A body of mass m kg collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. What is the mass of the target body ?

Options

A. m kg
B. 2/3 m kg
C. m/3 kg
D. m/2 kg

Answer

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

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Let the mass of the first body be \( m \) and its initial velocity be \( v_0 \).
- The second body (target body) has a mass \( m' \) and is initially at rest, so its initial velocity is \( 0 \).

2. **Determine the Final Velocities:**
- After the collision, the first body moves with half of its original speed, so its final velocity is \( \frac{v_0}{2} \).
- Let the final velocity of the second body be \( v' \).

3. **Apply Conservation of Momentum:**
- The total initial momentum before the collision is:
\[ P_{\text{initial}} = mv_0 + 0 = mv_0 \]
- The total final momentum after the collision is:
\[ P_{\text{final}} = m \left(\frac{v_0}{2}\right) + m'v' \]
- Setting initial momentum equal to final momentum gives:
\[ mv_0 = m \left(\frac{v_0}{2}\right) + m'v' \]
- Rearranging this equation, we have:
\[ mv_0 - m \left(\frac{v_0}{2}\right) = m'v' \]
- Simplifying further:
\[ mv_0 - \frac{mv_0}{2} = m'v' \]
\[ \frac{mv_0}{2} = m'v' \]
- This is our first equation (1).

4. **Apply Conservation of Kinetic Energy:**
- The total initial kinetic energy is:
\[ KE_{\text{initial}} = \frac{1}{2}mv_0^2 \]
- The total final kinetic energy is:
\[ KE_{\text{final}} = \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 + \frac{1}{2}m'v'^2 \]
- Setting initial kinetic energy equal to final kinetic energy gives:
\[ \frac{1}{2}mv_0^2 = \frac{1}{2}m\left(\frac{v_0^2}{4}\right) + \frac{1}{2}m'v'^2 \]
- Simplifying this equation:
\[ mv_0^2 = m\left(\frac{v_0^2}{4}\right) + m'v'^2 \]
\[ mv_0^2 - \frac{mv_0^2}{4} = m'v'^2 \]
\[ \frac{3mv_0^2}{4} = m'v'^2 \]
- This is our second equation (2).

5. **Substituting for \( v' \):**
- From equation (1), we have:
\[ v' = \frac{mv_0}{2m'} \]
- Substitute \( v' \) in equation (2):
\[ \frac{3mv_0^2}{4} = m' \left(\frac{mv_0}{2m'}\right)^2 \]
- Simplifying gives:
\[ \frac{3mv_0^2}{4} = m' \cdot \frac{m^2v_0^2}{4m'^2} \]
- Cancel \( v_0^2 \) from both sides:
\[ \frac{3m}{4} = \frac{m^2}{4m'} \]
- Rearranging gives:
\[ 3m' = m^2 \implies m' = \frac{m^2}{3} \]

### Final Answer:
The mass of the target body \( m' \) is \( \frac{m^2}{3} \).