A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body?

Options

A. `1.0 kg`
B. `1.5 kg`
C. `1.8 kg`
D. `1.2 kg`

Answer

To solve the problem, we will use the principles of conservation of momentum and the properties of elastic collisions. Here are the steps to find the mass of the second body.

### Step 1: Define the variables
Let:
- Mass of the first body, \( m_1 = 2 \, \text{kg} \)
- Initial velocity of the first body, \( v_0 \)
- Final velocity of the first body after the collision, \( v_1 = \frac{v_0}{4} \)
- Mass of the second body, \( m_2 = m \)
- Final velocity of the second body after the collision, \( v_2 \)

### Step 2: Apply conservation of momentum
According to the law of conservation of momentum:
\[ \text{Initial momentum} = \text{Final momentum} \]
Before the collision, the momentum is:
\[ m_1 v_0 + m_2 \cdot 0 = 2 v_0 + 0 = 2 v_0 \]
After the collision, the momentum is:
\[ m_1 v_1 + m_2 v_2 = 2 \left(\frac{v_0}{4}\right) + m v_2 = \frac{v_0}{2} + m v_2 \]
Setting the initial momentum equal to the final momentum:
\[ 2 v_0 = \frac{v_0}{2} + m v_2 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ m v_2 = 2 v_0 - \frac{v_0}{2} \]
\[ m v_2 = \frac{4 v_0}{2} - \frac{v_0}{2} = \frac{3 v_0}{2} \]

### Step 4: Use the property of elastic collisions
In an elastic collision, the relative velocity of approach equals the relative velocity of separation:
\[ v_0 = v_1 + v_2 \]
Substituting \( v_1 \):
\[ v_0 = \frac{v_0}{4} + v_2 \]
Solving for \( v_2 \):
\[ v_2 = v_0 - \frac{v_0}{4} = \frac{3 v_0}{4} \]

### Step 5: Substitute \( v_2 \) back into the momentum equation
Now substitute \( v_2 \) into the momentum equation:
\[ m \left(\frac{3 v_0}{4}\right) = \frac{3 v_0}{2} \]
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ m \left(\frac{3}{4}\right) = \frac{3}{2} \]
Now, solving for \( m \):
\[ m = \frac{3/2}{3/4} = \frac{3}{2} \cdot \frac{4}{3} = 2 \cdot 2 = 2 \, \text{kg} \]

### Step 6: Conclusion
Thus, the mass of the second body is:
\[ \boxed{1.2 \, \text{kg}} \]