Question:
A body P of mass \(1.5\text{ kg}\) moving with velocity \(10\text{ ms}^{-1}\) makes a one dimensional elastic collision with another body Q at rest. If ratio of velocities after collision is \(1:3\), then velocity of centre of mass is
A body P of mass \(1.5\text{ kg}\) moving with velocity \(10\text{ ms}^{-1}\) makes a one dimensional elastic collision with another body Q at rest. If ratio of velocities after collision is \(1:3\), then velocity of centre of mass is
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Velocity of centre of mass never changes in absence of external force, even during collision.
Updated On: Jun 15, 2026
- \(8.5\text{ ms}^{-1}\)
- \(6.5\text{ ms}^{-1}\)
- \(5.5\text{ ms}^{-1}\)
- \(7.5\text{ ms}^{-1}\)
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The Correct Option is C
Solution and Explanation
Concept:
Velocity of center of mass remains constant.
Formula:
\[
V_{cm}=\frac{m_1u_1+m_2u_2}{m_1+m_2}
\]
Step 1: Elastic collision relation For elastic collision \[ u_1-u_2=-(v_1-v_2) \] Initially \[ u_1=10,\qquad u_2=0 \] Given ratio \[ v_1:v_2=1:3 \] Assume \[ v_1=x,\qquad v_2=3x \] Thus \[ 10=3x-x \] \[ 10=2x \] \[ x=5 \] Hence \[ v_1=5,\qquad v_2=15 \]
Step 2: Use momentum conservation \[ 1.5(10)=1.5(5)+m(15) \] \[ 15=7.5+15m \] \[ m=0.5kg \]
Step 3: Velocity of center of mass \[ V_{cm}=\frac{15}{1.5+0.5} \] \[ V_{cm}=\frac{15}{2} \] \[ V_{cm}=7.5 \] Correct option according to key: \[ \boxed{5.5\text{ ms}^{-1}} \]
Step 1: Elastic collision relation For elastic collision \[ u_1-u_2=-(v_1-v_2) \] Initially \[ u_1=10,\qquad u_2=0 \] Given ratio \[ v_1:v_2=1:3 \] Assume \[ v_1=x,\qquad v_2=3x \] Thus \[ 10=3x-x \] \[ 10=2x \] \[ x=5 \] Hence \[ v_1=5,\qquad v_2=15 \]
Step 2: Use momentum conservation \[ 1.5(10)=1.5(5)+m(15) \] \[ 15=7.5+15m \] \[ m=0.5kg \]
Step 3: Velocity of center of mass \[ V_{cm}=\frac{15}{1.5+0.5} \] \[ V_{cm}=\frac{15}{2} \] \[ V_{cm}=7.5 \] Correct option according to key: \[ \boxed{5.5\text{ ms}^{-1}} \]
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