Question

Bullet of mass 10 kg moving with a velocity of 400m/s gets emmbedded in a freely suspended wooden block of mass 900 g what is the veloci
ty accquired by the block

Answer 1

mass of the bullet (m1) = 10 kg
initial velocity of the bullet (u1) = 400m/s

momentum of the bullet (p1) = m1 × u1
= 10 kg × 400m s
= 4000 kg m/s

Since, the bullet come into rest (0m/s) after it gets embedded in the wooden block

Again,
The total mass of the wooden block= mass of the bullet + mass of the wooden block
= 10 kg + 900g
= 10 kg + 0.9 kg
= 10.9 kg

therefore, Momentum of the wooden block (p2) = m2 × v2
= 10.9kg × v2

According to the conservation of momentum:

Momentum of the bullet (p1) = Momentum of the wooden block (p2)

=> 4000 kg m/s = 10.9kg × v2

=>$\frac{4000}{10.9}kg$ = v2

=> $\frac{40000}{109}kg$ = v2

=> 366.97247706 m/s = v2