Question

A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if the shoots 40 bullets a second at the speed of 500 m/s. If the mass of a bullet is 49 gm, what is the mass of the man with the gun ? Ignore resistance due to air and assume the acceleration due to gravity g = 9.8 $m/s^2$.

• A. 60 kg
• B. 75 kg
• C. 100 kg
• D. 125 kg

Answer 1

Answer: (C)

100 kg

We know

$F=\frac{\Delta P}{\Delta t}$

$\Rightarrow \Delta P=F.\Delta t$ . . . (i)

where $\Delta P$ is change in momentum and $\Delta t$ is the time interval

Let the mass of man with gun be $M$.

The force acting on the man alongwith the gun is $F=Mg$ downwards

Now this force is balanced by the change in momentum in a given time say $\Delta t=1sec$

The mass of one bullet $m=49gm=0.049kg$

Velocity of each bullet $v=500m/s$

The momentum change due to one bullet $\Delta p=p_f-p_i=mv-0=mv$

If $n$ bullets are fired per sec,

then the total momentum change $\Delta P=P_f-P_i=(n\times mv)-(n\times 0)=nmv$

Given $n=40$

We consider a time interval of $\Delta t=1sec$

So using (i),

$\Delta P=F.\Delta t$

$nmv=Mg\times \left(1\right)$

$M\times 9.8=40\times 0.049\times 500$
$M=\frac{40\times 49\times 500}{9.8\times 1000}$
$M=100kg$
So the mass of the man along with the gun is 100 kg.