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JEE PYQ: Laws Of Motion Question 34

Question 34 - 2019 | 12 Jan Shift 2] (MCQ)

A simple pendulum, made of a string of length $l$ and a bob of mass $m$, is released from a small angle $\theta_0$. 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 $\theta_1$. Then $M$ is given by :

(1) $\frac{m}{2}\left(\frac{\theta_0 + \theta_1}{\theta_0 - \theta_1}\right)$

(2) $m\left(\frac{\theta_0 - \theta_1}{\theta_0 + \theta_1}\right)$ [Correct]

(3) $m\left(\frac{\theta_0 + \theta_1}{\theta_0 - \theta_1}\right)$

(4) $\frac{m}{2}\left(\frac{\theta_0 - \theta_1}{\theta_0 + \theta_1}\right)$

Show Answer

Answer: (2)

Solution

For elastic collision with small angles: $\frac{m-M}{m+M} = \frac{\theta_1}{\theta_0}$. So $M = m\frac{\theta_0 - \theta_1}{\theta_0 + \theta_1}$.


Learning Progress: Step 34 of 34 in this series