Question

10 gram Bullet travelling at 200 m per second strikes and remains embedded in 2 kg Target which is original at rest but free to move at what speed does the target move off

Answer 1

mass of bullet=10g,10/100=.01kg,

we know that ,

p=mv,

p=0.1×200=20kg.m/sec.

now, mass of embedded=2kg,

so,total mass=2+0.1=2.1kg,

again,p=mv,

p=2.1×v=2.1vkg.m/sec.

now,total momentum=20+2.1v.

so,v=20/2.1=9.57m/sec.

hence,v=9.57m/sec.

Answer 2

HERE MOMENTUM IS CONSERVED

we will take target and bullet as one system

m OF BULLET - 10 g

                          10 / 1000 kg

                         = 0.01 kg

speed - 200 m/s

Pi(initial momentum)-mv

                                 - 0.01 x 200

                                 - 2 kg m/s + 0 (as velocity of target is 0 in starting )

Pf(final momentum)-mv

                               - (mass of bullet + mass of target) x (velocity)

                               - (0.01+2) x v

                               - 2.01 v

Pi=Pf

2 = 2.01 v

v = 2/2.01 m/s