Question

a machine gun having weight 20 kg fire bullet of 35g at rate of 400 bullet per minute at speed of 400 m per second. now what is force required to hold the gun at the same place.

Answer 1

machine gun weight = 20 kg

weight of bullet = 35g = 0.035 kg
speed of bullet = 400 m/s
impulse from one bullet = mv = 0.035×400 = 14 kg.m/s

number of bullets per second = 400/60 = 20/3
impulse per second = 14 × 20/3 = 280/3 = 93.33 kg.m/s/s = 93.33 kg.m/s² = 93.33 N

(Impulse per second is nothing but force)

You need 93.33 N force to hold the gun in place.

Answer 2

Heya.....!!!!

-Given in the question :—

⇒ Mass of machine gun => 20 kg
⇒ Speed of bullet. => 400m/sec
⇒ Weight of bullet. => 35 gm ( changing it to Kg ) =>> 0.035 Kg


From this we can take out impulse from 1 bullet = change in momentum =>> mv = 0.035 × 400 ⇒ 14 kg.m/s

— it is given that 400 bullets are fired per min Then ,, number of bullets per sec is. => 400/60 = 20/3

Now ,, Impulse per sec => 14×20/3 = 93.33 Kgm/sec² that is equal to => 93.33 N

→ Required force => 93.33 N .

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Hope It Helps You. ^_^