How Does an Exploding Firework Affect Projectile Landing Distance?

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The discussion focuses on the mechanics of a firework shell that explodes at the peak of its trajectory, resulting in two pieces landing at different distances. The smaller piece lands very close to the mortar, while the larger piece lands at a distance d. The center of mass of the shell continues along its initial trajectory, leading to the equation d = (5/4)cm, where cm represents the position it would have landed without the explosion. The participants clarify that d should be expressed in terms of r, which is the hypothetical landing position without the explosion. The conversation emphasizes the relationship between the explosion's effect on landing distance and the conservation of momentum.
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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 the figure) as designed. However, rather than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass 1/5 m and a larger piece of mass 4/5 m. 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 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 .
 

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The center of mass of the shell will continue on the initial trajectory:

cm = ((1/5)m*0 + (4/5)m*d)m

cm = (4/5)d

d = (5/4)cm where cm is the position it owuld have landed at
 
it says express d in terms of r. is this d is in terms of r?
 
kenau_reveas said:
it says express d in terms of r. is this d is in terms of r?

if r is the position it would have landed at
 
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