Question

A bullet of mass 20 grams is horizontally fired with the velocity of 400 m/s from a gun of 50 kg. what is recoil velocity of gun?​

Answer 1

Given :

mass of bullet (m_2,) = 20g = 0.2 kg = 20 x 10^-3 kg

Mass of gun (m_1) = 50 kg

and final velocity (v_2) = 400 ms^-1

According to law of conservation of momentum,

➸ m_1v_1 + m_2V_2 = m_1v_1 + m_2v_2

➸ -m_1v_1 = m_2,V_2

➸ -50 x V_1 = 20 x 10^-3 x 400

➸ v_1 = 20 x 10^-3 x 400/50

➸ 0.16

∴ Therefore, recoil velocity of gun (v) = 0.16 ms^-1

Answer 2

Answer:

Given :

• Mass of the bullet = 25 g = 0.025 kg

• Velocity of the bullet = 500 m/s

• Mass of the gun = 5 kg

To Find :

• Recoil velocity of the gun.

Solution :

Let's find the momentum of gun and the bullet individually

✒ Momentum of bullet = Mass of bullet × Velocity of bullet

→ p = 0.025 × 500

→ p = 12.5 kg m/s

✒Momentum of gun = Mass of gun × Velocity of gun

→ p = 5 × v

→ p = 5v

According to Law of Conservation of Momentum :

$★ \sf\: Momentum_{gun}\: =\: Momentum_{bullet}</p><p>$

→ 5v = 12.5

→ v =

$\sf\dfrac{12.5}{5}$

→ v = 2. 5 m/s

∴ Recoil velocity of the gun, v = 2.5 m/s