Question

A machine gun fires 10 bullets per second into a target.Each bullet weight 20 gm and had a speed of 1500 m/s.Find the force necessary to hold the gun in position?

Answer 1

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Mass of a bullet, m = 40 g = 0.04 kg

Velocity of a bullet, v = 1200 m/s

So, momentum of each bullet, p = mv = 48 Ns

Total time = 1 second

Force applied by 1 bullet per second Fb = p/t = 48/1 = 48 N

Maximum force exerted by man Fm = 144 N

Let n bullets are fired per second

then

Fm  > n Fb 

or

144 > n 48

n < 144/48

n < 3

Thus he can fire 3 bullets at max per second.

Answer 2

Answer:

The force necessary to hold the gun in position is equal to 300N.

Explanation:

We have given, the number of bullets, n= 10

The mass of one bullet = 20g = 20×10⁻³Kg

The speed of each bullet, v= 1500m/s

We know that, the rate of change of momentum is equal to force applied to that object.

So, $F=\frac{dP}{dt}$

All 10 bullets fired in one second, so dt = 1sec

$F=\frac{\big n\big P}{\big 1\big s\big e\big c}$

We know that momentum of an object is product of mass of the object and its velocity.

$P=mv$

$P=(20*10^{-3}Kg)(1500ms^{-1})$

$P=30Kgms^{-1}$

Therefore, force for 10 bullets will be given by: $F=\frac{(10)(30Kgms^{-1})}{1s}=300Kgms^{-2}=300N$

Therefore, the required force to keep the gun at its place is equal to 300N.