A ball moving with velocity $2 ext{ m s}^{-1}$ collides head on with another stationary ball of double the ma

Question

A ball moving with velocity $2	ext{ m s}^{-1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities (in $	ext{m s}^{-1}$) after collision will be:

Accepted Answer

1. **Identify Given Values:**
   - Ball 1: Mass $m_1 = m$, Initial velocity $u_1 = 2	ext{ m s}^{-1}$.
   - Ball 2: Mass $m_2 = 2m$, Initial velocity $u_2 = 0	ext{ m s}^{-1}$ (stationary).
   - Coefficient of restitution: $e = 0.5$.
2. **Apply Conservation of Linear Momentum:** Total momentum before collision equals total momentum after collision [Class 11 Physics, Ch 6, Sec 5.11.1].
   $$ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 $$
   $$ m(2) + 2m(0) = m v_1 + 2m v_2 $$
   $$ 2m = m v_1 + 2m v_2 $$
   $$ v_1 + 2v_2 = 2 \quad 	ext{--- (i)} $$
3. **Apply Coefficient of Restitution Formula:** The coefficient of restitution ($e$) relates the relative velocity of separation to the relative velocity of approach.
   $$ e = \frac{v_2 - v_1}{u_1 - u_2} $$
   $$ 0.5 = \frac{v_2 - v_1}{2 - 0} $$
   $$ 1 = v_2 - v_1 \implies v_1 = v_2 - 1 \quad 	ext{--- (ii)} $$
4. **Solve for Final Velocities:**
   Substitute (ii) into (i):
   $$ (v_2 - 1) + 2v_2 = 2 $$
   $$ 3v_2 = 3 \implies v_2 = 1	ext{ m s}^{-1} $$
   Substitute $v_2$ back into (ii):
   $$ v_1 = 1 - 1 = 0	ext{ m s}^{-1} $$
   Thus, the velocities after collision are $0	ext{ m s}^{-1}$ and $1	ext{ m s}^{-1}$.

Step-by-Step Solution

Practice All PHYSICS Questions