A body of mass 1 kg collides head on elastically with a stationary body of mass 3 kg. After collision, the smaller body reverses its direction of motion and moves with a speed of 2m ∕ s. The initial speed of the smaller body before collision is _______ ms−1.

JEE Mains - Elastic Collision

Question: A body of mass \( m_1 = 1\,\text{kg} \) collides head on elastically with a stationary body of mass \( m_2 = 3\,\text{kg} \). After the collision, the smaller body reverses its direction and moves with a speed of \( 2\,\text{m/s} \). Find the initial speed of the smaller body before the collision.
[25-Jan-2023 Shift 2]

Select the correct answer:

(A) \(2\,\text{m/s}\)

(B) \(4\,\text{m/s}\)

(C) \(6\,\text{m/s}\)

(D) \(8\,\text{m/s}\)

Solution

Let the smaller mass be \( m_1 = 1\,\text{kg} \) and the larger mass be \( m_2 = 3\,\text{kg} \). Let the initial speed of \( m_1 \) be \( v_{1i} \) and \( m_2 \) is initially at rest (\( v_{2i} = 0 \)). After the collision, the smaller body reverses its direction with speed \( v_{1f} = -2\,\text{m/s} \) (the negative sign indicates the reversal), and let the final speed of \( m_2 \) be \( v_{2f} \).

For an elastic collision, the relative speed of separation equals the relative speed of approach: \[ v_{1i} - v_{2i} = -(v_{1f} - v_{2f}). \] Since \( v_{2i} = 0 \), this simplifies to: \[ v_{1i} = -v_{1f} + v_{2f}. \]

Also, using the standard formula for elastic collisions for the smaller mass: \[ v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i}. \] Plugging in the values: \[ v_{1f} = \frac{1 - 3}{1 + 3}\,v_{1i} = \frac{-2}{4}\,v_{1i} = -\frac{1}{2}\,v_{1i}. \] Given \( v_{1f} = -2\,\text{m/s} \): \[ -\frac{1}{2}\,v_{1i} = -2 \quad \Longrightarrow \quad v_{1i} = 4\,\text{m/s}. \]

Hence, the initial speed of the smaller body is \( 4\,\text{m/s} \).