# Physics Questions and Answers – System of Particles – Motion of Centre of Mass

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This set of Physics Multiple Choice Questions & Answers (MCQs) focuses on “System of Particles – Motion of Centre of Mass”.

1. If forces are acting on a rigid body so that it has zero kinetic energy, then all forces will pass through the centre of mass and have a vector sum of zero. True or False?
a) True
b) False

Explanation: If a rigid body has zero kinetic energy means that vector sum of all forces will be zero and the torque on it will be zero. For torque to be zero the vector cross product of force and distance should be zero. It is not necessary for all forces to pass through the centre.

2. In a projectile motion, a particle breaks due to internal forces. We can say that its total momentum is conserved. True or False?
a) True
b) False

Explanation: For momentum to be conserved its value should remain constant, but in projectile motion the force of gravity is always changing its momentum. The internal forces will break the particle and the momentum of different particles can be conserved in a direction perpendicular to direction of gravity.

3. A ball of mass 1kg is thrown vertically upwards from a building with a speed of 10m/s, and another ball of mass 3kg is dropped from the same point towards the ground with the same speed. At what time will the centre of mass have the maximum height w.r.t the ground? Assume balls are left at t=0.
a) 0s
b) 1s
c) 2s
d) 2.8s

Explanation: When the balls are released in their respective directions, the ball going downwards is heavier and is covering more distance per unit time. This is because both the balls have the same initial speed and the heavier ball is gaining speed downwards while the other ball is losing speed as it moves upwards. Their mid-point will be below the point of release and centre of mass, will therefore, be even below that point, since heavier mass is moving downwards.

4. A boy of mass 50kg is standing on a frictionless surface. He throws a ball of mass 2kg away from him with a speed of 10m/s. Find the final speed of the centre of mass.
a) 0m/s
b) 20m/s
c) 10m/s
d) 0.4m/s

Explanation: Initially the boy was at rest and there is no friction, so the net force on the ball-boy system is zero. Hence, their momentum change will be zero and final velocity of centre will also be zero. Note that boy will move backwards with a speed of 2*10/50 = 0.4m/s.

5. A ball of mass 3kg is thrown at an angle of 30° with the horizontal & a speed of 10m/s. At the highest point the ball breaks into two parts, having mass ratio 2:1, due to internal forces. If the heavier part falls at a distance of 10m from the start, find the x coordinate of the second part.
a) 5.98m
b) 6.34m
c) 8.66m
d) 7.99m

Explanation: There is no net force on the particle in the x-direction, so the x-coordinate of centre of mass will be equal to the range of projectile motion.
The parts will have masses 2kg and 1kg.
The time of motion ‘t’ = 2uy/g = 1s.
The range = uxt = 5√3*1 = 8.66m.
The centre of mass will land at 8.66m.
So,
(m1x1 + m2x2)/(m1 + m2) = 8.66.
m1 = 2kg, m2 = 1kg, x1 = 10m, we have to calculate x2.
∴ m2x2 = (8.66*3) – 2*10
= 25.98 – 20 = 5.98.
∴ x2= 5.98/1 = 5.98m. 