Question

A gun of mass 3 kg fires a bullet of mass 30 g. the bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. calculate: a)The velocity with which the gun recoils. b)The force exerted on gunman due to recoil of the gun. who ever answers will be marked as brainliest!!! pakka

Answer 1

Given :

• Mass of the gun, $\sf\: m_1$ = 3kg

• Mass of the bullet, $\sf\: m_2$ = 30 g = 0.03 kg

• Time taken, t = 0.003 seconds

• Velocity of bullet, $\sf\: v_2$ = 100 m/s

To Find :

• The velocity with which the gun recoils

• The force exerted on the gun due to recoility

Solution :

1. Calculation of the recoil velocity

According to law of conservation of momentum

$\longrightarrow \sf \: m_1v_1 = m_2v_2$

$\longrightarrow \sf \: 3 \times v_1 =0.03\times 100$

$\longrightarrow \sf \: v_1 = \dfrac{0.03 \times 100}{3}$

$\longrightarrow \sf \: v_1 = 0.01 \times 100$

$\longrightarrow \sf \: v_1 = 1m {s}^{ - 1}$

$\therefore$ The velocity with which the gun recoils is 1 m/s.

2. Calculation of the force exerted.

• Initial Velocity of the gun, u = 0 m/s

• Final Velocity of the gun, v = 1 m/s

• Time, t = 0.03 seconds

Let's find acceleration first :

★ a = v - u/t

→ a = 1 - 0/0.03

→ a = $\sf\dfrac{1000}{3}$ m/s²

From Newton's 2nd Law of motion :

★ Force = Mass × Acceleration

$\longrightarrow \sf \: F= 3 \times \dfrac{1000}{3}$

$\longrightarrow \sf \: F= \dfrac{3000}{3}$

$\longrightarrow \sf \: F= 1000 \: Newtons$

$\therefore$ The force exerted on the gun due to recoility is 1000 Newtons.