Question

A body of mass 'm' moving with a velocity of 'v' collides head on with another body of mass '2m' at rest. If the coefficient of restitution between the two bodies is 'e', then the ratio of the velocities of the two bodies after collision is

• A. \(\frac{1+e}{1-2e}\)
• B. \(\frac{1+2e}{1-e}\)
• C. \(\frac{1-e}{1+2e}\)
• D. \(\frac{1-2e}{1+e}\)

Hint:

For any 1D collision problem, the process is standard: write the momentum conservation equation and the coefficient of restitution equation. This always gives you two equations to solve for the two unknown final velocities. Be careful with signs.

Answer 1

The Correct Option is D

Step 1: Set up the problem: Mass 1: \( m_1 = m \), Initial velocity \( u_1 = v \). Mass 2: \( m_2 = 2m \), Initial velocity \( u_2 = 0 \). Let final velocities be \( v_1 \) and \( v_2 \) respectively. Coefficient of restitution is \( e \).
Step 2: Apply Conservation of Linear Momentum: \[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \] \[ m(v) + 2m(0) = m(v_1) + 2m(v_2) \] Dividing by \( m \): \[ v = v_1 + 2v_2 \quad \dots(1) \]
Step 3: Apply Coefficient of Restitution Formula: \[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} = \frac{v_2 - v_1}{u_1 - u_2} \] \[ e = \frac{v_2 - v_1}{v - 0} \] \[ ev = v_2 - v_1 \implies v_1 = v_2 - ev \quad \dots(2) \]
Step 4: Solve for \( v_1 \) and \( v_2 \): Substitute eq(2) into eq(1): \[ v = (v_2 - ev) + 2v_2 \] \[ v + ev = 3v_2 \] \[ v_2 = \frac{v(1+e)}{3} \] Now find \( v_1 \) using eq(2): \[ v_1 = \frac{v(1+e)}{3} - ev \] \[ v_1 = \frac{v + ve - 3ve}{3} \] \[ v_1 = \frac{v(1 - 2e)}{3} \]
Step 5: Find the ratio \( v_1 : v_2 \): \[ \frac{v_1}{v_2} = \frac{ \frac{v(1-2e)}{3} }{ \frac{v(1+e)}{3} } \] \[ \frac{v_1}{v_2} = \frac{1-2e}{1+e} \]