Question

. A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest
but free to move. At what speed does the target move off ?

Answer 1

Answer:Answer: 0.99m/s

Step-by-step explanation:

According to the law of conservation of momentum,

For a collision occurring between two bodies or objects, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

(The body is in isolated system)

Therefore

Total momentum before collision = Total momentum after collision

LHS=

Total momentum before collision

= m1u1+m2u2

=(0.01*200)+(2*0)

= 2kgm/s

Total momentum after collision

= m1v1+m2v2

= (0.01+2)*v

According to the law,

2kgm /s = (2.01) v

Therefore v= 2/2.01

= 0.99m/s

The target will move at a speed off 0.99m/s

Hope it helps...marks as the brainliest.