After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley.
Find the fraction of kinetic energy lost when the body of mass `M` is jerked into motion

Options

A. `M/(M+m)`
B. `M/(M-m)`
C. `(2M)/(M+m)`
D. `M/(2(M+m))`

Explanation

To solve the problem, we need to analyze the motion of two bodies connected by a pulley system. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the velocity of mass `m` just before the string becomes taut
When the body of mass `m` falls from rest through a height `h`, we can use the equation of motion to find its velocity just before the string becomes taut.

Using the equation:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

### Step 2: Apply conservation of momentum when the string becomes taut
When the string becomes taut, the body of mass `M` is jerked into motion. At this point, we can apply the principle of conservation of momentum. Let \( v' \) be the common speed of both masses after the string is taut.

Before the string becomes taut, the momentum is:
\[ p_{\text{before}} = mv \]
After the string becomes taut, the momentum is:
\[ p_{\text{after}} = (M + m)v' \]
Setting these equal gives us:
\[ mv = (M + m)v' \]
From this, we can solve for \( v' \):
\[ v' = \frac{mv}{M + m} \]

### Step 3: Calculate the acceleration of the system
The acceleration of the system can be determined by considering the forces acting on it. The net force acting on the system is:
\[ F_{\text{net}} = Mg - mg \]
According to Newton's second law, this net force is equal to the total mass times acceleration:
\[ Mg - mg = (M + m)a \]
Rearranging gives:
\[ a = \frac{Mg - mg}{M + m} \]

### Step 4: Find the time taken for the system to return to its original position
Let the system return to its original position at time \( t \). Using the kinematic equation:
\[ s = vt + \frac{1}{2}at^2 \]
Since the displacement \( s \) is zero when the system returns to its original position, we can set up the equation:
\[ 0 = v't - \frac{1}{2}at^2 \]
Solving for \( t \) gives:
\[ t = \frac{2v'}{a} \]

### Step 5: Calculate the fractional loss of kinetic energy
The initial kinetic energy (just before the string becomes taut) is:
\[ KE_{\text{initial}} = \frac{1}{2}mv^2 \]
The final kinetic energy (after the string becomes taut) is:
\[ KE_{\text{final}} = \frac{1}{2}(M + m)v'^2 \]
The fractional loss of kinetic energy is given by:
\[ \text{Fractional loss} = \frac{KE_{\text{initial}} - KE_{\text{final}}}{KE_{\text{initial}}} \]

Substituting the expressions for kinetic energy:
\[ \text{Fractional loss} = \frac{\frac{1}{2}mv^2 - \frac{1}{2}(M + m)v'^2}{\frac{1}{2}mv^2} \]
This simplifies to:
\[ \text{Fractional loss} = 1 - \frac{(M + m)v'^2}{mv^2} \]

### Step 6: Substitute \( v' \) and simplify
Substituting \( v' = \frac{mv}{M + m} \) into the equation gives:
\[ \text{Fractional loss} = 1 - \frac{(M + m)\left(\frac{mv}{M + m}\right)^2}{mv^2} \]
This simplifies to:
\[ \text{Fractional loss} = 1 - \frac{m}{M + m} \]
Thus, the final result for the fractional loss of kinetic energy is:
\[ \text{Fractional loss} = \frac{M}{M + m} \]

### Final Answer
The fraction of kinetic energy lost when the body of mass `M` is jerked into motion is:
\[ \frac{M}{M + m} \]