>Mass of the moving block: $$\mathrm{m}$$

Velocity of the moving block: $$\mathrm{V}$$

Mass of the block at rest: $$4 \mathrm{~m}$$

Velocity of the block at rest: $$0$$

After the collision:

The block of mass $$\mathrm{m}$$ comes to rest, so its final velocity is $$0$$.
Let $$V_2$$ be the final velocity of the block with mass $$4 \mathrm{~m}$$ after the collision.

By the law of conservation of linear momentum, the total momentum before and after the collision must be equal:

$$ \mathrm{m} \cdot \mathrm{V} + 4 \mathrm{~m} \cdot 0 = \mathrm{m} \cdot 0 + 4 \mathrm{~m} \cdot V_2 $$

This simplifies to:

$$ \mathrm{V} = 4 V_2 \implies V_2 = \frac{\mathrm{V}}{4} $$

Next, we use the coefficient of restitution (e), which is defined as the relative speed of separation divided by the relative speed of approach. Mathematically, for a one-dimensional collision:

$$ e = \frac{\text{Relative speed after collision}}{\text{Relative speed before collision}} $$

Before the collision, the relative speed of approach is:

$$ \mathrm{V} - 0 = \mathrm{V} $$

After the collision, the relative speed of separation is:

$$ 0 - \left( \frac{\mathrm{V}}{4} \right) = - \frac{\mathrm{V}}{4} $$

Taking magnitudes (since speeds are always positive):

$$ e = \frac{\left| - \frac{\mathrm{V}}{4} \right|}{\mathrm{V}} = \frac{\frac{\mathrm{V}}{4}}{\mathrm{V}} = \frac{1}{4} = 0.25 $$

Thus, the coefficient of restitution for this collision is:

Answer: Option B (0.25)

Center Of Mass

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