Question

A ball moving with velocity $2\, m/s$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is $0.5$, then their velocities $(in\, m/s)$ after collision will be

• A. 0, 1
• B. 44562
• C. 1, 0.5
• D. 0, 2

Answer 1

The Correct Option is A

 To solve this problem, we need to understand the concepts of the collision, specifically using the coefficient of restitution and the conservation of momentum.

We have two balls:

1. Ball A with mass \(m\) moving with a velocity of \(2 \, \text{m/s}\).
2. Ball B with mass \(2m\) initially at rest.

 

Given:

Initial velocity of Ball A, \(u_1 = 2 \, \text{m/s}\)
Initial velocity of Ball B, \(u_2 = 0 \, \text{m/s}\)
Coefficient of restitution, \(e = 0.5\)

 

The coefficient of restitution formula is given by: \(e = \frac{v_2 - v_1}{u_1 - u_2}\)where:

• \(v_1, v_2\) are the final velocities of balls A and B respectively.

Substituting the known values in the equation: \(0.5 = \frac{v_2 - v_1}{2 - 0} = \frac{v_2 - v_1}{2}\)

Solving for \(v_2 - v_1\), we get: \(v_2 - v_1 = 1 \quad \Rightarrow \quad v_2 = v_1 + 1\)

Next, we use the principle of conservation of momentum, which states that: \(m \cdot u_1 + 2m \cdot u_2 = m \cdot v_1 + 2m \cdot v_2\)Simplifying with given values: \(m \cdot 2 + 2m \cdot 0 = m \cdot v_1 + 2m \cdot v_2 \quad \Rightarrow \quad 2 = v_1 + 2v_2\)

Now, substituting \(v_2 = v_1 + 1\) from our previous finding into the momentum equation: \(2 = v_1 + 2(v_1 + 1) = v_1 + 2v_1 + 2 = 3v_1 + 2\)Solving for \(v_1\): \(3v_1 = 0 \quad \Rightarrow \quad v_1 = 0\)

Thus, substituting \(v_1 = 0\) back into the equation for \(v_2\): \(v_2 = v_1 + 1 = 0 + 1 = 1\)

Therefore, the velocities after the collision are:

• Velocity of Ball A, \(v_1 = 0 \, \text{m/s}\)
• Velocity of Ball B, \(v_2 = 1 \, \text{m/s}\)

The correct option is 0, 1.