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$(A)$ In a head-on elastic collision,the relative velocity of approach is equal to the relative velocity of separation.Let the masses of the two parti

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$6 \,m/s$ in the same direction.

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$(A)$ In a head-on elastic collision,the relative velocity of approach is equal to the relative velocity of separation.Let the masses of the two particles be $m_1$ and $m_2$.Initial velocities are $u_1 = 5 \,m/s$ and $u_2 = 3 \,m/s$.Final velocities are $v_1 = 4 \,m/s$ and $v_2 = ?$.The relative velocity of approach is $u_1 - u_2 = 5 - 3 = 2 \,m/s$.The relative velocity of separation is $v_2 - v_1 = v_2 - 4$.Since the collision is elastic,$u_1 - u_2 = v_2 - v_1$.Substituting the values: $2 = v_2 - 4$.Therefore,$v_2 = 2 + 4 = 6 \,m/s$.Since the result is positive,the second particle moves in the same direction.

There is a head-on elastic collision between two particles moving in the same direction with speeds $5 \,m/s$ and $3 \,m/s$ respectively. After the collision,the velocity of the first particle becomes $4 \,m/s$ in the same direction. What is the velocity of the second particle?

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