Question

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?

• A. $\frac{1-e}{1+e}$
• B. $\frac{1+e}{1-e}$
• C. $\frac{e+1}{e-1}$
• D. $\frac{e-1}{e+1}$

Hint:

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!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This is a 1D elastic/inelastic collision problem between two identical masses. We need to use conservation of momentum and the coefficient of restitution to find the final velocities, then take their ratio.

Step 2: Detailed Explanation:
Let the mass of both spheres be $m$.
Initial velocity of first sphere ($u_1$) = $3u$
Initial velocity of second sphere ($u_2$) = $0$
Let their final velocities after collision be $v_1$ and $v_2$.
1. Apply Conservation of Linear Momentum:
$m u_1 + m u_2 = m v_1 + m v_2$
Since the masses are identical, $m$ cancels out:
$u_1 + u_2 = v_1 + v_2$
$3u + 0 = v_1 + v_2$
$v_1 + v_2 = 3u$ --- (Equation 1)
2. Apply the Coefficient of Restitution ($e$):
$e = \frac{\text{Velocity of Separation}}{\text{Velocity of Approach}}$
$e = \frac{v_2 - v_1}{u_1 - u_2}$
$e = \frac{v_2 - v_1}{3u - 0}$
$v_2 - v_1 = 3eu$ --- (Equation 2)
3. Solve for $v_1$ and $v_2$:
Add Equation 1 and Equation 2:
$(v_1 + v_2) + (v_2 - v_1) = 3u + 3eu$
$2v_2 = 3u(1 + e)$
$v_2 = \frac{3u(1 + e)}{2}$
Subtract Equation 2 from Equation 1:
$(v_1 + v_2) - (v_2 - v_1) = 3u - 3eu$
$2v_1 = 3u(1 - e)$
$v_1 = \frac{3u(1 - e)}{2}$
4. Find the ratio of $v_2$ to $v_1$:
$\text{Ratio} = \frac{v_2}{v_1} = \frac{ \frac{3u(1 + e)}{2} }{ \frac{3u(1 - e)}{2} }$
The terms $\frac{3u}{2}$ perfectly cancel out:
$\text{Ratio} = \frac{1 + e}{1 - e}$

Step 3: Final Answer:
The ratio is $\frac{1+e}{1-e}$, matching option (b).