Question 343: Centre of Mass and Rotational Motion

Question

A wooden black of mass '$$\mathrm{m}$$' moves with velocity '$$\mathrm{V}$$' and collides with another block of mass '$$4 \mathrm{~m}$$', which is at rest. After collision the block of mass '$$\mathrm{m}$$' comes to rest. The coefficient of restitution will be

Accepted Answer

To determine the coefficient of restitution, let's analyze the given problem using the principles of momentum conservation and the definition of the coefficient of restitution.
First, we note the initial conditions:
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:
$$ e = \frac{	ext{Relative speed after collision}}{	ext{Relative speed before collision}} $$
Before the collision, the relative speed of approach is:
$$ \mathrm{V} - 0 = \mathrm{V} $$
After the collision, the relative speed of separation is:
$$ 0 - \left( \frac{\mathrm{V}}{4} ight) = - \frac{\mathrm{V}}{4} $$
Taking magnitudes (since speeds are always positive):
$$ e = \frac{\left| - \frac{\mathrm{V}}{4} ight|}{\mathrm{V}} = \frac{\frac{\mathrm{V}}{4}}{\mathrm{V}} = \frac{1}{4} = 0.25 $$
Thus, the coefficient of restitution for this collision is:
Answer: Option B (0.25)

Answer

0.7

Answer

0.25

Answer

0.4

Answer

0.5

Centre of Mass and Rotational Motion

A wooden black of mass '$\mathrm{m}' moves with velocity '\mathrm{V}' and collides with another block of mass '4 \mathrm{~m}', which is at rest. After collision the block of mass '\mathrm{m}$' comes to rest. The coefficient of restitution will be

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