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, we will use the principles of conservation of momentum and the fact that the collision is elastic.

### Step 1: Define the variables
- Let the mass of the first body (m1) = 3 kg
- Let the initial velocity of the first body (u1) = v (unknown)
- Let the mass of the second body (m2) = m (unknown)
- Let the initial velocity of the second body (u2) = 0 (since it is at rest)
- After the collision, the velocity of the first body (v1) = v/2

### Step 2: Apply the conservation of momentum
According to the conservation of momentum:
\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Substituting the known values:
\[ 3v + 0 = 3 \left(\frac{v}{2}\right) + mv_2 \]

This simplifies to:
\[ 3v = \frac{3v}{2} + mv_2 \]

### Step 3: Rearranging the equation
Now, we can rearrange the equation to isolate \( mv_2 \):
\[ mv_2 = 3v - \frac{3v}{2} \]
\[ mv_2 = \frac{6v}{2} - \frac{3v}{2} \]
\[ mv_2 = \frac{3v}{2} \]

### Step 4: Use the elastic collision formula
In an elastic collision, the relative speeds of approach and separation are equal. Thus:
\[ u_1 - u_2 = v_2 - v_1 \]

Substituting the known values:
\[ v - 0 = v_2 - \frac{v}{2} \]
This simplifies to:
\[ v = v_2 - \frac{v}{2} \]
\[ v + \frac{v}{2} = v_2 \]
\[ v_2 = \frac{3v}{2} \]

### Step 5: Substitute \( v_2 \) back into the momentum equation
Now we can substitute \( v_2 \) back into the momentum equation:
\[ m \left(\frac{3v}{2}\right) = \frac{3v}{2} \]

### Step 6: Solve for mass \( m \)
From the equation:
\[ m \cdot \frac{3v}{2} = \frac{3v}{2} \]
Dividing both sides by \( \frac{3v}{2} \) (assuming \( v \neq 0 \)):
\[ m = 1 \text{ kg} \]

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

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