A bullet of mass m is fired from a gun of mass M. The recoiling gun compresses a spring of force constant k by a distance d. Then the velocity of the bullet is

Options

A. `kdsqrt(M//m)`
B. `(d)/(M)sqrt(km)`
C. `(d)/(m)sqrt(kM)`
D. `(kM)/(m)sqrt(d)`

Answer

To find the velocity of the bullet fired from a recoiling gun that compresses a spring, we can follow these steps:

### Step 1: Understand the Conservation of Momentum
When the bullet is fired, the total momentum before firing is equal to the total momentum after firing. Initially, both the bullet and the gun are at rest, so the initial momentum is zero.

\[ P_{\text{initial}} = 0 \]

After firing, let \( v \) be the velocity of the bullet and \( V \) be the velocity of the recoiling gun. The final momentum can be expressed as:

\[ P_{\text{final}} = mv - MV \]

Setting the initial momentum equal to the final momentum gives us:

\[ 0 = mv - MV \]

### Step 2: Solve for the Velocity of the Gun
From the conservation of momentum equation, we can express the velocity of the gun \( V \):

\[ MV = mv \implies V = \frac{m}{M}v \]

### Step 3: Relate Kinetic Energy to Potential Energy
The kinetic energy lost by the gun is converted into the elastic potential energy stored in the spring when it is compressed by a distance \( d \). The kinetic energy of the gun is given by:

\[ KE_{\text{gun}} = \frac{1}{2}MV^2 \]

The elastic potential energy stored in the spring is given by:

\[ PE_{\text{spring}} = \frac{1}{2}kd^2 \]

Setting these two energies equal gives us:

\[ \frac{1}{2}MV^2 = \frac{1}{2}kd^2 \]

### Step 4: Substitute for \( V \)
Substituting \( V = \frac{m}{M}v \) into the kinetic energy equation:

\[ \frac{1}{2}M\left(\frac{m}{M}v\right)^2 = \frac{1}{2}kd^2 \]

This simplifies to:

\[ \frac{1}{2}M \cdot \frac{m^2}{M^2}v^2 = \frac{1}{2}kd^2 \]

Cancelling \( \frac{1}{2} \) from both sides:

\[ \frac{m^2}{M}v^2 = kd^2 \]

### Step 5: Solve for the Velocity of the Bullet
Rearranging the equation to solve for \( v^2 \):

\[ v^2 = \frac{kd^2M}{m^2} \]

Taking the square root of both sides gives us the velocity of the bullet:

\[ v = \sqrt{\frac{kd^2M}{m^2}} = \frac{d}{m} \sqrt{kM} \]

### Final Answer
Thus, the velocity of the bullet is:

\[ v = \frac{d}{m} \sqrt{kM} \]