Surprising Exploding Firework?

Question

A mortar fires a shell of mass m at speed v_0. The shell explodes at the top of its trajectory as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass 1/5m and a larger piece of mass 4/5m. Both pieces land at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands a distance d from the mortar. If there had been no explosion, the shell would have landed a distance r from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.

Find the distance d from the mortar at which the larger piece of the shell lands.

Express d in terms of r.

Answer 1

If the shell had traveled the distance r, then the horizontal displacement was v_0*cos(th)*t=r

since the explosion, which caused an inelastic collision, occurred at apogee, we use conservation of momentum:
v_0*cos(th)*m=v1*m/5-v2*4*m/5

and, since the small piece lands basically on the mortar,
v2=v_0*cos(th)
plugging in
v_0*cos(th)=v1/5-v_0*cos(th)*4/5
or
v_0*cos(th)*9/5=v1/5
from above
v_0*cos(th)*t=r
and, since the two fragments land at exactly the same time,
v1/5*t/2+v_0*cos(th)*t/2=d

using the expression from above that
v_0*cos(th)*9/5=v1/5
v_0*cos(th)*9/5*t/2+v_0*cos(th)*t/2=d
or
v_0*cos(th)*t*(9/10+1/2)=d
v_0*cos(th)*t*14/10=d

now use
v_0*cos(th)*t=r
r*14/10=d
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