Question

A bullet with a mass of 0.005 kg is shot at a speed of 40

m/s towards a 6 kg stationary block. The bullet becomes
embedded in the block and the two fly off together. Find
the speed with which they fly off. (m1v1 + m2v2=
(m1+m2)Vf) *

Answer 1

combined velocity of the system = 0.033 m/s.

Explanation:

taking the bullet and block as single system since there is no external force involved momentum is conserved .

initial momentum of the bullet = mv

where m is the mass of bullet and v is velocity of bullet.

and initial momentum of block = 0 since its velocity is 0.

therefore initial momentum of the system = mv

= 0.005×40

= 0.2

when the bullet stick of together with the block let the combined velocity of the system be V

therefore applying conservation of momentum

mv = (M+m)V

0.2=(6+0.005)V

therefore V = 0.033 m/s