Question

A bullet of mass 20g is fired from gun of mass 10kg . If speed of bullet is 200m/s . Calculate recoil velocity of the gun.

Answer 1

Heya !

In your question,

• Mass of bullet (m₁) = 20 g
• Initial Velocity of bullet (u₁) = Initial Velocity of gun (u₂) = 0 ms⁻¹ - because both the bullet and gun were at rest when the bullet were fired.
• Mass of gun (m₂) = 10 kg = 10*1000 = 10,000 g
• Final Velocity of bullet (v₁) = 200 ms⁻¹
• Final / Recoil Velocity of gun = ?

Let The recoil velocity of the gun be (v₂).

So, before starting with the numerical, I'd like to tell what is

$\textbf{\textit{The Law Of Conservation Of Linear Momentum}}$

So,

The Law Of Conservation Of Linear Momentum states that whenever two or more masses or matter interact with each other, the sum of their total momenta remains the same before and after they interact.

Numerically, it can be expressed as

$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$

Putting the above values, we have,

$(20 \times 0) + (10000 \times 0) = (20 \times 200) + (10000 \times v_{2}) \\ \\ \implies 0 + 0 = 4000 + (10000 \times v_{2}) \\ \\ \implies 0 = 4000 + (10000 \times v_{2}) \\ \\ \implies -4000 = (10000 \times v_{2}) \\ \\ \implies \frac{-4000}{10000} = v_{2} \\ \\ \implies \frac{-4}{10} = v_{2} \\ \\ \implies v_{2} = -0.4$

Therefore, the recoil velocity of the gun will be -0.4 ms⁻¹

Hope it helps !!