Grade Select GradeMechanics

A gun fires 'n' bullets with a speed of 1800m/sec in 3 sec. Mass of each bullet is 25gm. If 450N force is required to hold the gun in position. Find number of bullets fired in each second?

Shanzay

6 Years agoGrade Select Grade

2 Answers

Arun

6 Years ago

Force exerted by n bullets is 450 N (as the system remains stable).

Force exerted by n bullets: F=ma 450=25*10-3*a a=18000 m/s2

v=u+at

Initial velocity of a bullet is u=0 m/s

Final velocity of a bullet v=1800 m/s

1800=18000·t

t = 0.1 s (time taken for 1 bullet to be fired)

Number of bullets: 3/0.1=30

Vikas TU

6 Years ago

Dear student

Here is the complete explanation.

As we know the rate of change of momentum of bullets is equal to force applied to hold the gun because from the law of conservation of momentum if M is mass of gun and V is backward velocity of gun when shots are fired the gun recoiled back and m is the mass of one bullet then “nm” will be the mass of “n” bullets . If “v” is velocity of each bullet

Then law of momentum says momentum before and after collision remain same . I.e

0=-MV+mnv ( initial momentum was zero when gun was not firing)

-MV means that gun recoiled back and mnv is momentum of shotted bullets

So MV=mnv

Dividing both sides by t

MV/t=mnv/t

As from the definition we Know that rate of change of momentum is equal to force applied

So MV/t =450 N. (1)

Because this 450 N is the force that hold the momentum of gun that is pushing backwards .

So from (1)

450=(mnv)/3

As from given mass of each bullet is 25gm or 0.025 kg and v is the velocity of each bullet v=1800m/s