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

12th Grade > Physics

CENTER OF MASS MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. Four particles of masses 'm', 'm', '2m' and '2m' are placed at the four corners of a square of side a as shown in diagram The (x, y) coordinates of the centre of mass are

(a2,2a)
(a2,a)
(a2,2a3)
(a,a3)

Discuss Question

Answer: Option C. -> (a2,2a3)
:
C
The (x, y) co-ordinates of the centre of mass are

x = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}

and

y = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}

It is easy to show that

x = \frac{a}{2}

and

y = \frac{2 a}{3}

, which is choice (c).

Question 12. Two children A and B of same mass (including their caps) M are sitting on a see – saw as chown in figure. Initially, the beam is horizontal. At Once, child B throws away his cap (mass M/25) which falls at point Q, midpoint of the left half of the beam, due to this the balance of beam is disturbed. To balance it again what is the mass m required to be put at point P on the right half of the beam?

0.16M
0.12M
0.04 M
0.1M

Discuss Question

Answer: Option B. -> 0.12M
:
B
Initially, the beam was horizontal because the centre of mass of the system was at the centre (pivot of see-saw) of the system. When child B threw his cap to point Q, centre of mass of the system shifted to the left of the pivot and that’s why the balance got disturbed and beam started rotating in anticlockwise sense. To balance it again, we must put some mass to the right of it so as to displace the centre of mass of the system again at the centre of the system.
If a mass m is put at point P, to bring the centre of mass of the system at the centre, we have

M \frac{L}{2} + \frac{M}{25} \frac{L}{4} = \frac{24 M}{25} \frac{L}{2} + m \frac{L}{4}

or m = 0.12 M

Question 13. An explosion breaks a rock into three parts. Two pieces go off at right angles to each other with 1.0 kg piece with a velocity of 12 m/s and other 2.0 kg piece with a velocity of 8 m/s. If the third piece flies off with a velocity of 40 m/s, compute the mass of the third piece.

0.5 kg
3 kg
0.7 kg
0.25kg

Discuss Question

Answer: Option A. -> 0.5 kg
:
A

An Explosion Breaks A Rock Into Three Parts. Two Pieces Go O...

Let

m_{1}, m_{2} a n d m_{3}

be the masses of the three pieces.

m_{1} = 1.0 k g, m_{2} = 2.0 k g

Let

v_{1} = 12 m / s, v_{2} = 8 m / s, v_{3} = 40 m / s . L e t v_{1} a n d v_{2}

be directed along x-and y-axes, respectively, and

v_{3}

be directed as shown.
By the principle of conservation of momentum, initial momentum is zero. Hence,
along x-axis:

0 = m_{1} v_{1} - m_{3} v_{3} c o s θ

Along y - axis:

0 = m_{2} v_{2} - m_{3} v_{3} s i n θ

\Rightarrow m_{1} v_{1} - m_{3} v_{3} c o s θ a n d m_{2} v_{2} - m_{3} v_{3} s i n θ

By squaring and adding, we get

m_{1}^{2} v_{1}^{2} + m_{2}^{2} v_{2}^{2} = m_{3}^{2} v_{3}^{2}

\Rightarrow m_{3}^{2} = \frac{1^{2} (12)^{2} + (2)^{2} (8)^{2}}{(40)^{2}} \Rightarrow m_{3} = 0.5 k g

Question 14. Mr. Verma (50 kg) and Mr. Mathur (60 kg) are sitting at the two extremes of a 4 m long boat (40 kg) standing still in water. To discuss a mechanics problem, they come to the middle of the boat. Neglecting friction with water, how far does the boat move in the water during the process?

2 m
1/15 m
2/15 m
3/15 m

Discuss Question

Answer: Option C. -> 2/15 m
:
C

Mr. Verma (50 Kg) And Mr. Mathur (60 Kg) Are Sitting At The ...

Let displacement of the boat is

X_{b} = x

towards right
The displacement of Mr. Verma:

X_{v} = (x + 2)

m
The displacement of Mr. Mathur:

X_{m} = (x - 2) m

As no external forces are acting on the system, the displacement of the centre of mass of the system should be zero.

