JEE Main201912 Jan 2019Morning ShiftPhysicsCenter of Mass, Momentum and CollisionActual
A simple pendulum, made of a string of length l and a bob of mass m , is released from a small angle θ 0 . It strikes a block of mass M , kept on horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ 1 . Then M is given by:
Options
- Am θ 0 - θ 1 θ 0 + θ 1
- Bm θ 0 + θ 1 θ 0 - θ 1
- Cm 2 θ 0 + θ 1 θ 0 - θ 1
- Dm 2 θ 0 - θ 1 θ 0 + θ 1
Correct answer
B. m θ 0 + θ 1 θ 0 - θ 1
Step-by-step solution
u = 2 g l 1 - c o s θ 0 .......(i) v = velocity of ball after collision v = m - M m + M u Since ball rises up to angle θ 1 v = 2 g l 1 - c o s θ 1 = m - M m + M u ......(ii) From (i) and (ii) m - M m + M = 1 - c o s θ 1 1 - c o s θ 0 = s i n θ 1 2 s i n θ 0 2 ⇒ M m = θ 0 - θ 1 θ 0 + θ 1 ⇒ M = θ 0 - θ 1 θ 0 + θ 1 m