Question

2. A 35-g bullet moving at 450 m/s strikes a 2.5-kg wooden block that is at rest. The bullet passes through the block, leaving at 250 m/s. How fast is the block moving when the bullet leaves?

Answer 1

By momentum conservation,

m*u1 + M*u2 = m*v1 + M*v2

here,

m = mass of bullet = 35 gm = 0.035 Kg

M = mass of block = 2.5 Kg

u1 = initial velocity of bullet = 450 m/sec.

u2 = initial velocity of block = 0

v1 = final velocity of bullet = 250 m/sec.

v2 = final velocity of block = ??

So,

0.035*450 + 2.5*0 = 0.035*250 + 2.5*v2

v2 = (0.035*450 + 2.5*0 - 0.035*250)/(2.5)

v2 = 2.8 m/sec.