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Question: A machine gun fires a bullet of mass \(65g\) with a velocity of \(1300m{{s}^{-1}}\). The man holding...

A machine gun fires a bullet of mass 65g65g with a velocity of 1300ms11300m{{s}^{-1}}. The man holding it can exert a maximum force of 169N169N on the gun. The number of bullets he can fire per second will be
A)1 B)2 C)3 D)4 \begin{aligned} & A)1 \\\ & B)2 \\\ & C)3 \\\ & D)4 \\\ \end{aligned}

Explanation

Solution

When a person fires a gun, an impulse is generated on the gun. Impulse refers to the product of force and the time interval, during which the force acts on an object. Impulse is also equal to the change in linear momentum of an object. Linear momentum is defined as the product of mass of an object and velocity of the object.
Formula used:
1)J=Fdt1)J=Fdt
2)J=mv2mv12)J=m{{v}_{2}}-m{{v}_{1}}

Complete answer:
When a bullet is fired by a person using a gun, an impulse is generated on the gun. Impulse is defined as the product of force acting on an object and the time interval, during which the force acts on the object. Mathematically, impulse is given by
J=FdtJ=Fdt
where
JJ is the impulse generated on an object
FF is the force acting on the object over a time interval dtdt
Let this be equation 1.
Impulse on an object is also equal to the change in linear momentum of the object. Linear momentum refers to the product of mass of an object and velocity of the object. Mathematically, impulse can also be represented as
J=mv2mv1J=m{{v}_{2}}-m{{v}_{1}}
where
JJ is the impulse acting on an object
mv2mv1m{{v}_{2}}-m{{v}_{1}} is the change in linear momentum of the object
Let this be equation 2.
Combining equation 1 and equation 2, we have
Fdt=mv2mv1Fdt=m{{v}_{2}}-m{{v}_{1}}
Let this be equation 3.
Coming to our question, we are provided that a machine gun is used to fire a bullet of mass 65g65g with a velocity of 1300ms11300m{{s}^{-1}}. It is also said that the man holding it can exert a maximum force of 169N169N on the gun. We are required to find the number of bullets he can fire per second.
Let us assume that the person can fire nn bullets per second. If mm represents the mass of a bullet, the total mass of bullets, the person can fire per second is equal to nmnm. Using equation 1, let us calculate the impulse generated per second. Clearly, using equation 1, we have
J=Fdt=165N×1s=165NsJ=Fdt=165N\times 1s=165Ns
where
JJ is the impulse generated in dt=1sdt=1s
F=165NF=165N is the maximum force, the person exerts on the gun
Let this be equation 4.
Now, let us also use equation 2 to determine the impulse generated per second on nn bullets. Clearly, using equation 2, we have
J=n(mv2mv1)=nm(v2v1)=nmv2=n×65g×1300ms1J=n(m{{v}_{2}}-m{{v}_{1}})=nm({{v}_{2}}-{{v}_{1}})=nm{{v}_{2}}=n\times 65g\times 1300m{{s}^{-1}}
where
JJ is the impulse generated per second on nn bullets
nm=n×65gnm=n\times 65g is the total mass of the bullets fired in one second
v1=0ms1{{v}_{1}}=0m{{s}^{-1}} is the initial velocity of the bullets
v2=1300ms1{{v}_{2}}=1300m{{s}^{-1}} is the final velocity of the bullets
Let this be equation 5.
Substituting equation 4 and equation 5 in equation 3, we have
Fdt=n(mv2mv1)165Ns=n×65g×1300ms1Fdt=n(m{{v}_{2}}-m{{v}_{1}})\Rightarrow 165Ns=n\times 65g\times 1300m{{s}^{-1}}
On further simplification, we have
165Ns=n×0.065kg×1300ms1n=165Ns0.065×1300kgms12165Ns=n\times 0.065kg\times 1300m{{s}^{-1}}\Rightarrow n=\dfrac{165Ns}{0.065\times 1300kgm{{s}^{-1}}}\approx 2
Therefore, the man can fire 22 bullets per second.

Hence, the correct answer is option BB.

Note:
Students need to be thorough with conversion formulas. Conversion formulas used in the above solution are given below.
1kg=1000g 1Ns=1kgms2×1s=1kgms1 \begin{aligned} & 1kg=1000g \\\ & 1Ns=1kgm{{s}^{-2}}\times 1s=1kgm{{s}^{-1}} \\\ \end{aligned}
Students need not get worried if they do not arrive at the exact number of bullets given in the options. But it should be made sure that the value closest to the calculated answer is selected without any confusion.