# Water Melon (momentum problem)

1. Jun 13, 2007

### ClzdCaptions

1. The problem statement, all variables and given/known data
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.

2. Relevant 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

3. 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

2. Jun 13, 2007

### Dick

Good job so far! And well presented! The problem is your assumption that all of the energy going in (KE bullet initial) is transferred to (KE jet) and (KE bullet final). If this were the case then WHERE DID THE FINAL KINETIC ENERGY OF THE MELON COME FROM??? That can't work. I think you'll have to leave the final velocity of the bullet and of the melon both as unknowns and solve your momentum and KE relations simultaneously. This means you are assuming the negligible energy went into actually tearing the melon apart - but that might be ok. Melons are pretty soft.

3. Jun 13, 2007

### ClzdCaptions

I think I understand what you're saying...The first assumption neglected the final velocity of the second melon- but didn't I account for that with what I did under the label "(second assumption...)?"

4. Jun 13, 2007

### Dick

Come to think of it, it does look like it does. So you have answers of 8.762 and 8.71. They aren't that far off from each other. Do you happen to know what the answer you are looking for is? I can try and check it tomorrow.

5. Jun 13, 2007

### ClzdCaptions

I don't know the answer- but the last part of the question states:
"Compare this to a typical measured recoil speed of about 1.60 ft/s."
So perhaps something around that?

6. Jun 13, 2007

### Dick

Ok, I can confirm that if I follow your second approach (which seems right to me), I get 8.706 for the final velocity of the melon. Go figure.

7. Jun 13, 2007

### Dick

BTW, using 8 ft/sec or so as the final velocity of the melon tells you that the kinetic energy of the melon is less than 100 kg*ft^2/sec^2 (gotta hate those units). Which is pretty negligible compared with KE of the bullet and jet. Which is why your two approaches come pretty close to agreeing.

8. Jun 13, 2007

### ClzdCaptions

The answer is -5.98×10 ^-1 m/s or -.598 m/s...
I don't know how to get that though.

9. Jun 13, 2007

### Dick

I don't either. Why is the answer stated in m/sec when all of the distance units in the problem are in ft?

10. Jun 13, 2007

### ClzdCaptions

Not sure- probably just standard for the system.

11. Jun 14, 2007

### ClzdCaptions

Well, I found out how to do it...
Apparently it's just P(bullet) = P(melon) + P(jet)
where P(bullet) = (.0107)(1766)
P(melon) = (2.73-.129)(V2f)
P(jet) = (.129)(186.183)

I'm just not sure why the momentum of the bullet afterwards is neglected.

12. Jun 14, 2007

### Dick

It looks like the assumption is being made that the bullet doesn't exit the melon and that the 86.6% of it's kinetic energy went into scrambling the melon (ultimately - heat). I don't think this solution is any more correct than yours until the question is rewritten to include info about the fate of the bullet.