# Free Response Momentum Problem

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1. May 20, 2015

### Puddles

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

(a.)Apply the law of conservation of momentum to the perfectly inelastic collision of a moving object of mass m1 and velocity vi with a stationary object of mass m2
(b.) Solve this for final velocity vf
(c.) Write a formula for the initial kinetic energy (KEi)
(d.) Write a formula for the final kinetic energy (KEf) in terms of vf
(e.) Substitute your formula for vf from (a.) into your equation above to get a formula for KEf in terms of m1, m2, and vi alone
(f.) Find the ratio KEf/KEi in terms of m1 and m2 alone
(g.) Express the final kinetic energy in terms of the initial kinetic energy KEi
(h.) For the case of a moving object much, much more massive than the stationary object (m1 >> m2), how much kinetic energy is lost in the collision?
(i.) For the case of a stationary object much, much more massive than the moving object (m2 >> m1), how much kinetic energy is lost in the collision?
(j.) Given two projectiles fired with the same KE, fired at a target which is free to move, which will cause the most destruction (that is, result in the greatest loss of kinetic energy), a small high-speed projectile or a massive slow-moving one?

2. Relevant equations

Momentum: P = m x v
Kinetic Energy: KE = 1/2mv^2

3. The attempt at a solution

(a.) P = M1V1 + M2V2 = (M1+M2)Vf

(b.) Vf = M1V1/(M1+M2)

(c.) Cons. Energy, KEi = KE1 + KE2, KEi = 1/2(M1V1^2) + 1/2(M2V2^2)
KEi = 1/2(M1V1^2)

(d.) 1/2 (M1 + M2) Vf^2

(e.) Here's where my confusion starts, (M1+M2)Vf = M1V1 + M2V2 and 1/2(M1+M2)Vf^2, I tried
1/2(M1+M2)((M1V1+M2V2)/(M1+M2))^2, substituting in for Vf but I'm not sure if that's right

(f.) I'm not even sure how to resolve this from the last piece of information, because I can't set M1 and M1 or M2 and M2 equal to each other from my previous formulas, can I?

Since I'm stuck there I couldn't make an attempt on the rest yet.

2. May 20, 2015

### grzz

Note that, since M2 is initially at rest, V2 = 0.
Hence use this in part (e).

3. May 20, 2015

### Puddles

Okay, so then I've got...
.5(M1 + M2)((M1V1)/(M1 + M2))^2

So then (f.) is...
(.5(M1 + M2)((M1V1)/(M1 + M2))^2) /
(.5(M1V1^2)), which can be simplified, cancel the .5's, a (M1 + M2), and (M1V1)

Okay I got (M1)/(M1 + M2)

How do I use that to get KEf in terms of KEi now? It's KEi times ((M1V1)/(M1 + M2))^2?

4. May 20, 2015

### grzz

You got
KEf/KEi = M1/(M1 + M2)

hence

KEf = ( M1/(M1 + M2) )KEi.

5. May 21, 2015

### Puddles

Oh, right, thank you. Sometimes I fail to make simple connections in physics, I really appreciate your help.

For (h.) and (i.) then should I just be treating M1 or M2 as zero depending on which part I'm doing because it's got comparatively insignificant mass to the other? So for (h.) for example I would use KEf = (M1(M1 + M2))KEi = (M1(M1 + M2)(.5M1V1^2), where if M1 >> M2, M2 = 0, then compare it to the same formula when M1 = M2? As I'm reading that I know it's not right…

Okay, if I'm trying to find lost KE I should set KEi = KEf and then look at the difference between the two to find what's lost right? Using the idea from above where if one mass is smaller I'm treating it as insignificant? Or I feel like that's no right either, but how do I go about working with one mass as though its smaller than the other without assigning values to the masses?

6. May 21, 2015

### grzz

For part (h),

imagine a very large heavy truck moving and colliding with a very small car. Do you think the initial KE of the truck will decrease much?

7. May 21, 2015

### Puddles

Well no, I'd imagine there would be basically no energy lost, and that the reverse would be true for (i.), but I wasn't sure if it was necessary to show some kind of work behind this thought, maybe just plug in 10^5 vs. 1, or for a question like that is it unnecessary to prove this since it's a simple deduction? I guess that's dependent on my teacher unless it's a point most would consider rudimentary to the point a proof is unnecessary.

Also for (j.) I believe it would be a small-fast moving projectile but I'm not sure if I need to do something to show this.

8. May 21, 2015

### grzz

You imagined correctly.

I think that as for (h) and (i) you do not have to substitute any numbers. Just consider KEf = ( M1/(M1 + M2) )KEi.

Then ignore the smaller mass when compared to the bigger one in (h) and in (i) and make the obvious deductions as regards what happens to the KEf and KEi.

9. May 21, 2015

### Puddles

Okay, awesome. Thank you so much for your help! I really appreciate it.