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Question
Consider a head-on collision between two particles of masses m1 and m2. The initial speeds of the particles are u1 and u2 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.
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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:
m1u1 + m2u2 = m1v(t) + m2v'
\[\Rightarrow m_1 u_1 + m_2 u_2 = m_1 u_1 + m_1 \times \left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + m_2 v'\]
\[\text{ On dividing both the sides by m} _2 , \text{ we get: }\]
\[ u_2 = \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 ) + v'\]
\[\Rightarrow v' = u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\]
The speed of the second particle during collision can be written as a function of time and is given by the expression,\[u_2 - \frac{m_1}{m_2}\left( \frac{t}{∆ t} \right)( v_1 - u_1 )\] .
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