Question

Two balls \(A\) and \(B\), of masses \(M\) and \(2M\) respectively collide each other. If the ball \(A\) moves with a speed of \(150\;\text{m s}^{-1}\) and collides with ball \(B\), moving with speed \(v\) in the opposite direction. After collision if ball \(A\) comes to rest and the coefficient of restitution is \(1\), then the speed of ball \(B\) before it collides with ball \(A\) is

• A. \(37.5\;\text{m s}^{-1}\)
• B. \(12.5\;\text{m s}^{-1}\)
• C. \(75\;\text{m s}^{-1}\)
• D. \(25\;\text{m s}^{-1}\)

Hint:

For head-on collisions, use conservation of momentum along with coefficient of restitution. Always assign signs carefully according to the chosen positive direction.

Answer 1

The Correct Option is A

Step 1: Assign symbols and directions.
Let ball $A$ (mass $M$) move in the positive direction with $u_1 = 150\ \text{m s}^{-1}$. Ball $B$ (mass $2M$) moves the opposite way with speed $v$, so $u_2 = -v$. After the hit, $A$ stops: $v_1 = 0$. Restitution coefficient $e = 1$. We want $v$.
Step 2: Apply conservation of momentum.
\[ M u_1 + 2M u_2 = M v_1 + 2M v_2 \] Dividing through by $M$: \[ 150 + 2(-v) = 0 + 2v_2 \] So \[ 150 - 2v = 2v_2 \quad (\text{i}) \]
Step 3: Apply the restitution relation.
For $e = 1$, the speed of separation equals the speed of approach: \[ v_2 - v_1 = u_1 - u_2 \] \[ v_2 - 0 = 150 - (-v) = 150 + v \] So \[ v_2 = 150 + v \quad (\text{ii}) \]
Step 4: Substitute (ii) into (i).
\[ 150 - 2v = 2(150 + v) \] \[ 150 - 2v = 300 + 2v \]
Step 5: Solve for $v$.
\[ 150 - 300 = 2v + 2v \] \[ -150 = 4v \] \[ v = -37.5\ \text{m s}^{-1} \]
Step 6: Interpret the magnitude.
The negative sign simply confirms $B$ was moving opposite to $A$; its speed before collision is $37.5\ \text{m s}^{-1}$, matching option (1). \[ \boxed{37.5\ \text{m s}^{-1}} \]