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Homework Statement
A counterintuitive physics demonstration can be performed by firing a rifle bullet into a melon. (Don't try this at home!) When hit, nine times out of ten the melon will jump backward, toward the rifle, opposite to the direction in which the bullet was moving. (The tenth time, the melon simply explodes.) Doesn't this violate the laws of conservation of momentum? It doesn't, because we're not dealing simply with a two-body collision. Instead, a significant fraction of the energy of the bullet can be dumped into a jet of melon that is violently ejected out of the front of the melon. This jet can have a momentum greater than the momentum of the bullet, so that the rest of the melon must jump backward to conserve momentum. Let's make the following assumptions:
1. The mass of the melon is 2.73 kg.
2. The mass of the rifle bullet is 10.7 g and its velocity is 1766 ft/s.
3. 13.4 percent of the energy of the bullet is deposited as kinetic energy into a jet flying out of the front of the melon.
4. The mass of the matter in the jet is 0.129 kg.
5. All collisions occur in a straight line. What would be the speed of the melon's recoil? Compare this to a typical measured recoil speed of about 1.60 ft/s.
Homework Equations
Pi = Pf = m1v1i + m2v2i = m1v1f + m2v2f
Ei = Ef = 1/2 m1v1i^2 + 1/2 m2v2i^2 = 1/2 m1v1f^2 + 1/2 m2v2f^2
The Attempt at a Solution
In my attempt, I identified the 3 objects- the bullet, the melon, and the jet of melon- as m1, m2, and m3 respectively.
Data:
m1 (bullet) = .0107 kg v1i = 1766 ft/s v1f = ??
m2i (melon) = 2.73 kg m2f = 2.601 v2i = 0 ft/s v2f = ??
m3f (jet) = .129 kg v3f = ?? (initial values = 0)
(conservation of momentum...melon and jet have no momentum before...)
m1v1i = m1v1f + m2fv2f + m3fv3f
(.0107)(1766) = (.0107)(v1f) + (2.73)(v2f) + (.129)(v3f)
(13.4% of the bullet's energy is deposited as kinetic energy into the jet)
KE(bullet)i = (1/2)(0.0107)(1766^2) = 16685.34
KE(jet) = (.134)(1/2)(0.0107)(1766^2) = 2235.84
KE(jet) = (1/2)(.129)(v3f^2) = 2235.84 ; v3f^2 = 186.183
(first I assumed...)
KE(bullet)f = KE(bullet)i - KE(jet) = 14449.51
14449.51 = (1/2)(.0107)(v1f^2) ; v1f = 1643.42
(so...)
m1v1i = m1v1f + m2fv2f + m3fv3f
(.0107)(1766) = (.0107)(1643.42) + (2.601)(v2f) + (.129)(186.183)
v2f = -8.762 ft/s
...which is wrong. (it asks for speed, but 8.762 is wrong as well)
(second assumption...)
KE(total) = KE(bullet)i = KE(bullet)f + KE(melon) + KE(jet)
/ KE(bullet)f + KE(melon) + (.134)KE(bullet)i
/ (.866)KE(bullet)i = KE(bullet)f + KE(melon)
/ 14449.51 = (1/2)(.0107)(v1f^2) + (1/2)(2.601)(v2f^2)
m1v1i = m1v1f + m2fv2f + m3fv3f
(.0107)(1766) = (.0107)(v1f) + (2.601)(v2f) + (.129)(186.183)
(I combine the two above equations to solve for v2f...)
where... v1f = [(.0107)(1766) - (.129)(186.183) - (2.601)(v2f)] / (.0107)
/ = -478.636 - 243.084(v2f)
v2f = -8.71 ft/s
so...I'm getting nowhere...
Thanks in advance for any help