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 \[v(t) = u_1 + \frac{t}{∆ t}( v_1 - u_1 )\]

Find the speed of the second particle as a function of time during the collision.

#### Solution

It is given that:

Speed of the first particle during collision, \[v(t) = u_1 + \frac{t}{∆ t}( v_1 - u_1 )\]v

Let v' be the speed of the second particle, during collision.

On applying the law of conservation of linear momentum on both particles, we get:

m_{1}u_{1} + m_{2}u_{2} = m_{1}v(t) + m_{2}v'

\[\Rightarrow m_1 u_1 + m_2 u_2 = m_1 u_1 + m_1 \times \left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + m_2 v'\]

\[\text{ On dividing both the sides by m} _2 , \text{ we get: }\]

\[ u_2 = \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + v'\]

\[\Rightarrow v' = u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\]

The speed of the second particle during collision can be written as a function of time and is given by the expression,\[u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\] .