Question

6 A bullet of mass 0.02 kg is horizontally fired with a velocity 150 ms from
mass 2 kg
(a) What is the momentum of the pistol before the fire?
(b) What is the momentum of the bullet before the fire?
(c) What is the momentum of the pistol after the fire?
(d) Find the value of u.​

Answer 1

A) 0

B) 0

C) m= 2kg

V = 150m/s

MV = 300kg•m/s

D) law of conservation of momentum

(Momentum before after remains conserved)

So, 150×3 = 0.02×u

300/0.02 = u

u = 15,000 m/s