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# A machine gun fires a bullet of mass 40g with a velocity 1200 m/s. the man holding it can exert a maximum force 144 N on the gun. How many bullets can he fire per second at the most.

neeraj agarwal
34 Points
7 years ago
here, *The gun will exert an impulsive force on the man when it will be fired. * As force will be of impulsive nature I=MV * 40*10^-3*1200=48 N-s * the man can hold with a force of 144 N for 1 sec. * The force must be of impulsive nature. * no. of bullets=144/48=3.
Jai shiv naman
13 Points
5 years ago
Given that m= 40*10^-3,f=144N v=1200m/s We know that force = dp/dt means that change in momentum is called force Then p=mv hence f= d(mv)/dt but find out n bullets per seconds Then f=mvn/t 144=n *40*10^-3*1200/1 hence n=3
saa
13 Points
2 years ago
mass=40g=40*10^-3 kg
v=1200 m/s
F=144N
F is directly proportinal to dp/dt
f is equal to k*dp/dt
let k be no of bullets fired per second =n
t=1s
F=n*dp/dt
144=n*(40*10*-3*1200)/1
144/48000*10^-3=n
n=3

ankit singh
one year ago
Mass of each bullet = 40 g = .040 kg
Velocity of each bullet = 1200 m/s
Let 'n' be the number of bullets fired per second. Then, change in momentum of the bullets coming out per second is,
= force on the man
=> n (0.040)(1200) = 144
=> n = 3