 # A machine gun fires 25g bullates at the rate of 600 per minute with a speed of 400 m/s . Calculate the force required to keep the gun in position. plz explain thanks in advance.  AKASH GOYAL AskiitiansExpert-IITD
420 Points
12 years ago
Dear Jayati

Force=rate of change of momentum= dP/dt=d(mv)/dt

F=v*dm/dt + m*dv/dt

here v is constant hence dv/dt=0

F=v*dm/dt

600 bullets are fired in 60sec

hence in one second 600/60=10 bullets

10 bullets=10x25=250 g= 0.25kg

dm/dt=0.25 kg/sec

F=400x0.25= 100N

hence 100N force is to be applied on gun to keep it in the position

All the best.
AKASH GOYAL

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

12 years ago

hi,

force required to keep the gun in position will be rate of change of momentum

600 bullets are fired in in 1minute the in one second 10 bullets are fired.

hence rate of change of momentum of bullets is 25*10*400/1000 newtons

by calculation we get,

force required=rate of change of momentum of bullet=100N

12 years ago

F = MvdN/dt

dN/dt = 600 per minute = 600 = 600/60 per second

=10 per sec

F = 25*10-3 *400*10

=100N

thus 100N force is required

6 years ago
By the law of conservation of momentum20*V=600*. 025*400V=300 m/sUsing equationV=u+atU=0m/s. V=300m/s300=a*60. 1min=60seca=5m/s^2F=maF=20*5=100N