Question

5) A bullet of mass 3og is fired horizontally with
Velocity of 120 m/s from a pistol of mass 2kg,
determine the recoil Velocity of Pistol.​

Answer 1

Given :

Mass of the bullet = 30g = 0.03kg

Velocity of bullet = 120 m/s

Mass of pistol = 2 kg

To Find :

Recoil velocity of pistol.

Solution :

❖ Since no external force acts on the whole system, linear momentum is conserved.

• We know that momentum is measured as the product of mass and velocity.

It is a vector quantity having both magnitude as well as direction.

SI unit : N s

♦ Initial momentum of the system will be zero$.$

➙ 0 = p + p'

➙ 0 = MV + mv

• M denotes mass of pistol
• V denotes recoil velocity
• m denotes mass of bullet
• v denotes velocity of bullet

By substituting the given values;

➙ 0 = (2)(V) + (0.03)(120)

➙ 0 = 2V + 3.6

➙ V = -3.6 / 2

➙ V = -1.8 m/s

[Negative sign shows opposite direction.]

Answer 2

Answer:

Given :

Mass of the bullet = 30g = 0.03kg

Velocity of bullet = 120 m/s

Mass of pistol = 2 kg

To Find :

Recoil velocity of pistol.

Solution :

❖ Since no external force acts on the whole system, linear momentum is

conserved.

We know that momentum is measured as the product of mass and velocity.

It is a vector quantity having both magnitude as well as direction.

SI unit : N s

♦ Initial momentum of the system will be zero..

➙ 0 = p + p'

➙ 0 = MV + mv

M denotes mass of pistol

V denotes recoil velocity

m denotes mass of bullet

v denotes velocity of bullet

By substituting the given values;

➙ 0 = (2)(V) + (0.03)(120)

➙ 0 = 2V + 3.6

➙ V = -3.6 / 2

➙ V = -1.8 m/s

[Negative sign shows opposite direction.]