when the spring restores its natural length and

m_{2}

will start going towards right. At the time of breaking

m_{1}

m_{2}

will be going towards right with a velocity v, which is given as

\frac{1}{2} k x^{2} = \frac{1}{2} m_{2} v^{2} \Rightarrow v = (\sqrt{\frac{k}{m_{2}}}) x

and the velocity of the centre of mass at this instant is

v_{C M} = \frac{m_{1} \times 0 + m_{2} \times v}{m_{1} + m_{2}} = \frac{(\sqrt{m_{2} k}) x}{m_{1} + m_{2}}

Question 7.

Two masses $m_{1}$ and $m_{2}$ placed at a distance 'l' apart, let the centre of mass of this system be at a point named C. If $m_{1}$ is displaced by $l_{1}$ towards C and $m_{2}$ is displaced by $l_{2}$ away from C, find the distance from C where the new centre of mass will be located.

$\frac{m_{1} l_{1} + m_{2} l_{2}}{m_{1} + m_{2}}$
$\frac{m_{1} l_{1} - m_{2} l_{2}}{m_{1} + m_{2}}$
$l_{1} + l_{2}$
$l_{1} - l_{2}$

Discuss Question

Answer: Option A. ->

\frac{m_{1} l_{1} + m_{2} l_{2}}{m_{1} + m_{2}}

:
A
If

m_{1}

and

m_{2}

are placed at a distance l apart, their centre of mass will be located at a distance x from

m_{1}

, where

x = \frac{m_{2} l}{m_{1} + m_{2}}

m_{1}

is displaced by

l_{1}

towards C and

m_{2}

is displaced by

l_{2}

away from C, the new centre of mass

C^{^{'}}

now will be located at a distance

x^{'}

from

m_{1}

, where

x^{^{'}} = \frac{m_{2} (l - l_{1} + l_{2})}{m_{1} + m_{2}}

Displacement of the centre of mass is

Δ x = x^{^{'}} + l_{1} - x = \frac{m_{1} l_{1} + m_{2} l_{2}}{m_{1} + m_{2}}

Question 8.

Figure below shows a flat car of mass 'M' on a frictionless road. A small massless wedge is fitted on it as shown. A small ball of mass 'm' is released from the top of the wedge, it slides over it and falls in the hole at distance 'l' from the initial position of the ball. Find the distance the flat car moves till the ball gets into the hole.

$\frac{m l}{M + m}$
$\frac{M l}{M + m}$
$\frac{m}{M} l$
$\frac{M}{m} l$

Discuss Question

Answer: Option A. ->

\frac{m l}{M + m}

:
A
When the ball falls into the hole, with respect to the flat car, the ball travels as horizontal distance l. During this motion, to conserve momentum and to maintain the position of centre of mass, the car moves towards left, say by a distance x. Thus, the total distance travelled by the ball towards right is (l – x). As centre of mass remains at rest, the change in mass moments of the two (ball and car) about any point must be equal to zero. Hence

m (l - x) = M x \Rightarrow m = \frac{m l}{M + m}

Question 9.

Figure below shows a fixed wedge on which two blocks of masses 2 kg and 3 kg are placed on its smooth inclined surfaces. When the two blocks are released from rest, find the acceleration of centre of mass of the two blocks.