JEE MAIN - Physics (2004 - No. 46)
A machine gun fires a bullet of mass $$40$$ $$g$$ with a velocity $$1200m{s^{ - 1}}.$$ The man holding it can exert a maximum force of $$144$$ $$N$$ on the gun. How many bullets can he fire per second at the most?
Two
Four
One
Three
Explanation
Assume the man can fire $$n$$ bullets in one second.
$$\therefore$$ change in momentum per second $$ = n \times mv = F$$
[ $$m=$$ mass of bullet, $$v=$$ velocity, $$F$$ = force) ]
$$\therefore$$ $$n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3$$
$$\therefore$$ change in momentum per second $$ = n \times mv = F$$
[ $$m=$$ mass of bullet, $$v=$$ velocity, $$F$$ = force) ]
$$\therefore$$ $$n = {F \over {mv}} = {{144 \times 1000} \over {40 \times 1200}} = 3$$
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