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

Question 1. 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.

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

mhM+m
mhcotθM+m
mhcosθM+m
mhsinθM+m

Explanation:-

Answer: Option B. -> mhcotθM+m
:
B

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 2. Four Particles of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed at the corners of a square of side 2 m in the x – y plane in anticlockwise sense. If the origin of the co-ordinate system is taken at the mass of 1 kg, the (x, y) co-ordinates of the centre of mass, expressed in 'metre' are

(1,75)
(2,75)
(3,75)
(4,75)

Explanation:-

Answer: Option A. -> (1,75)
:
A
The (x, y) co - ordinates of masses 1 kg, 2kg, 3kg and 4kg respectively are

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

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

x_{C M} = \frac{1 k g \times 0 + 2 k g \times 2 m + 3 k g \times 2 m + 4 k g \times 0}{1 k g + 2 k g + 3 k g + 4 k g} = 1 m

y_{C M} = \frac{1 k g \times 0 + 2 k g \times 0 + 3 k g \times 2 m + 4 k g \times 2 m}{1 k g + 2 k g + 3 k g + 4 k g} = \frac{7}{5} m

Question 3. Find the displacement of the wedge when the mass 'm' has comes out of the wedge. There is no friction anywhere.

mlM+2m
mlM+m
ml−mlcosθM+2m
ml+mlcosθM+2m

Explanation:-

Answer: Option C. -> ml−mlcosθM+2m
:
C
Let wedge move towards right by distance x; then

Δ x_{C M} = \frac{m (l c o s θ + x) + M x + m (x - l)}{m + M + m}

0 = m l c o s θ - m l + (m + M + m) x

x = \frac{m l - m l c o s θ}{2 m + M}

Question 4. Two masses

m_{1}

and

m_{2}

are moving with velocities

v_{1}

and

v_{2}

respectively. Find their total kinetic energy in the reference frame of centre of mass.

m1m2m1+m2(v1−v2)2
12m1m2m1+m2(v1−v2)2
12m1m2m1+m2(v1+v2)2
12(m1m2)(v1−v2)2

Explanation:-

Answer: Option B. -> 12m1m2m1+m2(v1−v2)2
:
B

K = \frac{1}{2} m_{1} v_{1 c}^{2} + \frac{1}{2} m_{2} v_{2 c}^{2}

…… (i)
Where

v_{1 c}

and

v_{2 c}

are velocities relative to the CM.

v_{1 c} = v_{1} - v_{C M} = v_{1} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{2} (v_{1} - v_{2})}{m_{1} + m_{2}})

v_{2 c} = v_{2} - v_{C M} = v_{2} - (\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}) = (\frac{m_{1} (v_{2} - v_{1})}{m_{1} + m_{2}})

Putting these in Equation (i),

K = \frac{1}{2} \frac{m_{1} m_{2}}{(m_{1} + m_{2})} (v_{1} - v_{2})^{2}

Note: For a system of two particles of masses

m_{1}

and

m_{2}

, the total kinetic energy is

k = (1 / 2) μ v^{2} + (1 / 2) (m_{1} + m_{2}) v_{C M}^{2}

where

μ = \frac{m_{1} m_{2}}{m_{1} + m_{2}}

(known as reduced mass of the system and

\to v = {\to v}_{1} - {\to v}_{2}

(relative velocity).

Question 5. Consider a rectangular plate of dimensions

a \times b

. If the plate is considered to be made up of four rectangles of dimensions

\frac{a}{2} \times \frac{b}{2}

and we now remove one (the lower right) out of the four rectangles, find the position where the centre of mass of the remaining system will be (considering the center of the rectangular plate as the origin)

Consider A Rectangular Plate Of Dimensions A×b. If The Plat...

−a12 ,b12
a12,−b12
−a12,−b12
a12,b12

Explanation:-

Answer: Option A. -> −a12 ,b12
:
A
The rectangular plate is shown in the figure of which one part is removed. We can find the x and y coordinates of the centre of mass of this system, taking origin at the centre of the plate. The coordinates of the three remaining rectangles are (a/4, b/4), (-a/4, +b/4) and (-a/4, -b/4). By geometry, masses of these rectangles can be taken as M/4. Now x-coordinate of the centre of mass:

x_{C M} = \frac{\frac{M a}{16} - \frac{M a}{16} - \frac{M a}{16}}{3 \frac{M}{4}} = - \frac{a}{12}

and y-coordinate of the centre of mass:

y_{C M} = \frac{\frac{M b}{16} - \frac{M b}{16} - \frac{M b}{16}}{3 \frac{M}{4}} = \frac{b}{12}

Question 6. A 30 kg projectile moving horizontally with a velocity

{\to v}_{0} = (120 m / s)^i

explodes into two fragments A and B of masses 12 kg and 18 kg, respectively. Taking point of explosion as origin and knowing that 3 s later the position of fragment A is (300, 24, –48) m, determine the position of fragment B at the same instant.

