Question

A bullet of mass 120 grams is fired into a target with a velocity of 400 m/s. The mass of the target is 10kg and it is free to move Find the loss of kinetic energy?​

Answer 1

Answer:

why this question

Explanation:

sorry I have no answer in this question

Answer 2

Answer:

Momentum is conserved. Initial momentum p is:

p=mv0=(0.12)(400)=48 kg m/s

Final momentum is the same, therefore:

p=(M+m)vf=48

Therefore final velocity vf is:

vf=4810+0.12=4.74308 m/s

Initial kinetic energy Ek0 is:

Ek0=12mv20=(0.5)(0.12)(4002)=9600 joules

Final kinetic energy Ekf is:

Ekf=12(M+m)v2f=(0.5)(10.12)(4.743082)=113.83 joules

Therefore the loss in kinetic energy ΔEk is:

ΔEk=Ek0−Ekf=9600−113.83

= 9486.17 joules