Δ x_{C M} = \frac{M_{b} x_{b} + M_{v} x_{v} + M_{m} x_{m}}{(M_{b} + M_{v} + M_{m})}

∴

0 = \frac{40 x + 50 (x + 2) + 60 (x - 2)}{150} \Rightarrow x = 2 / 15

Question 15. Figure below shows two blocks of masses 5 kg and 2 kg placed on a frictionless surface and connected with a spring. An external kick gives a velocity of 14 m/s to the heavier block in the direction of lighter one. Deduce (a) velocity gained by the centre of mass and (b) the separate velocities of the two blocks in the centre of mass coordinates just after the kick.

10 m/s, 14 m/s, 10 m/s
10 m/s, 4 m/s, 10 m/s
10 m/s, 4 m/s, -10 m/s
10 m/s, 10 m/s, 10 m/s

Discuss Question

Answer: Option C. -> 10 m/s, 4 m/s, -10 m/s
:
C
a. Velocity of centre of mass is

v_{C M} = \frac{5 \times 14 + 2 \times 0}{5 + 2} = 10 m / s

b. Due to the kick on the 5 kg block, it starts moving with a velocity 14 m/s immediately, but due toinertia, 2 kg block remains at rest, at that moment. Thus, velocity of 5 kg block with respect tocentre of mass is

v_{1} = 14 - 10 = 4 m / s

, and velocity of 2 kg block with respect to centre of mas is

v_{2} = 0 - 10 = - 10 m / s

Question 16. What is the net displacement of the wedge when the ball reaches the bottom of the groove.

mRM+m to the left
mRM+m to the right
m(R−r)M+m to the left
m(R−r)M+m to the right

Discuss Question

Answer: Option C. -> m(R−r)M+m to the left
:
C

Δ x_{C M} = \frac{M x + m (R - r + x)}{M + m}

0 = M x + m x + m (R - r)

x = \frac{- m (R - r)}{M + m}

Question 17. A cannon and a supply of cannon balls are inside a sealed rail road car. The cannon fires to the right, and the car recoils to the left. The cannon balls remain in the car after hitting the far wall. Find the maximum distance by which the rail road car move assuming it starts from the rest.

L2
L
0
Can't determine

Discuss Question

Answer: Option B. -> L
:
B
Initially, the whole system is at rest, so

v_{C M} = 0

. As there is no external force acting on the system,

{\to v}_{C M}

= constant = 0. So position of centre of mass of system remains fixed.

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

…… (i)
where m is mass of the cannon balls and M that of the (car + cannon) system.
As

Δ x_{C M} = 0

, therefore m

Δ x_{1} + M Δ x_{2} = 0

…… (ii)
As cannon balls cannot leave the car, so maximum displacement of the balls relative to the car is L and in doing so the car will shift a distance

Δ x_{2} = D

(say) relative to the ground, opposite to the displacement of the balls; then the displacement of balls relative to ground will be

Δ x_{1} = L - D

…… (iii)
Substituting the value of

Δ x_{1}

from Equation (iii) in Equation (ii), we get

m (L - D) - M D = 0

\Rightarrow D = \frac{m L}{M + m} = \frac{L}{1 + \frac{M}{m}}

\Rightarrow D < L

i.e. rail road car cannot travel more than L.

Question 18. 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 19. 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

mRcosθM+m to the right
mRsinθM+m to the right
mRcosθM+m to the left
mRsinθM+m to the left

Discuss Question

Answer: Option D. -> mRsinθM+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 20. 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.

Figure Below Shows The System Is At Rest Initially With X = ...

m1x1+m2lm1+m2+m0
m1x1−m2lm1+m2+m0
m1x1+m2(l−x1)m1+m2+m0
m1x1−m2(l−x1)m1+m2+m0

Discuss Question

Answer: Option D. -> m1x1−m2(l−x1)m1+m2+m0
:
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,