A shell of mass `200g` is ejected from a gun of mass `4 kg` by an explosion that generate `1.05 kJ` of energy. The initial velocity of the shell is

Options

A. `100ms^(-1)`
B. `80ms^(-1)`
C. `40ms^(-1)`
D. `120ms^(-1)`

Answer

To find the initial velocity of the shell ejected from the gun, we can use the principles of conservation of momentum and kinetic energy. Here’s a step-by-step solution:

### Step 1: Understand the System
- We have a shell (bullet) of mass \( m_1 = 200 \, \text{g} = 0.2 \, \text{kg} \) and a gun of mass \( m_2 = 4 \, \text{kg} \).
- The total energy generated by the explosion is \( E = 1.05 \, \text{kJ} = 1050 \, \text{J} \).

### Step 2: Apply Conservation of Momentum
- Initially, both the shell and the gun are at rest, so the initial momentum of the system is \( 0 \).
- After the explosion, let the velocity of the shell be \( v_1 \) and the velocity of the gun be \( v_2 \).
- According to the conservation of momentum:
\[ m_1 v_1 + m_2 (-v_2) = 0 \]
This simplifies to:
\[ 0.2 v_1 = 4 v_2 \quad \text{(1)} \]
From this, we can express \( v_2 \) in terms of \( v_1 \):
\[ v_2 = \frac{0.2}{4} v_1 = \frac{v_1}{20} \quad \text{(2)} \]

### Step 3: Apply Conservation of Energy
- The total kinetic energy after the explosion is equal to the energy generated by the explosion:
\[ \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = 1050 \]
- Substituting \( m_1 \) and \( m_2 \):
\[ \frac{1}{2} (0.2) v_1^2 + \frac{1}{2} (4) v_2^2 = 1050 \]
- Substitute \( v_2 \) from equation (2):
\[ \frac{1}{2} (0.2) v_1^2 + \frac{1}{2} (4) \left(\frac{v_1}{20}\right)^2 = 1050 \]
- Simplifying:
\[ 0.1 v_1^2 + 0.5 \left(\frac{v_1^2}{100}\right) = 1050 \]
\[ 0.1 v_1^2 + 0.005 v_1^2 = 1050 \]
\[ 0.105 v_1^2 = 1050 \]

### Step 4: Solve for \( v_1 \)
- Rearranging gives:
\[ v_1^2 = \frac{1050}{0.105} \]
\[ v_1^2 = 10000 \]
\[ v_1 = 100 \, \text{m/s} \]

### Final Answer
The initial velocity of the shell is \( v_1 = 100 \, \text{m/s} \).

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