Question

A bullet of mass 20g moves with a velocity of 200m/s, and gets embeded in a stationary wooden block of 980g. With what velocity will the block move?

Answer 1

By using conservation of momentum equation
mu = (M + m)v
v = mu / (M + m)
= (20 × 10^-3 × 200) / [(980 + 20) × 10^-3]
= 4 m/s

Block will move with velocity equal to 4 m/s

Answer 2

M1(mass of bullet) = 20 g= 0.02kg

m2(mass of wooden block)=980 g= 0.98 kg

u1= 200 m/s (Initial velocity of the bullet)

u2= 0 m/s (Initial velocity of the block of wood)

After the collision, the block as well as the bullet move witha common welocity. Let this velocity be v.

By Law of Conservation of Momentum

m1u1 + m2u2 = m1v + m2v

= 0.02*200 + 0*0.98 = v(m1+m2)

= 4 = v(0.98 + 0.02)

= 4 = v

Therefore the velocity with which both the wooden block and the bullet move is 4m/s