Question

A 750kg gun fires a 2.0 kg shell with a muzzle velocity of 500m/s.The gun recoils against a resisting force of 2500N.
i)Determine the recoil velocity of the gun.
ii)What is the recoil time of the gun?

Answer 1

@zarkon can you help me?

Answer 2

This question involves both the balance of forces and conservation of momentum. Since initially the system has 0 momentum, the net momentum must remain 0 even after the gun is fired. The momentum of the bullet in one direction must exactly balance the momentum of the gun in the other direction.

The momentum of the bullet is p = 2*500 = 1000 kg m/s

The gun therefore also recoils with this momentum. However, due to its greater mass, its velocity is smaller: 1000 = 750*v --> v = 1.33 m/s

The recoil doesn't last forever since there is a resisting force that opposes the gun's momentum and so slows it down. Recall that p = F*t, so we can say F*t = mv.

2500*t = 750*1.33 --> t = 0.4 s

Answer 3

i think velocity will be \[1.87\times10^{5}\]

Answer 4

@muhammad9t5 I think you're right...

Answer 5

Is that 1.87*10^5 m/s? How did you arrive at that answer?

Answer 6

by
\[m1v1=m2v2 \]
\[v2=m1v1\div m2 \]

Answer 7

@matt102 I didn't get what you did for "t"

Answer 8

@muhammad9t5 I did exactly that above (just split it into two steps to explain it more clearly) and got 1.33 m/s:

v = (2)(500)/750 = 1.33

There are no unit conversions involved so I don't see how you can get an answer on the order of 10^5 m/s. 1.33 m/s is the sort of answer you would expect intuitively. If you've ever seen an artillery gun firing a shell, you never see the shell because it moves so fast, but you do observe the gun roll backwards a little in recoil.

@supercrazy92 using the above value will give 0.4 s. Do you have a different way of finding t?