Question

A bullet of mass 25 g is horizontally fired with a velocity of 500 m/s from a gun of mass 5 Kg. The magnitude of the recoil velocity of the gun is ?​

Answer 1

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 :

★ $\sf\: Momentum_{gun}\: =\: Momentum_{bullet}$

→ 5v = 12.5

→ v = $\sf\dfrac{12.5}{5}$

→ v = 2. 5 m/s

$\therefore$ Recoil velocity of the gun, v = 2.5 m/s

Answer 2

⚘ Given :-

• Mass of bullet = 25g = 0.025kg.

• Velocity Of Bullet = 500m/s.

• Mass Of Gun = 5kg.

⚘ To Find :-

• The magnitude of the recoil velocity of the gun.

⚘ Solution :-

We know that,

According to Law Of Conservation Of Momentum,

Momentum Of Gun = Momentum Of Bullet.

So,

Momentum Of Gun = Mass × Velocity = 5v.

Momentum Of Bullet = Mass × Velocity = 12.5kgm/s.

Now, Put them Equal.

⇢ 5v = 12.5 kgm/s

⇢ v = 12.5/5 = 2.5kgm/s

Hence, The magnitude of the recoil velocity of the gun = 2.5kgm/s.