Question

If a bullet of mass 5 g moving with velocity 300m/s in a horizontal direction gets embedded in a freely suspended wooden block of mass 1.2 kg. what is the velocity of the block after the bullet is embedded in it

Answer 1

Given :

Mass of bullet = 5g = 0.005kg

Initial velocity of bullet = 300m/s

Mass of block = 1.2kg

To Find :

Velocity of combined mass (bullet + block) after collision$.$

Concept :

Since no external force is involved, we can easily apply concept of linear momentum conservation to solve this problem.

Initially block is at rest so initial momentum of block will be zero.

→ mu = (m + M)v

m denotes mass of bullet

u denotes velocity of bullet

M denotes mass of block

v denotes mass of combined mass

Calculation :

→ 0.005 × 300 = (1.2 + 0.005) × v

→ 1.5 = 1.205v

→ v = 1.5/1.205

→ v = 1.24 m/s