A block of mass m moving with speed v , collides head on with another block of mass 2m at rest .
If coefficient of restitution is `1/2 ` then what is velocity of first block after the impact ?

Options

A. Zero
B. v/2
C. v/4
D. v/6

Explanation

To solve the problem, we will use the principles of conservation of momentum and the definition of the coefficient of restitution.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of the first block, \( m_1 = m \)
- Mass of the second block, \( m_2 = 2m \)
- Initial velocity of the first block, \( u_1 = v \)
- Initial velocity of the second block, \( u_2 = 0 \) (at rest)
- Coefficient of restitution, \( e = \frac{1}{2} \)

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:
\[ mv + 2m \cdot 0 = mv_1 + 2m v_2 \]
This simplifies to:
\[ mv = mv_1 + 2m v_2 \quad \text{(Equation 1)} \]
Dividing through by \( m \):
\[ v = v_1 + 2v_2 \]

3. **Use the Coefficient of Restitution:**
The coefficient of restitution is defined as the ratio of the relative velocity of separation to the relative velocity of approach.
\[ e = \frac{v_2 - v_1}{u_1 - u_2} \]
Substituting the known values:
\[ \frac{1}{2} = \frac{v_2 - v_1}{v - 0} \]
Rearranging gives:
\[ v_2 - v_1 = \frac{1}{2} v \quad \text{(Equation 2)} \]

4. **Solve the Equations:**
We now have two equations:
- Equation 1: \( v = v_1 + 2v_2 \)
- Equation 2: \( v_2 - v_1 = \frac{1}{2} v \)

From Equation 2, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = v_1 + \frac{1}{2} v \]

Substitute \( v_2 \) into Equation 1:
\[ v = v_1 + 2\left(v_1 + \frac{1}{2} v\right) \]
Simplifying this:
\[ v = v_1 + 2v_1 + v = 3v_1 + v \]
Rearranging gives:
\[ 0 = 3v_1 \]
Thus, we find:
\[ v_1 = 0 \]

5. **Conclusion:**
The velocity of the first block after the impact is \( v_1 = 0 \).