As another example of conservation of momentum, consider the recoil of guns.
ANALYSIS
Initially the projectile is at rest in the gun. The charge explodes, pushing the projectile from the gun and at the same time pushing back with the same force on the gun. Since no external forces act during this explosion, the sum of the momenta of the projectile forward and the gun backward must be the same as the initial momentum of the system, i.e., zero. Thus,
m_{p} v_{p}+m_{g} v_{g}=0
where m_{p}, m_{g}= masses of projectile and gun, respectively
Solving for speed of recoil, v_{g}=-\left(m_{p} / m_{g}\right) v_{p} . The minus sign indicates that the gun moves in the opposite direction to the projectile.
The greater the mass of the gun, the less will be the speed of recoil. Hence, the energy to be absorbed by the recoil spring or other devices will be correspondingly less. Of course, the weight is limited by the need for mobility.