An elastic ball of mass m falls from a height h on an Aluminium disc of area A floating in a mercury pool. If the collision is perfectly elastic, the momentum transferred to the disc is :

Options

A. `sqrt(2mgh)`
B. `2sqrt(mgh)`
C. `m sqrt(2gh)`
D. `2m sqrt(2gh)`

Answer

To solve the problem of finding the momentum transferred to the disc when an elastic ball of mass \( m \) falls from a height \( h \) and collides with it, we will follow these steps:

### Step 1: Calculate the velocity of the ball just before the collision
Using the third equation of motion, we can find the velocity of the ball just before it hits the disc. The equation is given by:

\[ v^2 = u^2 + 2gh \]

where:
- \( v \) is the final velocity just before impact,
- \( u \) is the initial velocity (which is 0, since it starts from rest),
- \( g \) is the acceleration due to gravity,
- \( h \) is the height from which the ball falls.

Since \( u = 0 \), the equation simplifies to:

\[ v^2 = 2gh \implies v = \sqrt{2gh} \]

### Step 2: Determine the velocity after the collision
Since the collision is perfectly elastic, the velocity of the ball after the collision (\( v_0 \)) can be determined using the principle of conservation of momentum and the fact that the relative velocity of separation is equal to the relative velocity of approach. For a perfectly elastic collision, we have:

\[ \frac{v_0 - 0}{0 - v} = 1 \]

This implies:

\[ v_0 = -v \]

Substituting \( v \):

\[ v_0 = -\sqrt{2gh} \]

### Step 3: Calculate the change in momentum of the ball
The change in momentum (\( \Delta p \)) of the ball can be calculated as follows:

\[ \Delta p = p_{\text{final}} - p_{\text{initial}} = mv_0 - mv \]

Substituting the values we found:

\[ \Delta p = m(-\sqrt{2gh}) - m(\sqrt{2gh}) = -m\sqrt{2gh} - m\sqrt{2gh} = -2m\sqrt{2gh} \]

### Step 4: Determine the momentum transferred to the disc
The momentum transferred to the disc is equal in magnitude but opposite in direction to the change in momentum of the ball. Therefore, the momentum transferred to the disc is:

\[ \text{Momentum transferred to the disc} = 2m\sqrt{2gh} \]

### Final Answer
Thus, the momentum transferred to the disc is:

\[ \boxed{2m\sqrt{2gh}} \]

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