A mortar fires a shell of mass m at speed v0. The shell explodes at the top of i

Question

A mortar fires a shell of mass m 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, a smaller piece of mass 15m and a larger piece of mass 45m. 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.

Explanation / Answer

Considering the effect of conservation of momentum, the centre of mass, COM , of the two pieces os shell will reach the ground at some distance ,say r from the mortoar location.
Since the shell broke off at its heighest point, the shell(COM) has covered half the horizontal distance of the total distance it will cover.
So we can say that shell broke apart at the apex of trajectory, which is a parabola. hence,
v = d/t
and . p = mv :
p_1 = 1/5m*(-0.5r/t)
p_2 = 4/5m *(d-0.5r)/t
Note the sign of p_1 is negative , this is because the smaller fragment travelled backward.

now, p_1 + p_2 = 1/2 *m r/t
-1/10 mr/t + 4/5md/t - 2/5mr/t = 0.5 m r/t
-1/10 r + 4/5d - 2/5 r = 0.5 r
4/5 d = r
d = 5/4 r

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