
A gun fires a bullet of mass \[50{\text{ }}g\] with a velocity of \[30{\text{ }}m\;{s^{ - 1}}\]. Because of this, the gun is pushed back with a velocity of \[{\text{1 }}m\;{s^{ - 1}}\]. What is the mass of the gun?
A. \[{\text{3}}{\text{.5 k}}g\]
B. \[{\text{30 k}}g\]
C. \[{\text{1}}{\text{.5 k}}g\]
D. \[{\text{20 k}}g\]
Answer
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Hint:In this question, we are given the mass of the bullet \[50{\text{ }}g\] and the velocity of the bullet and the gun i.e., \[30{\text{ }}m\;{s^{ - 1}}\], \[{\text{1 }}m\;{s^{ - 1}}\] respectively. We have to find the mass of the gun. To calculate the mass, equate the total momentum of the gun and bullet after the fire and before the fire.
Formula used:
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Here, ${m_1},{m_2}$ are the masses of the body, ${u_1},{u_2}$ is the initial velocities of the bodies and ${v_1},{v_2}$ are the final velocities.
Complete step by step solution:
Given that,
Mass of the bullet, ${M_B} = 50{\text{ }}g = 0.05{\text{ }}kg$
Mass of the gun $ = {M_G}$
Initial velocity of the bullet and the gun is equal to zero (rest). Therefore, the total momentum of the bullet and the gun before the fire will be zero (Initial velocity is zero).
Now, the final velocity of the bullet,
\[{v_B} = 30{\text{ }}m\;{s^{ - 1}}\]
Final velocity of the gun,
\[{v_G} = - 1{\text{ }}m\;{s^{ - 1}}\]
Total momentum after the fire (bullet and gun) $ = {M_B} \times {v_B} + {M_G} \times {v_G}$
$\Rightarrow 0.05 \times 30 + {M_G} \times \left( { - 1} \right)$
$\Rightarrow 1.5 - {M_G}$
Now, the momentum will be conserved. So, the total momentum before the fire and after the fire will be equal
$\text{Total momentum before the fire = Total momentum after the fire}$
$\Rightarrow 1.5 - {M_G} = 0$
$\therefore {M_G} = 1.5{\text{ kg}}$
Hence, option C is the correct answer.
Note: When you discharge a bullet from the gun, the gun experiences a backward force. The gun gains velocity in the reverse direction as a result of this force. This is referred to as recoil velocity. Also, if the gun is pushed back when the gun fires the bullet then its velocity will be negative.
Formula used:
Conservation of total momentum –
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]
Here, ${m_1},{m_2}$ are the masses of the body, ${u_1},{u_2}$ is the initial velocities of the bodies and ${v_1},{v_2}$ are the final velocities.
Complete step by step solution:
Given that,
Mass of the bullet, ${M_B} = 50{\text{ }}g = 0.05{\text{ }}kg$
Mass of the gun $ = {M_G}$
Initial velocity of the bullet and the gun is equal to zero (rest). Therefore, the total momentum of the bullet and the gun before the fire will be zero (Initial velocity is zero).
Now, the final velocity of the bullet,
\[{v_B} = 30{\text{ }}m\;{s^{ - 1}}\]
Final velocity of the gun,
\[{v_G} = - 1{\text{ }}m\;{s^{ - 1}}\]
Total momentum after the fire (bullet and gun) $ = {M_B} \times {v_B} + {M_G} \times {v_G}$
$\Rightarrow 0.05 \times 30 + {M_G} \times \left( { - 1} \right)$
$\Rightarrow 1.5 - {M_G}$
Now, the momentum will be conserved. So, the total momentum before the fire and after the fire will be equal
$\text{Total momentum before the fire = Total momentum after the fire}$
$\Rightarrow 1.5 - {M_G} = 0$
$\therefore {M_G} = 1.5{\text{ kg}}$
Hence, option C is the correct answer.
Note: When you discharge a bullet from the gun, the gun experiences a backward force. The gun gains velocity in the reverse direction as a result of this force. This is referred to as recoil velocity. Also, if the gun is pushed back when the gun fires the bullet then its velocity will be negative.
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