A sphere of mass 'm', moving with velocity '3u' collides head-on with another identical sphere at rest. If 'e' is coefficient of restitution then what will be the ratio of velocity of the second sphere to that of first sphere after collision?
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For identical masses colliding where the second is at rest, the final velocities are always proportional to $(1-e)$ for the first body and $(1+e)$ for the second body. Memorizing this saves time!
Step 1: Understanding the Concept:
For a head-on collision between two identical masses ($m_1 = m_2 = m$), we use the conservation of momentum and the definition of the coefficient of restitution ($e$). Step 2: Formula Application:
Conservation of Momentum: $m(3u) + m(0) = mv_1 + mv_2 \implies v_1 + v_2 = 3u$.
Coefficient of Restitution: $e = \frac{v_2 - v_1}{u_1 - u_2} = \frac{v_2 - v_1}{3u} \implies v_2 - v_1 = 3ue$. Step 3: Explanation:
Adding the two equations: $2v_2 = 3u(1+e) \implies v_2 = \frac{3u(1+e)}{2}$.
Subtracting the equations: $2v_1 = 3u(1-e) \implies v_1 = \frac{3u(1-e)}{2}$.
The ratio $\frac{v_2}{v_1} = \frac{1+e}{1-e}$. Step 4: Final Answer:
The ratio of the velocity of the second sphere to the first is $\frac{1+e}{1-e}$.
