A ball of mass `m=6kg` is shot with speed `v=20m//s` into a barrel of spring gun of mass `M=24kg` initially at rest. All surfaces are smooth & spring is massless. Speed of gun after ball stops relative to gun is

Options

A. `2m//s`
B. `4m//s`
C. `6m//s`
D. `8m//s`

Explanation

To solve the problem, we will use the principle of conservation of momentum. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the System
We have a ball of mass \( m = 6 \, \text{kg} \) moving with a speed \( v = 20 \, \text{m/s} \) and a spring gun of mass \( M = 24 \, \text{kg} \) that is initially at rest. The ball is shot into the gun, and we need to find the speed of the gun after the ball stops relative to it.

### Step 2: Apply Conservation of Momentum
According to the law of conservation of momentum, the total momentum before the event must equal the total momentum after the event.

**Initial Momentum:**
- The initial momentum of the system (ball + gun) is given by the momentum of the ball since the gun is at rest.
\[ \text{Initial Momentum} = m \cdot v + M \cdot 0 = 6 \cdot 20 + 24 \cdot 0 = 120 \, \text{kg m/s} \]

**Final Momentum:**
- After the ball comes to rest relative to the gun, both the ball and the gun move together with a common speed \( V \).
\[ \text{Final Momentum} = (m + M) \cdot V = (6 + 24) \cdot V = 30V \, \text{kg m/s} \]

### Step 3: Set Initial Momentum Equal to Final Momentum
Now we can set the initial momentum equal to the final momentum:
\[ 120 = 30V \]

### Step 4: Solve for \( V \)
To find \( V \), we rearrange the equation:
\[ V = \frac{120}{30} = 4 \, \text{m/s} \]

### Conclusion
The speed of the gun after the ball stops relative to the gun is \( 4 \, \text{m/s} \).

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