Question

A body 'A' of mass $ m_1 $ moving with a velocity of 6 m/s collides with another body 'B' of mass $ m_2 $ which is at rest. After collision if 'A' moves with 2 m/s and 'B' moves with 3 m/s in the initial direction of A, then the ratio of $ \frac{m_1}{m_2} $ is

• A. \( \frac{4}{3} \)
• B. \( \frac{3}{4} \)
• C. \( \frac{2}{3} \)
• D. \( \frac{3}{2} \)

Hint:

In collision problems, always apply the conservation of momentum to find unknown masses or velocities.

Answer 1

The Correct Option is A

Since momentum is conserved, we use the principle of conservation of momentum: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] where \( u_1 = 6 \, \text{m/s} \), \( u_2 = 0 \, \text{m/s} \), \( v_1 = 2 \, \text{m/s} \), and \( v_2 = 3 \, \text{m/s} \).
Substituting these values into the equation: \[ m_1 \times 6 + m_2 \times 0 = m_1 \times 2 + m_2 \times 3 \] \[ 6m_1 = 2m_1 + 3m_2 \] \[ 4m_1 = 3m_2 \]
Thus, the ratio \( \frac{m_1}{m_2} = \frac{4}{3} \).