An artillery piece of mass M₁ fires a shell of mass M₂ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

Solution

Physics sir · 3 steps

Both get the same momentum, so write KE as p²/2m and the heavier one ends up with less energy.

1. Momentum is shared. Before firing, the gun and shell are at rest, so the total momentum is zero. Just after firing, the gun recoils with momentum equal in size to the shell's: M₁ v₁ = M₂ v₂ = p.
2. Write KE using p. Kinetic energy is 1/2mv² = p²/2m. Since p is the same for both, each kinetic energy is inversely proportional to its mass.

   K_(artillery)/Kₛₕₑₗₗ = (p²/2M₁)/(p²/2M₂) = M₂/M₁

3. So. The ratio is M₂/M₁, which is option (b). The heavy gun takes only a small share of the energy, and the light shell carries almost all of it.

The idea to keep. Whenever two bodies push apart from rest, their momenta are equal and opposite. Then K = p²/2m tells you the lighter body always carries more kinetic energy, in the inverse ratio of the masses.

Answer from the NTA final answer key. Solution by PrepWiser.