Question

There is head-on elastic collision between the two particles moving in the same direction with speeds \(5\text{ m/s}\) and \(3\text{ m/s}\) respectively. After collision, the velocity of the first particle becomes \(4\text{ m/s}\) in the same direction. The velocity of the second particle should be

• A. \(6\text{ m/s}\) in the same direction.
• B. \(4\text{ m/s}\) in the same direction.
• C. \(2\text{ m/s}\) in the opposite direction.
• D. \(3\text{ m/s}\) in the same direction.

Hint:

Relative velocity reverses in elastic collision.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
In an elastic collision, both linear momentum and total kinetic energy are conserved.
A more direct property is that the coefficient of restitution \(e = 1\), meaning the relative velocity of separation equals the relative velocity of approach.
Step 2: Key Formula or Approach:
For head-on elastic collision: \(v_2 - v_1 = e(u_1 - u_2)\)
where \(e = 1\).
Step 3: Detailed Explanation:
Given initial velocities: \(u_1 = 5\text{ m/s}\), \(u_2 = 3\text{ m/s}\) (same direction).
Relative velocity of approach = \(u_1 - u_2 = 5 - 3 = 2\text{ m/s}\).
Given final velocity of the first particle: \(v_1 = 4\text{ m/s}\) (same direction).
Let the final velocity of the second particle be \(v_2\).
Apply the property of elastic collision: \[ v_2 - v_1 = 1 \cdot (u_1 - u_2) \] \[ v_2 - 4 = 2 \] \[ v_2 = 6\text{ m/s} \] Since \(v_2>v_1\), the second particle is moving in the same direction as the first to separate from it.
Step 4: Final Answer:
The velocity of the second particle is \(6\text{ m/s}\) in the same direction.