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`        A gun of mass 'M' fires a bullet of mass 'm' with a velocity 'v' relative to the gun. The average force required to bring the gun to rest in 0.5 seconds is`
7 years ago

## Answers : (2)

147 Points
```										Dear ramya
conservation of linear momentum
let the velocity of gun is V1 wrt earth
so velocity of bullet wrt eart is v-V1
m(v-v1) =Mv1
v1=mv/(m+M)
now distance travel by gun in 0.5 sec
0=v1-a*0.5
a= 2v1     wher a is retardation of gun
so S= v1 *0.5 -1/2 *a *(0.5)2
=v1 /4
now from work energy theorem total work done is equal to change in kinetic energy
F.S =1/2 M *v12    -0
F.v1 /4 = 1/2 M *v12
F=2Mv1
=2Mmv/(m+M)

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.  All the best.  Regards, Askiitians Experts Badiuddin

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7 years ago
Ishana
13 Points
```										This is not the correct answer for this question. I asked you to find total kinetic energy in the question but you have found something else.????????!!!!!!!!!!!
```
23 days ago
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