Question

A mortar fires a shell of mass m = 1.79 kg at speed v0. The shell explodes at the top of its trajectory (shown by a star in (Figure 1)) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, the smaller one of which has a mass 0.240 kg . If there had been no explosion, the shell would have landed a distance r =  172 m  from the mortar. Both pieces land at exactly the same time. The smaller piece lands at a distance 30.8 m  . The larger piece lands a distance d from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.

You'll get a rating if you complete the problem and don't just post the formulas. I know the formulas involved but I'm still stuck. Thanks in advance

Answer 1

m = 1.79 kg

m₁ = mass of smaller piece = 0.240 kg

m₂ = mass of larger piece = m - m₁ = (1.79 - 0.240) kg = 1.55 kg

r = 172 m (from mortar)

d₁ = 30.8 m (from mortar) = distance of smaller piece from mortar

d₂ = the distance of larger piece from mortar.

The shell explodes into two pieces of which smaller one lands 30.8 m from the mortar and the other lands at a distance d₂.

However, the center of mass of the two pieces will follow the original trajectory and we can safely assume that the center of mass will be at a distance r from the mortar which is supposed to be the distance without explosion. This is so because the the explosion involved no external forces and the energy was internal and hence center of mass will follow the the original trajectory which shell would have followed if there had been no explosion.

Let us assume the mortar firing position as origin.

Therefore, X_cm = [(m₁*d₁)+(m₂*d₂)]/(m₁+m₂) = position of center of mass = r

d₂ = 193.8632 m