(360, -45, 32)
(400, -91, 32)
(400, -91, 48)
(360, -91, 48)

Explanation:-

Answer: Option B. -> (400, -91, 32)
:
B
Initial coordinates of CM = (0, 0, 0)
x coordinates of CM after 3 s,

x_{C M} = 120 \times 3 = 360 m

y coordinates of CM after 3 s,

y_{C M} = - \frac{1}{2} g t^{2} = - 45 m

After 3 s,

(m_{1} + m_{2}) x_{C M} = m_{1} x_{1} + m_{2} x_{2}

x_{2} = \frac{30 \times 360 - 12 \times 300}{18} = 400 m

Similarly,

(m_{1} + m_{2}) y_{C M} = m_{1} y_{1} + m_{2} y_{2}

- 30 \times 45 = 12 \times 24 + 18 \times y_{2}

y_{2} = - 91 m

and in z coordinate,

m_{1} z_{1} + m_{2} z_{2} = 0 \Rightarrow z_{2} = 32 m

Question 7.

lim x \to 2 \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}

is equal to

12
1
2
0

Explanation:-

Answer: Option A. -> 12
:
A

lim x \to 2 \frac{\sqrt{x - 2} + \sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}}

= lim x \to 2 (\frac{\sqrt{x - 2}}{\sqrt{x + 2} \sqrt{x - 2}} + \frac{\sqrt{x} - \sqrt{2}}{\sqrt{x^{2} - 4}})

On rationalisation -

= lim x \to 2 (\frac{1}{\sqrt{x + 2}} + \frac{x - 2}{\sqrt{x^{2} - 4} (\sqrt{x} + \sqrt{2})})

= lim x \to 2 \frac{1}{\sqrt{x + 2}} + lim x \to 2 \sqrt{\frac{x - 2}{x + 2}} \times \frac{1}{\sqrt{x} + \sqrt{2}}

= \frac{1}{2}

Question 8. Let there are three equal masses situated at the vertices of an equilateral triangle, as shown in figure. Now the particle A starts with a velocity

v 1

towards line AB, particle B starts with the velocity

v_{2}

towards line BC and particle starts with velocity

v_{3}

towards line CA. Find the displacement of the centre of mass of the three particles A, B and C after time t. What would it be the displacement of centre of mass if

v_{1} = v_{2} = v_{3}

?

Let There Are Three Equal Masses Situated At The Vertices Of...

(v1+v2+v3)t
(v1+v2+v3)t2
(v1+v2+v3)t3
'0'

Explanation:-

Answer: Option D. -> '0'
:
D
First we write the three velocities in vectorial form, taking right direction as positive x-axis and upwards as positive y-axis.

{\to v}_{1} = - \frac{1}{2} v_{1}^i - \frac{\sqrt{3}}{2} v_{1}^j

{\to v}_{2} = v_{2}^i, {\to v}_{3} = - \frac{1}{2} v_{3}^i + \frac{\sqrt{3}}{2} v_{3}^j

Thus the velocity of centre of mass of the system is

{\to v}_{C M} = \frac{{\to v}_{+} {\to v}_{2} + {\to v}_{3}}{3}

= (v_{2} - \frac{1}{2} v_{1} - \frac{1}{2} v_{3})^i + \frac{\sqrt{3}}{2} (V_{3} - V_{1})^j

Which can be written as

{\to v}_{C M} = v_{x}^i + v_{y}^j

Δ \to r = v_{x} t^i + v_{x} t^j

If

v_{1} = v_{2} = v_{3} = v

, we have

{\to v}_{C M} = 0

Therefore, there is no displacement of centre of mass of the system

Question 9. Four particles, each of mass 'm', are placed at the corners of 'a' square of side a in the x – y plane. If the origin of the co-ordinate system is taken at the point of intersection of the diagonals of the square, the co - ordinates of the centre of mass of the system are

(a, a)
(–a, a)
(a, –a)
(0,0)

Explanation:-

Answer: Option D. -> (0,0)
:
D

Four Particles, Each Of Mass 'm', Are Placed At The Corners ...

Given AB = BC = CD = AD =a. The (x, y) co-ordinates of the masses at A, B, C and D respectively are

(x_{1} = - \frac{a}{2} y_{1} = - \frac{a}{2}), (x_{2} = \frac{a}{2}, y_{2} = - \frac{a}{2})

(x_{3} = \frac{a}{2}, y_{3} = \frac{a}{2}), (x_{4} = - \frac{a}{2}, y_{4} = \frac{a}{2})

It is easy to see that the co-ordinates of the centre of mass are (0, 0). Hence the correct choice is (d)

Question 10. Two balls A and B of equal masses are projected upward simultaneously, one from the ground with speed 50

m s^{- 1}

and other from height 40 m above the first ball high tower with initial speed 30

m s^{- 1}

. Find the maximum height attained by their centre of mass.

Two Balls A And B Of Equal Masses Are Projected Upward Simul...

80 m
100 m
20 m
60 m

Explanation:-

Answer: Option B. -> 100 m
:
B
The initial position of centre of mass is

y_{c m} = \frac{m \times 0 + m \times 40}{2 m} = 20 m

Initial velocity of centre of mass

u_{c m} = \frac{m \times 50 + m \times 30}{2 m} = 40 m s^{- 1}

Acceleration of centre of mass,

a_{c m} = - g

Using kinematics equation:

v_{c m}^{2} = u_{c m}^{2} + 2 a_{c m} H

Here H is the maximum height reach by centre of mass of two balls from initial level.

∴

0^{2} = 40^{2} - 2 \times 10 \times H

\Rightarrow H = \frac{1600}{20} = 80 m

Hence, maximum height reach by centre of mass from ground level will be

h_{m a x} = (h_{c m})_{i n i t i a l} + H = 20 + 80 = 100 m