A small bucket of mass `M` is attached to a long inextensible cord of length `L`. The bucket is released from rest when the cord is in a horizontal position. In its lowest position the bucket scoops up m of water, what is the height of the swing above the lowest position?

Explanation

To solve the problem step by step, we will use the principles of conservation of energy and momentum.

### Step 1: Determine the velocity of the bucket at the lowest position
When the bucket is released from the horizontal position, it falls under the influence of gravity. We can use the work-energy theorem to find the velocity of the bucket at the lowest position.

The work done by gravity when the bucket falls a distance \( L \) is equal to the change in kinetic energy. Since the bucket starts from rest, the initial kinetic energy is zero.

\[ \text{Work done by gravity} = \Delta KE = KE_{final} - KE_{initial} \]
\[ mgh = \frac{1}{2} mv^2 \]

Here, \( h = L \) (the height fallen), and \( g \) is the acceleration due to gravity. Thus, we have:

\[ mgL = \frac{1}{2} mv^2 \]

Cancelling \( m \) from both sides (assuming \( m \neq 0 \)), we get:

\[ gL = \frac{1}{2} v^2 \]

Rearranging gives:

\[ v^2 = 2gL \]

Taking the square root:

\[ v = \sqrt{2gL} \]

### Step 2: Calculate the velocity after scooping up water
When the bucket scoops up \( m \) mass of water, we need to find the new velocity \( v' \) of the bucket. By conservation of momentum:

\[ \text{Initial momentum} = \text{Final momentum} \]

Before scooping, the momentum is:

\[ Mv \]

After scooping, the momentum is:

\[ (M + m)v' \]

Setting these equal gives:

\[ Mv = (M + m)v' \]

Solving for \( v' \):

\[ v' = \frac{Mv}{M + m} \]

Substituting \( v = \sqrt{2gL} \):

\[ v' = \frac{M\sqrt{2gL}}{M + m} \]

### Step 3: Determine the height \( h \) above the lowest position
Now, we will apply the work-energy theorem again to find the height \( h \) the bucket rises after scooping the water. The work done by gravity as the bucket moves up is equal to the change in kinetic energy.

\[ \text{Work done by gravity} = -mg h = \Delta KE = \frac{1}{2} (M + m)v'^2 - 0 \]

Substituting for \( v' \):

\[ -mgh = \frac{1}{2}(M + m)\left(\frac{M\sqrt{2gL}}{M + m}\right)^2 \]

Simplifying the right side:

\[ -mgh = \frac{1}{2}(M + m) \cdot \frac{M^2 \cdot 2gL}{(M + m)^2} \]

Cancelling \( (M + m) \) from both sides:

\[ -mgh = \frac{M^2 gL}{(M + m)} \]

Now, rearranging for \( h \):

\[ h = \frac{M^2 L}{(M + m)} \]

### Final Answer
The height of the swing above the lowest position is:

\[ h = \frac{M^2 L}{(M + m)} \]

---