A body of mass $10 g m$ moving with a velocity of $u_{1} c m / s$ collides with a stationary mass of $90 g m$ . The collision is perfectly elastic. Find the percentage loss of kinetic energy of the first body.

36

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48

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64

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0

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Solution

The correct option is A $36$
From conservation of Momentum principle,

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$

$∴ 10 u_{1} + 90 (0) = 10 v_{1} + 90 v_{2}$ (1)

Also,
The collision is perfectly elastic,

$∴ u_{1} - 0 = v_{2} - v_{1}$

$∴ u_{1} = v_{2} - v_{1}$ (2)

from (1) & (2)

$10 (v_{2} - v_{1}) = 10 v_{1} + 90 v_{2}$

$10 v - 2 - 10 v_{1} = 10 v_{1} + 90 v_{2}$

$∴ v_{1} = - 4 v_{2}$

Percentage loss of K.E. of first body,

$\frac{\frac{1}{2} m_{1} u_{1}^{2} - \frac{1}{2} m_{1} v_{1}^{2}}{\frac{1}{2} m_{1} u_{2}}$

$= \frac{u_{1}^{2} - v_{1}^{2}}{u_{1}^{2}}$

$= \frac{(v_{2} - v_{1})^{2} - v_{1}^{2}}{(v_{2} - v_{1})^{2}}$

$= \frac{v_{2}^{2} - 2 v_{1} v_{2} + / v_{1}^{2} - / v_{1}^{2}}{(v_{2} - v_{1})^{2}}$

$= \frac{v_{2}^{2} + 8 v_{2}^{2}}{(5 v_{2})^{2}} = \frac{9}{25} = 36$ %

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