JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Conservation of Momentum

A simple pendulum, made of a string of length I and a bob of mass m , is released from a small angle

θ_{0}

. It strikes a block of mass M , kept on a 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:

(A) $\frac{m}{2} (\frac{θ _{0} + θ _{1}}{θ _{0} - θ _{1}})$
(B) $m (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})$
(C) $m (\frac{θ _{0} + θ _{1}}{θ _{0} - θ _{1}})$ ✓ Correct
(D) $\frac{m}{2} (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})$

Answer: (C)

Solution

The bob reaches the bottom with speed

v = 2 g L (1 - cos θ_{0})

and, after the elastic hit, rebounds with speed

v_{1} = 2 g L (1 - cos θ_{1})

. For a one-dimensional elastic collision a mass

m

striking a stationary mass

M

bounces back with

v_{1} = \frac{M - m}{M + m} v

(the sign just signals the reversal). Taking the ratio of speeds and using

1 - cos θ = 2 sin^{2} \frac{θ}{2}

, then the small-angle approximation

sin \frac{θ}{2} \approx \frac{θ}{2}

, gives

\frac{θ _{1}}{θ _{0}} = \frac{M - m}{M + m}

. Cross-multiplying,

M θ_{0} - m θ_{0} = M θ_{1} + m θ_{1}

, and collecting

M

gives

M = m (\frac{θ _{0} + θ _{1}}{θ _{0} - θ _{1}})

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

Practise more: all JEE Main PYQs on Conservation of Momentum →

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Conservation of Momentum

Solution

More JEE Main questions on Conservation of Momentum