Question

A 10g bullet travelling at 200m/s strikes and remains embedded in a 2kg target which is free to move.At what speed the target moves of?​

Answer 1

Answer:

0.99m/s

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.

Answer 2

Mass of bullet, m1 = 10g = 0.01 kg Velocity of bullet v1 = 200 m/s Mass of block with the bullet as bullet gets embedded in it, m2 = 2+0.01 = 2.01 kg Recoil velocity v2 According to the law of conservation of momentum m1 x v1 = m2 x v2 0.01 x 200 = 2.01 x v2 Read more on Sarthaks.com - https://www.sarthaks.com/120480/bullet-travelling-strikes-and-remains-embedded-target-which-originally-rest-but-free-move