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Momentum, K.E. Relation

Consider a he...

Question

Consider a head-on collision between two particles of masses $m_{1}$ and $m_{2}$ .The initial speeds of the particles are $u_{1}$ and $u_{2}$ in the same direction.The collision starts at t=0 and the particles interact for a time interval $△ t$ .During the collision,the speed of the first particle varies as

$v (t) = u_{1} + \frac{t}{△ t} (v_{1} - u_{1}) .$

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Solution

Using law of consetrvation of momentum $m_{1} u_{1} + m_{2} u_{2} = m_{1} v (t) + m_{2} v^{1}$

where v=speed of 2nd particle during collision.

$\Rightarrow m_{1} u_{1} + m_{2} u_{2}$

$= m_{1} u_{1} + m_{1} . (\frac{t}{△ t}) (v_{1} - u_{1}) + m_{2} v^{1}$

$\Rightarrow \frac{m_{2} u_{2}}{m_{2}} = \frac{m_{1}}{m_{2}} (\frac{t}{△ t}) (v_{1} - u_{1}) + v^{1}$

$∴ v^{1} = u_{2} - \frac{m_{1}}{m_{2}} \frac{t}{△ t} (v_{1} - u_{1})$

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