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# A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resist the downward acceleration if he shoots 40 bullets a second at the speed of $500\,\,m/s$. If the weight of a bullet is $49\,g$, what is the weight of the man with the gun? Ignore resistance due to air and assumed the acceleration due to gravity $g = 9.8\,m/s^2$ A. $50\,N$B. $75\,N$C. $100\,N$D. $125\,N$

Last updated date: 16th Jul 2024
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Hint: Here if the man with the gun should stop accelerating downward, net downward force should be balanced by the force due to firing of the bullet. The force exerted on the gun due to firing of the bullet can be found using Newton's second law. If we equate this force with the net weight of the man and the gun acting downwards, we can find the answer.

It is given that a soldier is falling from an airplane with a machine gun and if he shoots 40 bullets a second with the speed $500\,\,m/s$ he will be able to resist the downward acceleration.
We need to find the weight of the man with the gun.
The weight of the bullet is given as $49\,g$.
${W_b} = {m_b} \times g = 49\,g$
So, the mass of the bullet will be the weight divided by the acceleration dur to gravity.
It is given that $g = 9.8\,m/s$ .
$\Rightarrow {m_b} = \dfrac{{49 \times {{10}^{ - 3}}}}{{9.8}} = 5 \times {10^{ - 3}}kg$
We can find the answer by balancing the forces. The force acting downward is the weight of the man with the gun.
That is
$F = \left( {{M_m} + {M_g}} \right)g$
Here, ${M_m}$ is the mass of the man and ${M_g}$ is the mass of the gun.
This force should be balanced by the force due to firing of the bullet.
Let's find the force on the gun due the firing of the bullet.
This is a case where the mass is changing.
As the bullets leave the gun the mass of the gun is being reduced.
From Newton's second law we know that force is the rate of change of momentum P.
$F = \dfrac{{dP}}{{dt}}$
Momentum is the product of mass and velocity.
$P = mv$
Since velocity is not changing, we get
$\Rightarrow F = \dfrac{{d\left( {mv} \right)}}{{dt}}$
$\Rightarrow F = v\dfrac{{d\left( m \right)}}{{dt}}$ (1)
Here, $\dfrac{{dm}}{{dt}}$ can be found as the product of the number of bullets fired in one second and the mass of each bullet.
$\dfrac{{dm}}{{dt}} = 40 \times 5 \times {10^{ - 3}}$
On substituting all the value in equation 1, we get the force acting on the gun as
$\Rightarrow F = 500 \times 40 \times 5 \times {10^{ - 3}}\,N$
This force is balanced by the total weight of the man and the gun so that there is no net downward acceleration.
On equating the weight of the gun and man with this force, we get
$\Rightarrow \left( {{M_m} + {M_g}} \right)g = 500 \times 40 \times 5 \times {10^{ - 3}}\,N$
$\Rightarrow \left( {{M_m} + {M_g}} \right)g = 100\,N$
This is the total weight of the man with the gun.
So, the correct answer is option C.

Note: Remember that to resist the downward acceleration the force acting downward should be balanced. When we fire the gun due to the conservation of momentum there will be recoil velocity on the gun. The force due to this recoil can cancel the net force downward. If the force of friction is also considered then the weight of the man and gun will be balanced by the total force due to firing of the gun and the force due to friction.