A machine gun fires a bullet of mass $40g$ with a velocity $1200m{s^{ - 1}}$. The man holding it can exert a maximum force of 144N on the gun. How many bullets can be fired per second at the most?
A) $1$
B) $2$
C) $3$
D) $4$
Answer
260.7k+ views
Hint: In order to find the solution of the given question, first of all, we need to find the change in momentum of the bullet. After that we need to use the relation that the change in momentum of the bullet per second will be equal to the force applied by the man. By using this relation, we can find out the number of bullets fired per second. Also, we need to relate this question with Newton’s third law of motion, which says that every action has an equal and opposite reaction.
Complete step by step solution:
It is given that the force exerted by the man on the gun, $F = 144N$
The mass of the bullet, $m = 40g = \dfrac{{40}}{{1000}}kg$
The velocity of the bullet,$v = 1200m{s^{ - 1}}$
Now, we need to find the change in momentum of the bullet.
From Newton’s third law of motion, we know that,
${F_{action}} = {F_{reaction}}$
${F_{reaction}} = $Change in momentum of the bullet
Now, Momentum, $p = {m_1}{v_1} - {m_2}{v_2}$
$ \Rightarrow p = \dfrac{{40}}{{1000}} \times 1200 - 0$
$\therefore p = 48kgm{s^{ - 1}}$
Now, let us assume there are ‘n’ numbers of bullets fired per second.
According to Newton’s third law, discussed in step two,
$144 = n \times 48$
$\therefore n = 3$
Therefore, $3$ bullets are fired per second.
Hence, option (C), i.e. $3$ is correct for the given question.
Note: Momentum of a body is the product of mass and velocity of the body. It is related to the quantity of motion of an object. It is given by the relation $p = mv$. It follows conservation law. According to Newton’s third law of motion, every action has an equal and opposite reaction. Action force and reaction force always act opposite to each other.
Complete step by step solution:
It is given that the force exerted by the man on the gun, $F = 144N$
The mass of the bullet, $m = 40g = \dfrac{{40}}{{1000}}kg$
The velocity of the bullet,$v = 1200m{s^{ - 1}}$
Now, we need to find the change in momentum of the bullet.
From Newton’s third law of motion, we know that,
${F_{action}} = {F_{reaction}}$
${F_{reaction}} = $Change in momentum of the bullet
Now, Momentum, $p = {m_1}{v_1} - {m_2}{v_2}$
$ \Rightarrow p = \dfrac{{40}}{{1000}} \times 1200 - 0$
$\therefore p = 48kgm{s^{ - 1}}$
Now, let us assume there are ‘n’ numbers of bullets fired per second.
According to Newton’s third law, discussed in step two,
$144 = n \times 48$
$\therefore n = 3$
Therefore, $3$ bullets are fired per second.
Hence, option (C), i.e. $3$ is correct for the given question.
Note: Momentum of a body is the product of mass and velocity of the body. It is related to the quantity of motion of an object. It is given by the relation $p = mv$. It follows conservation law. According to Newton’s third law of motion, every action has an equal and opposite reaction. Action force and reaction force always act opposite to each other.
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