Question

A gun of mass 3kg fires a bullet of mass 30kg. The bullet 0.003s to move through the barrel of the gun and acquires a velocity of 10m/s i) The velocity with the gun recoils ii) The force exerted on gunmam due to recoil of the gun​

Answer 1

Correct Question

A gun of mass 3kg fires a bullet of mass 30g The bullet takes 0.003s to move through the barrel of the gun and acquires a velocity of 10m/s. Find i) The velocity with the gun recoils ii) The force exerted on gunmam due to recoil of the gun

Given:-

• Mass of gun (m¹) = 3kg

• Mass of Bullet ( m² ) = 30g = 0.03kg

• Initial Velocity of gun ( u¹ ) = 0

• Initial Velocity of Bullet ( u² ) = 0

• Final Velocity of Bullet = ( v² ) = 10m/s

To Find:-

• The Velocity with which gun recoiles.

Formulae used:-

• Conservation of Momentum.

• v = u + at

• F = ma

where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• F = force
• m = Mass
• t = Time

Now,

Initially the bullet and gun was at rest

→ m¹ u¹ + m² u² = m¹v¹ + m² v²

→ 3 × 0 + 0.03 × 0 = 3 × 10 + 0.03 × v²

→ 0 = 30 + 0.03v²

→ -30 = 0.03v²

→ v² = -30/0.03

→ v² = 1000m/s.

Hence, The Recoil Velocity of gun is 1m/s.

Now, Coming to 2nd Question.

As, The Final Velocity of gun(v) is 1000m/s & Initial Velocity = 0.

So

→ v = u + at

→ -1000 = 0 + a × 0.03

→ -1000 = 0.03a

→a = -1000/0.03

→ a = -33,333.33m/s²

Therefore,

→ F = ma

→ F = 0.03 × -33,333.33

→ F = -1000N

Hence, The Force exerted on gunman is -1000N.

Answer 2

1.Answer:

50 m/s

2.3234N

Explanation: