Momentum Question 3

A cannon of mass 1200 kg stands at rest on smooth, level ground. It fires a shell of mass 15 kg horizontally at 240 ms^-1.(i) Calculate the recoil velocity of the cannon.(ii) Compare the kinetic energies of the shell and the cannon just after firing.(i) Before firing nothing is moving, so the total momentum is zero. No external horizontal force acts, so the total momentum after firing must also be zero. Take the direction of the shell as positive:0 = (15 × 240) + 1200vv = -3600 ÷ 1200 = -3.0 ms^-1The minus sign shows that the cannon recoils backwards, opposite to the shell, at 3.0 ms^-1. The shell and the cannon have equal and opposite momenta of 3600 kg ms^-1.(ii)Shell: E_k = ½(15)(240)² = 432 000 JCannon: E_k = ½(1200)(3.0)² = 5400 JThe momenta are equal in size, but the shell has 80 times the kinetic energy of the cannon. Since E_k = p²/2m, for the same momentum the lighter object gets far more kinetic energy.The total kinetic energy has increased from zero, so kinetic energy is not conserved in an explosion. The energy comes from the chemical energy stored in the explosive. Momentum, however, is still conserved.