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Standard XII

Physics

Perfectly Elastic Collision

Two pendulum ...

Question

Two pendulum bobs of mass $m$ and $2 m$ collide elastically at the lowest point in their motion. If both the balls are released from a height $H$ above the lowest point, to what heights do they rise for the time after collision.

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Solution

By energy conservation,
$\frac{1}{2} m_{1} v_{1}^{2} = m_{1} g H$
$v_{1} = \sqrt{2 g H}$
Similarly, $v_{2} = \sqrt{2 g H}$
After collision velocity will be $u_{1}$ and $u_{2}$
After momentum conservation,
$m_{1} v_{1} = m_{2} u_{2}$
$m \sqrt{2 g H} = 2 m u_{2}$
$u_{2} = \sqrt{\frac{g H}{2}}$
Similarly, $u_{1} = \sqrt{8 g H}$
Therefore ,by energy conservation
$m_{1}$ goes height $H_{1}$
$\frac{1}{2} m_{1} u_{1}^{2} = m_{1} g H_{1}$
$\frac{1}{2} \times 8 g H = g H_{1}$
$H_{1} = 4 H$
Similarly, $H_{2} = \frac{H}{2}$

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Two pendulum bobs of mass $m$ and $2 m$ collide elastically at the lowest point in their motion. If both the balls are released from a height $H$ above the lowest point, to what heights do they rise for the time after collision.