Question

A man fires a bullet of mass 200g at a speed of \[5m{{s}^{-1}}\]. The gun is of 1 kg mass. By what velocity the gun rebounds backwards?
A) 0.1\[m{{s}^{-1}}\]
B) 10\[m{{s}^{-1}}\]
C) 1\[m{{s}^{-1}}\]
D) 0.01\[m{{s}^{-1}}\]

Answer 1

Hint: We need to find the relation between the masses of the gun and the bullet and the velocities of the bullet and gun before and after firing the gun to find the required solution. The conservation of momentum is to be considered for solving this.

Complete step-by-step answer:
We are given a system of a gun which is firing a bullet of mass 200 g with a speed of \[5m{{s}^{-1}}\]. The mass of the gun is given as 1 kg. The gun is fired and the bullet moves out of the gun. According to Newton's third law of motion, the forward movement of the bullet from the gun causes the gun to recoil. We can use the conservation of linear momentum to find the velocity of the gun moving backwards.
Initially, the gun is at rest and the bullet is only moving. Finally, once the bullet is out of the gun there is no forward movement, so the final velocity of the bullet will be zero. It is given as –
\[mu+MU=mv+MV\]
Where, m is the mass of the bullet,
M is the mass of the gun,
u and U are the initial velocity of the bullet and gun respectively,
v and V are the final velocities of the bullet and gun respectively.
We can substitute the values as –
\[\begin{align}
  & mu+MU=mv+MV \\
 & \Rightarrow (0.2)(5)+0=0+(1)V \\
 & \therefore V=1m{{s}^{-1}} \\
\end{align}\]
The velocity of the gun when it recoils is 1 \[m{{s}^{-1}}\].

So, the correct answer is “Option C”.

Note: The Newton’s third law of motion states that every action has an equal and opposite reaction. The recoiling of the gun is the direct consequence of the forward movement of the bullet. The conservation of linear momentum is valid for such situations also.