A body of mass `3 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. `1kg`
B. `1.5kg`
C. `2kg`
D. `5kg`

Explanation

To solve the problem of a body of mass \(3 \, \text{kg}\) colliding elastically with another body at rest and then moving with half its original speed, we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let the mass of the first body \(m_1 = 3 \, \text{kg}\).
- Let the initial velocity of the first body \(u_1 = v\).
- The second body (target body) has mass \(m_2 = M\) and is initially at rest, so \(u_2 = 0\).
- After the collision, the first body moves with half its original speed, so its final velocity \(v_1 = \frac{v}{2}\).
- Let the final velocity of the second body be \(v_2\).

2. **Apply Conservation of Momentum:**
The total momentum before the collision must equal the total momentum after the collision:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
Substituting the known values:
\[ 3v + 0 = 3 \left(\frac{v}{2}\right) + M v_2 \]
Simplifying this gives:
\[ 3v = \frac{3v}{2} + M v_2 \]

3. **Rearranging the Equation:**
Rearranging the equation to isolate \(M v_2\):
\[ 3v - \frac{3v}{2} = M v_2 \]
Converting \(3v\) to a common denominator:
\[ \frac{6v}{2} - \frac{3v}{2} = M v_2 \]
This simplifies to:
\[ \frac{3v}{2} = M v_2 \]
Thus, we have:
\[ M = \frac{3v}{2v_2} \]

4. **Apply Conservation of Kinetic Energy:**
For an elastic collision, the kinetic energy before and after the collision is conserved:
\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]
Substituting the known values:
\[ \frac{1}{2} (3) v^2 + 0 = \frac{1}{2} (3) \left(\frac{v}{2}\right)^2 + \frac{1}{2} M v_2^2 \]
This simplifies to:
\[ \frac{3v^2}{2} = \frac{3v^2}{8} + \frac{1}{2} M v_2^2 \]

5. **Rearranging the Kinetic Energy Equation:**
Rearranging gives:
\[ \frac{3v^2}{2} - \frac{3v^2}{8} = \frac{1}{2} M v_2^2 \]
Converting to a common denominator:
\[ \frac{12v^2}{8} - \frac{3v^2}{8} = \frac{1}{2} M v_2^2 \]
This simplifies to:
\[ \frac{9v^2}{8} = \frac{1}{2} M v_2^2 \]

6. **Solving for \(M\):**
Now we have two equations:
- From momentum: \(M = \frac{3v}{2v_2}\)
- From kinetic energy: \(M = \frac{9v^2}{4v_2^2}\)

Setting these equal to each other:
\[ \frac{3v}{2v_2} = \frac{9v^2}{4v_2^2} \]
Cross-multiplying gives:
\[ 3v \cdot 4v_2^2 = 9v^2 \cdot 2v_2 \]
Simplifying leads to:
\[ 12v v_2 = 18v^2 \]
Dividing both sides by \(6v\) (assuming \(v \neq 0\)):
\[ 2v_2 = 3v \implies v_2 = \frac{3v}{2} \]

7. **Substituting Back to Find \(M\):**
Substitute \(v_2\) back into the equation for \(M\):
\[ M = \frac{3v}{2 \cdot \frac{3v}{2}} = \frac{3v}{3v} = 1 \, \text{kg} \]

### Final Answer:
The mass of the target body is \(1 \, \text{kg}\).