Question

An object of mass \(m_1\) collides with another object of mass \(m_2\), which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses \(m_2 : m_1\) is:

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 3 : 1

Hint:

In elastic collisions, the velocity of approach equals the velocity of separation.

Answer 1

The Correct Option is D

Step 1: Let initial velocity of \(m_1\) be \(u\) and \(m_2\) be 0. After collision, let their speeds be \(v\). Since they move in opposite directions, velocities are \(v\) and \(-v\).
Step 2: Apply Conservation of Linear Momentum: \[m_1 u + m_2(0) = m_2 v + m_1(-v) \implies m_1 u = (m_2 - m_1)v \quad .......(1)\]
Step 3: Assuming an elastic collision, \(e = 1\): \[e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} \implies 1 = \frac{v - (-v)}{u} \implies u = 2v\]
Step 4: Substitute \(u = 2v\) into (1): \[m_1(2v) = (m_2 - m_1)v \implies 2m_1 = m_2 - m_1 \implies 3m_1 = m_2 \implies \frac{m_2}{m_1} = 3\]