Question:
Two bodies A and B of masses 2 kg and 3 kg respectively are moving along the same straight line such that the linear momentum of body A is greater than the linear momentum of body B. The velocity of centre of mass of the system of the two bodies when they are moving in the same direction is 9 times the velocity of centre of mass when they are moving in opposite directions. The ratio of the velocities of the bodies A and B is
Two bodies A and B of masses 2 kg and 3 kg respectively are moving along the same straight line such that the linear momentum of body A is greater than the linear momentum of body B. The velocity of centre of mass of the system of the two bodies when they are moving in the same direction is 9 times the velocity of centre of mass when they are moving in opposite directions. The ratio of the velocities of the bodies A and B is
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Centre of mass velocity depends on algebraic sum of momenta.
Updated On: Jun 22, 2026
- 15 : 8
- 8 : 15
- 3 : 7
- 7 : 3 \bigskip
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The Correct Option is A
Solution and Explanation
Concept:
Velocity of centre of mass depends on vector sum of momenta:
\[
V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}
\]
Step 1: Let velocities be $v_A = x$, $v_B = y$.
Masses: $m_A = 2$, $m_B = 3$
Step 2: Same direction case.
\[ V_1 = \frac{2x + 3y}{5} \]
Step 3: Opposite direction case.
\[ V_2 = \frac{2x - 3y}{5} \]
Step 4: Given relation.
\[ \frac{2x + 3y}{2x - 3y} = 9 \] \[ 2x + 3y = 18x - 27y \] \[ 16x = 30y \] \[ \frac{x}{y} = \frac{15}{8} \] Final Answer: \[ (A)\ 15:8 \]
Step 1: Let velocities be $v_A = x$, $v_B = y$.
Masses: $m_A = 2$, $m_B = 3$
Step 2: Same direction case.
\[ V_1 = \frac{2x + 3y}{5} \]
Step 3: Opposite direction case.
\[ V_2 = \frac{2x - 3y}{5} \]
Step 4: Given relation.
\[ \frac{2x + 3y}{2x - 3y} = 9 \] \[ 2x + 3y = 18x - 27y \] \[ 16x = 30y \] \[ \frac{x}{y} = \frac{15}{8} \] Final Answer: \[ (A)\ 15:8 \]
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