1. Given

• Mass of first body: $m_1 = 2\,\text{kg}$
• Initial velocity of first body: $u_1 = u$
• Second body is initially at rest: $u_2 = 0$
• After elastic collision, first body continues in the same direction with one-fourth of original speed: $v_1 = \frac{u}{4}$

We need to find the mass $m_2$ of the second body.

2. Use the standard result for 1D elastic collision

For a head-on elastic collision where the second body is initially at rest, $v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1$

Substitute the given values: $\frac{u}{4} = \frac{2 - m_2}{2 + m_2} u$

Cancel $u$ from both sides: $\frac{1}{4} = \frac{2 - m_2}{2 + m_2}$

3. Solve for $m_2$

Cross-multiply: $2 - m_2 = \frac{1}{4}(2 + m_2)$

Multiply by $4$: $8 - 4m_2 = 2 + m_2$

Bring like terms together: $8 - 2 = 5m_2$ $6 = 5m_2$ $m_2 = \frac{6}{5} = 1.2\,\text{kg}$

4. Check with options

• A: $1.2\,\text{kg}$ ✅
• B: $1.0\,\text{kg}$
• C: $1.8\,\text{kg}$
• D: $1.5\,\text{kg}$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They match.