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
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
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For head-on collisions, use conservation of momentum along with coefficient of restitution. Always assign signs carefully according to the chosen positive direction.
Updated On: Jun 22, 2026
- \(37.5\;\text{m s}^{-1}\)
- \(12.5\;\text{m s}^{-1}\)
- \(75\;\text{m s}^{-1}\)
- \(25\;\text{m s}^{-1}\)
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The Correct Option is A
Solution and Explanation
Step 1: Assign directions and velocities.
Let the initial direction of ball \(A\) be positive.
Mass of ball \(A\): \[ m_1=M \] Mass of ball \(B\): \[ m_2=2M \] Initial velocity of ball \(A\): \[ u_1=150\;\text{m s}^{-1} \] Since ball \(B\) is moving in the opposite direction, \[ u_2=-v \] After collision, ball \(A\) comes to rest, so \[ v_1=0 \] Let final velocity of ball \(B\) be \(v_2\).
Step 2: Use coefficient of restitution.
Coefficient of restitution is \[ e=1 \] Using the formula, \[ e=\frac{v_2-v_1}{u_1-u_2} \] Substitute the known values: \[ 1=\frac{v_2-0}{150-(-v)} \] \[ v_2=150+v \]
Step 3: Apply conservation of linear momentum.
Initial momentum is \[ M(150)+2M(-v) \] \[ =150M-2Mv \] Final momentum is \[ M(0)+2M(v_2) \] \[ =2Mv_2 \] By conservation of momentum, \[ 150M-2Mv=2Mv_2 \] Dividing by \(M\), \[ 150-2v=2v_2 \] Substitute \[ v_2=150+v \] So, \[ 150-2v=2(150+v) \] \[ 150-2v=300+2v \] \[ -4v=150 \] \[ v=-37.5 \] Since \(v\) represents speed, we take magnitude: \[ v=37.5\;\text{m s}^{-1} \]
Step 4: Final conclusion.
Hence, the speed of ball \(B\) before collision is \[ \boxed{37.5\;\text{m s}^{-1}} \]
Let the initial direction of ball \(A\) be positive.
Mass of ball \(A\): \[ m_1=M \] Mass of ball \(B\): \[ m_2=2M \] Initial velocity of ball \(A\): \[ u_1=150\;\text{m s}^{-1} \] Since ball \(B\) is moving in the opposite direction, \[ u_2=-v \] After collision, ball \(A\) comes to rest, so \[ v_1=0 \] Let final velocity of ball \(B\) be \(v_2\).
Step 2: Use coefficient of restitution.
Coefficient of restitution is \[ e=1 \] Using the formula, \[ e=\frac{v_2-v_1}{u_1-u_2} \] Substitute the known values: \[ 1=\frac{v_2-0}{150-(-v)} \] \[ v_2=150+v \]
Step 3: Apply conservation of linear momentum.
Initial momentum is \[ M(150)+2M(-v) \] \[ =150M-2Mv \] Final momentum is \[ M(0)+2M(v_2) \] \[ =2Mv_2 \] By conservation of momentum, \[ 150M-2Mv=2Mv_2 \] Dividing by \(M\), \[ 150-2v=2v_2 \] Substitute \[ v_2=150+v \] So, \[ 150-2v=2(150+v) \] \[ 150-2v=300+2v \] \[ -4v=150 \] \[ v=-37.5 \] Since \(v\) represents speed, we take magnitude: \[ v=37.5\;\text{m s}^{-1} \]
Step 4: Final conclusion.
Hence, the speed of ball \(B\) before collision is \[ \boxed{37.5\;\text{m s}^{-1}} \]
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