Question

A body of mass 2 kg moving with a speed of 4 m/s makes an elastic collision with another body at rest and continues to move in the original direction but with one fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10}$ m/s. Then the value of $x$ is ________.

Hint:

Since external force is zero, \(V_{cm}\) before collision is equal to \(V_{cm}\) after collision. Calculating it before collision is usually faster.

Answer 1

Correct Answer: 25

Step 1: Understanding the Concept:
In an elastic collision, both momentum and kinetic energy are conserved. The velocity of the center of mass remains constant because there are no external forces.
Step 2: Key Formula or Approach:
1. Final velocity of first mass in elastic collision: \(v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1 + \frac{2m_2}{m_1 + m_2} u_2\).
2. Velocity of center of mass: \(V_{cm} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}\).
Step 3: Detailed Explanation:
Given: \(m_1 = 2 \text{ kg}\), \(u_1 = 4 \text{ m/s}\), \(u_2 = 0 \text{ m/s}\), \(v_1 = \frac{u_1}{4} = 1 \text{ m/s}\).
Using the velocity equation for \(v_1\):
\[ 1 = \frac{2 - m_2}{2 + m_2} (4) \implies 2 + m_2 = 8 - 4m_2 \]
\[ 5m_2 = 6 \implies m_2 = 1.2 \text{ kg} \]
Now, calculate the velocity of the center of mass:
\[ V_{cm} = \frac{(2 \times 4) + (1.2 \times 0)}{2 + 1.2} = \frac{8}{3.2} = 2.5 \text{ m/s} \]
According to the question:
\[ V_{cm} = \frac{x}{10} = 2.5 \implies x = 25 \]
Step 4: Final Answer:
The value of \(x\) is 25.