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

12th Grade > Physics

CENTER OF MASS MCQs

Total Questions : 30 | Page 3 of 3 pages

lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}

−1
1
16
32

Discuss Question

Answer: Option A. -> −1
:
A

lim x \to \infty \frac{(2 + x)^{40} (4 + x)^{5}}{(2 - x)^{45}}

= lim x \to \infty \frac{x^{45} {(\frac{2}{x} + 1)}^{40} {(\frac{4}{x} + 1)}^{5}}{x^{45} {(\frac{2}{x} - 1)}^{45}}

= \frac{(1)^{40} (1)^{5}}{(- 1)^{45}}

= \frac{(1)^{45}}{(- 1)^{45}}

= - 1

Question 22. 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.

√31g10
g
g/2
g/5

Discuss Question

Answer: Option A. -> √31g10
:
A
Block A slides with acceleration

\frac{g}{2}

and block B slides with acceleration

\frac{\sqrt{3} g}{2}

. Now the acceleration of centre of mass of the system of blocks A and B can be given both in x and in y directions as

a_{x} = \frac{3 \times \frac{\sqrt{3} g}{2} c o s 60^{\circ} - 2 \times \frac{g}{2} s i n 30^{\circ}}{5} = \frac{\sqrt{3} g}{20}

and

a_{y} = \frac{3 \times \frac{\sqrt{3} g}{2} s i n 60^{\circ} + 2 \times \frac{g}{2} s i n 30^{\circ}}{5} = \frac{11 g}{20}

Thus acceleration of the centre of mass of the system is given as

a_{C M} = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{{(\frac{\sqrt{3} g}{20})}^{2} + {(\frac{11 g}{20})}^{2}} = \frac{\sqrt{31}}{10} g

Question 23. 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.

m1l1+m2l2m1+m2
m1l1−m2l2m1+m2
l1+l2
l1−l2

Discuss Question

Answer: Option A. -> m1l1+m2l2m1+m2
:
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 24. 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.

Two Blocks Of Masses M1 And M2 Connected By A Massless Sprin...

√m2Km1+m2x
√m1Km1+m2x
√(m1+m2)Km1+m2x
zero

Discuss Question

Answer: Option A. -> √m2Km1+m2x
:
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 25. Two blocks A and B each of equal masses 'm' are released from the top of a smooth fixed wedge as shown in the figure. The magnitude of acceleration of the centre of mass of the two blocks is ___.

Discuss Question

Let us consider incline surfaces of wedge as x-axis and y-axis respectively. We can write accelerations of blocks A and B.

{\to a}_{A} = m g s i n 30^{\circ}^i a n d {\to a}_{B} = m g s i n 60^{\circ}^j

The acceleration of centre of mass of two blocks can be written as

{\to a}_{a m} = \frac{m_{A} {\to a}_{A} + m_{B} {\to a}_{B}}{m_{A} + m_{B}} = \frac{m g s i n 30^{\circ}^i + m g s i n 60^{\circ}^j}{m + m}

= \frac{g s i n 30^{\circ}^i + m g s i n 60^{\circ}^j}{m + m}

= \frac{g}{4}^i + \frac{\sqrt{3} g}{4}^j \Rightarrow | {\to a}_{a m} | = \frac{g}{2}

Question 26. Four particles, each of mass 'm', are placed at the corners of a square of side 'a' shown in the figure. The position vector of the centre of mass is

a(i+j)
a2(i+j)
a(i−j)
a2(i−j)

Discuss Question

Answer: Option B. -> a2(i+j)
:
B

The (x, y) co-ordinates of the masses at O, A, B and C respectively are

(x_{1} = 0, y_{1} = 0), (x_{2} = a, y_{2} = 0), (x_{3} = a, y_{3} = a), (x_{4} = 0, y_{4} = a) .

Therefore, the (x,y) co-ordinates of the centre of mass are

x_{C M} = \frac{a}{2} a n d y_{C M} = \frac{a}{2}

. The position vector of the centre of mass is

\frac{a}{2} (i + j)

, which is choice (b)

Question 27. The motion of centre of mass of a system of two particles is unaffected by their internal forces

Irrespective of the actual directions of the internal forces
Only if they are along the line joining the particles
Only if they are at right angles to the line joining the particles
Only if they are obliquely inclined to the line joining the particles

Discuss Question

Answer: Option A. -> Irrespective of the actual directions of the internal forces
:
A
The correct choice is (a) because the centre of mass is affected only be external forces; not by internal forces.

Question 28. A uniform rod of mass 'm' and length 'L' is tied to a vertical shaft. It rotates in horizontal plane about the vertical axis at angular velocity

ω

. How much horizontal force does the shaft excert on the rod?

A Uniform Rod Of Mass 'm' And Length 'L' Is Tied To A Vertic...

mw2L
Mw2L2
mw2L4
zero

Discuss Question

Answer: Option B. -> Mw2L2
:
B
Let T be the force applied horizontally on the rod. The acceleration of centre of mass is

ω^{2} \frac{L}{2}

, as it is moving in a circle of radius

\frac{L}{2}

.
Using

{\to F}_{e x t} = M {\to a}_{c m} \Rightarrow T = M ω^{2} (\frac{L}{2})

Thus, the horizontal force exerted is

\frac{1}{2} M ω^{2} L

Question 29. 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.

mlM+m to the right
mlM+m to the left
MlM+m to the right
MlM+m to the left

Discuss Question

Answer: Option B. -> mlM+m to the left
:
B
Let displacement of the cart be

x_{c} = - x

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 30. 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.

mlM+m
MlM+m
mMl
Mml

Discuss Question

Answer: Option A. -> mlM+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