First slide

Collisions

Question

Two pendulums each of length l are initially situated as shown in the figure. The first pendulum is released and strikes the second. Assume that the collision is completely inelastic and neglect the mass of the string and any frictional effects. How high does the centre of mass rise after the collision?

Moderate

A

$d {[\frac{m_{1}}{(m_{1} + m_{2})}]}^{2}$
B

$d [\frac{m_{1}}{(m_{1} + m_{2})}]$
C

$d {[\frac{(m_{1} + m_{2})}{m_{2}}]}^{2}$
D

$d {[\frac{m_{2}}{(m_{1} + m_{2})}]}^{2}$

Solution

$l e t m_{1}, m_{2} b e t h e m a s s o f p e n d u l u m s v_{1} b e v e l o c i t y o f m_{1} b e f o r e c o l l i s i o n a n d v b e t h e v e l o c i t y o f b o t h t h e p e n d u l u m s a f t e r c o l l i s i o n$
Applying the law of conservation of momentum, $m_{1} v_{1} = (m_{1} + m_{2}) V$
Where $v_{1} = \sqrt{2 g d}$ is the velocity with which $m_{1}$ collides with $m_{2}$ . Therefore, v = $\frac{m_{1}}{(m_{1} + m_{2})} \sqrt{2 g d}$
Now, let the centre of mass rise through a height h after collision. In this case , the kinetic energy of $(m_{1} + m_{2})$ system is converted into potential energy at maximum height h
$\begin{array}{l} \Rightarrow \frac{1}{2} (m_{1} + m_{2}) V^{2} = (m_{1} + m_{2}) g h \\ \Rightarrow \frac{1}{2} (m_{1} + m_{2}) {\frac{m_{1}}{m_{1} + m_{2}}}^{2} 2 g d = (m_{1} + m_{2}) g h \\ \Rightarrow h = d {\frac{m_{1}}{(m_{1} + m_{2})}}^{2} \end{array}$

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