A

Step 1: Understand the scenario
Let the mass of block A be represented as m, and thus the mass of block B will be 2m (since it is stated that the mass of B is twice that of A).

Step 2: Apply conservation of momentum
Before the collision, let's assume block A is moving with velocity V, and block B is at rest (velocity = 0). Therefore, the total momentum before the collision is:

$$ p_{initial} = mV + 2m(0) = mV $$
After the collision, block A comes to rest (final velocity = 0), and block B moves with a velocity (V'). Using conservation of momentum:

$$ mV = 2mV' $$
Simplifying this gives:

$$ V' = \frac{V}{2} $$

Step 3: Apply the coefficient of restitution (e)
The coefficient of restitution is defined as the relative velocity of separation to the relative velocity of approach. Mathematically, it is given as:

$$ e = \frac{V'_{B} - V'_{A}}{V_{A} - V_{B}} $$
Where:
- $V'_{A} = 0$, since A stops after collision.
- $V'_{B} = V' = \frac{V}{2}$, as derived.

The velocities before the collision are:
- $V_{A} = V$ (for A)
- $V_{B} = 0$ (for B)

Now substituting these values into the equation:

$$ e = \frac{\frac{V}{2} - 0}{V - 0} = \frac{\frac{V}{2}}{V} = \frac{1}{2} $$

Conclusion
Therefore, the coefficient of restitution is:
Answer: A: 0.5.