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 $Δ t$ . During the collision, the speed of the first particle varies as
$v (t) = u_{1} + \frac{t}{Δ t} (v_{1} - u_{1})$
Find the speed of the second particle as a function of time during the collision.

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Solution

Given mass of particle $1 = m_{1}; m a s s o f p a r t i c l e 2 = m_{2}$
Using law of conservation of momentum,
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v (+) + m_{2} V$
( $(v^{1} = S p e e d o f s e c o n d p a r t i c l e d u r i n g C o l l i s i o n)$
$V^{1} = (\frac{m_{1}}{m_{2}}) V_{1} + V_{2} - (\frac{m_{1}}{m_{2}}) V (t) V^{1} = (\frac{m_{1}}{m_{2}}) V_{1} + V_{2} - (\frac{m_{1}}{m_{2}}) . (U_{1} + (\frac{t}{Δ t}) (V_{1} - V_{1})}$
$V^{1} = U_{2} - (\frac{m_{1}}{m_{2}}) . (\frac{t}{Δ t}) (V_{1} - U_{1})$

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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