A simple pendulum,made of a string of length | vedclass

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(C) For small angles,$\cos 	heta \approx 1 - \frac{	heta^2}{2}$. The velocity of the bob of mass $m$ at the lowest point is $v = \sqrt{2gl(1 - \cos

Vedclass · Accepted Answer

$m\left( \frac{	heta_0 + 	heta_1}{	heta_0 - 	heta_1} ight)$

Vedclass · Answer

(C) For small angles,$\cos 	heta \approx 1 - \frac{	heta^2}{2}$. The velocity of the bob of mass $m$ at the lowest point is $v = \sqrt{2gl(1 - \cos 	heta_0)} \approx \sqrt{2gl(\frac{	heta_0^2}{2})} = 	heta_0 \sqrt{gl}$.After the elastic collision,the bob bounces back with velocity $v'$ and reaches angle $	heta_1$,so $v' = 	heta_1 \sqrt{gl}$.In an elastic collision with a stationary block of mass $M$,the velocity of the bob $m$ after collision is $v' = \left( \frac{m - M}{m + M} ight)v$.Since the bob bounces back,its velocity is negative relative to the initial direction,so $v' = -	heta_1 \sqrt{gl}$.Thus,$-	heta_1 \sqrt{gl} = \left( \frac{m - M}{m + M} ight) 	heta_0 \sqrt{gl}$.$-	heta_1 = \frac{m - M}{m + M} 	heta_0 \implies -	heta_1(m + M) = 	heta_0(m - M)$.$-m	heta_1 - M	heta_1 = m	heta_0 - M	heta_0$.$M(	heta_0 - 	heta_1) = m(	heta_0 + 	heta_1)$.$M = m\left( \frac{	heta_0 + 	heta_1}{	heta_0 - 	heta_1} ight)$.

$A$ 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

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