A body of mass 1 kg strikes elastically with another body at rest and continues to move in the same direction with one fourth of its initial velocity. The mass of the other bodyis-

Options

A. `0.6 kg`
B. `2.4 kg`
C. `3 kg`
D. `4 kg`

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and conservation of kinetic energy, as the collision is elastic.

### Step 1: Define the initial conditions
Let:
- Mass of the first body, \( m_1 = 1 \, \text{kg} \)
- Initial velocity of the first body, \( u \)
- Mass of the second body, \( m_2 = m \) (unknown)
- Initial velocity of the second body, \( u_2 = 0 \) (at rest)

### Step 2: Apply conservation of linear momentum
According to the conservation of momentum:
\[ m_1 u + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 1 \cdot u + m \cdot 0 = 1 \cdot \left(\frac{u}{4}\right) + m \cdot v_2 \]
This simplifies to:
\[ u = \frac{u}{4} + m v_2 \]
Rearranging gives:
\[ m v_2 = u - \frac{u}{4} = \frac{3u}{4} \tag{1} \]

### Step 3: Apply conservation of kinetic energy
For elastic collisions, kinetic energy is also conserved:
\[ \frac{1}{2} m_1 u^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the known values:
\[ \frac{1}{2} \cdot 1 \cdot u^2 + \frac{1}{2} m \cdot 0^2 = \frac{1}{2} \cdot 1 \cdot \left(\frac{u}{4}\right)^2 + \frac{1}{2} m v_2^2 \]
This simplifies to:
\[ \frac{1}{2} u^2 = \frac{1}{2} \cdot \frac{u^2}{16} + \frac{1}{2} m v_2^2 \]
Multiplying through by 2 to eliminate the fractions:
\[ u^2 = \frac{u^2}{16} + m v_2^2 \]
Rearranging gives:
\[ m v_2^2 = u^2 - \frac{u^2}{16} = \frac{16u^2 - u^2}{16} = \frac{15u^2}{16} \tag{2} \]

### Step 4: Relate equations (1) and (2)
From equation (1), we have:
\[ v_2 = \frac{3u}{4m} \]
Substituting \( v_2 \) into equation (2):
\[ m \left(\frac{3u}{4m}\right)^2 = \frac{15u^2}{16} \]
This simplifies to:
\[ m \cdot \frac{9u^2}{16m^2} = \frac{15u^2}{16} \]
Cancelling \( u^2 \) from both sides (assuming \( u \neq 0 \)):
\[ \frac{9}{16m} = \frac{15}{16} \]
Cross-multiplying gives:
\[ 9 = 15m \]
Thus:
\[ m = \frac{9}{15} = \frac{3}{5} \, \text{kg} = 0.6 \, \text{kg} \]

### Final Answer
The mass of the other body is \( 0.6 \, \text{kg} \).

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