Question

1) A 60 g Bullet travel at 20 metre per second strike and remain embedded in a 2kg target which is originally at rest but free to move at what speed does the target move off.​

Answer 1

Answer:

0.58 m/s

Explanation:

due to conservation of momentum,m1v1=m2v2.

Here: m1=bullet

v1=velocity of bullet

m2=mass of block

v2=required velocity

as the bullet is embedded after collision, m1v1=(m1+m2)v2

60*20 = (2000+60)*v2

so v2=0.58