A mortar fires a shell of mass at speed . The shell explodes at the top of its t

Question

A mortar fires a shell of mass at speed . The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass and a larger piece of mass . 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 from the mortar. If there had been no explosion, the shell would have landed a distance from the mortar. Assume that air resistance and the mass of the shell's explosive charge are negligible.
Find the distance from the mortar at which the larger piece of the shell lands.
Express in terms of r.

Explanation / Answer

Ans: We know this because it broke apart at the apex of the trajectory, which is a parabola. So we have v = d/t p = mv 1/5 m * -1/2 r/t = p_1 4/5 m * (d - 1/2 r)/t = p_2 The sign of p_1 is negative because the smaller fragment traveled backward. We have (d - 1/2 r) in the equation for p_2 because that's the distance from the explosion to the landing point of the larger fragment. p_1 + p_2 = p = 1/2 mr/t -1/10 mr/t + 4/5 md/t - 2/5 mr/t = 1/2 mr/t -1/2 r + 4/5 d = 1/2 r 4/5 d = r d = 5/4 r

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