Question

A bullet of mass 50 g is fired from a gun of mass 2 kg. If the total kinetic energy produced is 2050 J, the energy of the bullet and the gun separately are:

• A. \(200\text{ J}, 5\text{ J}\)
• B. \(2000\text{ J}, 50\text{ J}\)
• C. \(5\text{ J}, 200\text{ J}\)
• D. \(50\text{ J}, 2000\text{ J}\)

Hint:

When a system starts from rest and splits (gun–bullet system):
Momentum of both parts is equal
Kinetic energy is inversely proportional to mass
Lighter body always gets more kinetic energy

Answer 1

The Correct Option is B

Step 1: Understand the physical situation. When a bullet is fired from a gun:
The bullet moves forward with high velocity.
The gun recoils backward with a much smaller velocity.
By the law of conservation of momentum, both acquire equal and opposite momentum.
Step 2: Write the relation for kinetic energy using momentum. Kinetic energy in terms of momentum \(p\) and mass \(m\) is: \[ KE = \frac{p^2}{2m} \] Thus, for the same momentum: \[ \frac{KE_{\text{bullet}}}{KE_{\text{gun}}} = \frac{m_{\text{gun}}}{m_{\text{bullet}}} \]
Step 3: Substitute given masses. \[ m_{\text{bullet}} = 50\text{ g} = 0.05\text{ kg}, \quad m_{\text{gun}} = 2\text{ kg} \] \[ \frac{KE_{\text{bullet}}}{KE_{\text{gun}}} = \frac{2}{0.05} = 40 \] So, \[ KE_{\text{bullet}} = 40 \times KE_{\text{gun}} \]
Step 4: Use total kinetic energy. \[ KE_{\text{bullet}} + KE_{\text{gun}} = 2050 \] \[ 40KE_{\text{gun}} + KE_{\text{gun}} = 2050 \] \[ 41KE_{\text{gun}} = 2050 \] \[ KE_{\text{gun}} = 50\text{ J} \]
Step 5: Find kinetic energy of the bullet. \[ KE_{\text{bullet}} = 40 \times 50 = 2000\text{ J} \] Final Answer: \[ \boxed{KE_{\text{bullet}} = 2000\text{ J}, \quad KE_{\text{gun}} = 50\text{ J}} \]