Question

A bullet of 7 g is fired into a block of metal weighing of 7 kg. The block is free to move. After the impret. The velocity of the bullet and the block is 70m/s. What is the initial velocity of the bullet

Answer 1

Answer:

Given -

m1=7g=7/1000=0.007kg

m2=7kg

u2=0

v=70m/s

u1=?

According to the law of conservation of momentum -

m1u1+m2u2=m1v1+m2v2

(0.007×u1)+(7×0)=(0.007×70)+(7×70)

0.007×u1+0=0.49+490

0.007×u1=490.49

u1=490.49/0.007

u1=70070

If my answer help you then please mark as brainliest