A ball of mass `m` is projected with a speed `v` into the barrel of a spring gun of mass `M` initially at rest lying on a frictionless surface. The mass sticks in the barrel at the point of maximum compression in the spring. The fraction of kinetic energy of the ball stored in the spring is

Options

A. `m/M`
B. `M/(m+M)`
C. `m/(M+M)`
D. none of these

Explanation

To solve the problem, we need to find the fraction of kinetic energy of the ball that is stored in the spring when the ball sticks in the barrel of the spring gun. Here’s a step-by-step solution:

### Step 1: Understand the system
We have a ball of mass \( m \) projected with speed \( v \) into a spring gun of mass \( M \) that is initially at rest on a frictionless surface. When the ball sticks in the barrel, we need to consider the conservation of momentum.

### Step 2: Apply conservation of momentum
Before the collision, the momentum of the system is just the momentum of the ball:
\[ p_{\text{initial}} = mv \]
After the ball sticks to the spring gun, the total mass is \( m + M \) and let \( v' \) be the speed of the combined mass after the collision. By conservation of momentum:
\[ mv = (m + M)v' \]
From this, we can solve for \( v' \):
\[ v' = \frac{mv}{m + M} \]

### Step 3: Calculate initial and final kinetic energy
The initial kinetic energy \( K_i \) of the ball is:
\[ K_i = \frac{1}{2} mv^2 \]
The final kinetic energy \( K_f \) of the system (ball + spring gun) after the collision is:
\[ K_f = \frac{1}{2} (m + M) (v')^2 \]
Substituting \( v' \) from step 2:
\[ K_f = \frac{1}{2} (m + M) \left(\frac{mv}{m + M}\right)^2 \]
Simplifying this:
\[ K_f = \frac{1}{2} (m + M) \cdot \frac{m^2 v^2}{(m + M)^2} = \frac{1}{2} \cdot \frac{m^2 v^2}{m + M} \]

### Step 4: Calculate the energy stored in the spring
The energy stored in the spring is equal to the loss of kinetic energy:
\[ \text{Energy stored in spring} = K_i - K_f \]
Substituting the values:
\[ \text{Energy stored in spring} = \frac{1}{2} mv^2 - \frac{1}{2} \cdot \frac{m^2 v^2}{m + M} \]
Factoring out \( \frac{1}{2} v^2 \):
\[ \text{Energy stored in spring} = \frac{1}{2} v^2 \left(m - \frac{m^2}{m + M}\right) \]
Finding a common denominator:
\[ = \frac{1}{2} v^2 \left(\frac{m(m + M) - m^2}{m + M}\right) = \frac{1}{2} v^2 \left(\frac{mM}{m + M}\right) \]

### Step 5: Calculate the fraction of kinetic energy stored in the spring
The fraction of kinetic energy stored in the spring is given by:
\[ \text{Fraction} = \frac{\text{Energy stored in spring}}{K_i} = \frac{\frac{1}{2} v^2 \left(\frac{mM}{m + M}\right)}{\frac{1}{2} mv^2} \]
Simplifying this:
\[ = \frac{mM}{m + M} \cdot \frac{1}{m} = \frac{M}{m + M} \]

### Final Answer
The fraction of kinetic energy of the ball stored in the spring is:
\[ \frac{M}{m + M} \]