Question

A 3 kg ball and 4 kg ball having speed of 7 m/s and 5 m/s, are approaching each other. The speed of 1st ball after collision for coefficient of restitution 0.75 is

• A. 7 m/s
• B. 4 m/s
• C. 3 m/s
• D. 5 m/s

Answer 1

Answer: (D)

5 m/s

Answer 2

Let the masses be $m_1 = 3$ kg and $m_2 = 4$ kg. Taking rightward as positive, assume:

$$u_1 = +7\ \text{m/s}, \quad u_2 = -5\ \text{m/s}$$

The coefficient of restitution is given by:

$$e = \frac{v_2 - v_1}{u_1 - u_2} \quad \Rightarrow \quad v_2 - v_1 = e (u_1 - u_2)$$

Calculating the relative speed of approach:

$$u_1 - u_2 = 7 - (-5) = 12\ \text{m/s}$$

Thus,

$$v_2 - v_1 = 0.75 \times 12 = 9\ \text{m/s} \quad \Rightarrow \quad v_2 = v_1 + 9$$

Using conservation of momentum:

$$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$$ $$3(7) + 4(-5) = 3v_1 + 4(v_1 + 9)$$ $$21 - 20 = 3v_1 + 4v_1 + 36$$ $$1 = 7v_1 + 36$$ $$7v_1 = 1 - 36 = -35 \quad \Rightarrow \quad v_1 = -5\ \text{m/s}$$

The negative sign indicates that the 1st ball reverses direction. The speed is the magnitude, so:

$$\text{Speed} = 5\ \text{m/s}$$