A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 \(m\,s^{-1}\), what is the recoil speed of the gun ?
Solution and Explanation
Mass of the gun, M = 100 kg
Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = \(mv-MV\)
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
\(mv-MV\)= 0
\(\therefore V\) = \(\frac{mv}{M}\)
= \(\frac{0.020\times 80}{100\times 1000}\)
= 0.016 m/s
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.