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If similar bullets are fired from two guns out which one is lighter and another one is heavier. Which gun will give a more violent jerk?
A) Same by both
B) Lighter one
C) Heavier one
D) Will depend upon the type of gun

Answer
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Hint: To solve this question, we can use the concept, conservation of linear momentum in a system. Here, momentum of the system before firing and after firing should be the same. Initially, the bullet and gun are at rest. Then, according to the conservation of momentum, initial and final momentum will be zero. By converting the momentum in terms of velocity and mass, we can explain which gun will experience more jerk.
Formula used:
Vg=mbvbMg

Complete step by step solution:
Initially before firing, both the bullet and the gun are at rest. Hence, before firing, the initial momentum of the bullet and gun is zero. Since, the total momentum in a system is conserved; momentum of the bullet and the gun before firing and after firing should be zero.
Hence,
Momentum of bullet +Momentum of gun=0
MgVg+mbvb=0
Where,
Mg is the mass of gun
Vg is the velocity of gun
mb is the mass of bullet
vb is the velocity of a bullet.
Then,
Vg=mbvbMg
From the above equation we can see that, mass of gun and velocity of gun are inversely proportional.
Therefore, if the mass of the gun is less then, the velocity of the gun will be more and if the gun is heavier, then its velocity will be less.
Hence, when two similar bullets are fired from two lighter and heavier guns, the lighter gun will give a more violent jerk.

Answer is option B.

Note:
Recoil of a gun means sudden backward movement or jerk of gun when a bullet is fired from it. When a bullet is fired, the force sending the bullet forward is the same as the force sending the gun backward. But, due to the high mass of the gun, it moves only a small distance backward. This force gives a backward jerk or kick to the shoulder of the gunman. This is called the recoiling of the gun.
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