A gun of mass M fires a bullet of mass m, with a Kinetic energy E. The total kinetic energy in the firing process is

Options

A. `[(m+M)/M]E`
B. `([m+M]E)/m`
C. `(M/(M+m))E`
D. `(m/(M+m))E`

Answer

To find the total kinetic energy in the firing process when a gun of mass \( M \) fires a bullet of mass \( m \) with kinetic energy \( E \), we can follow these steps:

### Step 1: Understand the Kinetic Energy of the Bullet
The kinetic energy \( E \) of the bullet can be expressed as:
\[ E = \frac{1}{2} m v_B^2 \]
where \( v_B \) is the velocity of the bullet.

### Step 2: Apply Conservation of Momentum
According to the conservation of momentum, the momentum before firing is equal to the momentum after firing. Initially, both the gun and the bullet are at rest, so:
\[ 0 = M v_G + m v_B \]
where \( v_G \) is the recoil velocity of the gun. Rearranging gives:
\[ M v_G = -m v_B \quad \Rightarrow \quad v_G = -\frac{m}{M} v_B \]

### Step 3: Calculate the Kinetic Energy of the Gun
The kinetic energy of the gun can be expressed as:
\[ KE_{\text{gun}} = \frac{1}{2} M v_G^2 \]
Substituting \( v_G \) from Step 2:
\[ KE_{\text{gun}} = \frac{1}{2} M \left(-\frac{m}{M} v_B\right)^2 = \frac{1}{2} M \cdot \frac{m^2}{M^2} v_B^2 = \frac{m^2}{2M} v_B^2 \]

### Step 4: Relate \( v_B^2 \) to \( E \)
From Step 1, we know:
\[ E = \frac{1}{2} m v_B^2 \quad \Rightarrow \quad v_B^2 = \frac{2E}{m} \]
Substituting this into the kinetic energy of the gun:
\[ KE_{\text{gun}} = \frac{m^2}{2M} \cdot \frac{2E}{m} = \frac{mE}{M} \]

### Step 5: Calculate Total Kinetic Energy
The total kinetic energy in the firing process is the sum of the kinetic energy of the bullet and the kinetic energy of the gun:
\[ KE_{\text{total}} = KE_{\text{bullet}} + KE_{\text{gun}} = E + \frac{mE}{M} \]
Factoring out \( E \):
\[ KE_{\text{total}} = E \left(1 + \frac{m}{M}\right) = E \frac{M + m}{M} \]

### Final Answer
Thus, the total kinetic energy in the firing process is given by:
\[ \text{Total Kinetic Energy} = \frac{(M + m)}{M} E \]