# Homework Help: Cant get problem (momentum)

1. Nov 30, 2005

### B-80

An explosion breaks an object into two pieces. One of which has 1.5 times the mass of the other. If 7500 joules were released in the explosion, how much kenetic energey did each piece require?

ive neem working on this for two days now, apparently its not that hard, but i cant get it so far, can someone pleace explain this two me

Heavier Piece: 4500 Joules
Lighter Piece: 3000 Joules

2. Nov 30, 2005

### tony873004

3000 * 1.5 = 4500

4500 + 3000 = 7500

3. Dec 1, 2005

### B-80

i know but where do i getr 3000 from

4. Dec 1, 2005

### Staff: Mentor

Hint: Apply conservation of momentum.

5. Dec 1, 2005

### emptymaximum

i don't think momentum helps here because it's about kinetic energies, and i don't think it's at the level where they would take dKe/dv to get momentum. it's just as easy to use energy anyways.

you have two objects that have some kinetic energy. let's call it Ke1 and Ke2.
the total kinetic energy of the system is: KE = ?
since KE is given, you only need the forumals for Ke1 and Ke2.
how are Ke1 and Ke2 related?

6. Dec 1, 2005

### Staff: Mentor

conservation of momentum is required

Sorry... but you can't do this problem without using conservation of momentum. (That's how you find the relationship between the speeds of the two pieces.)

7. Dec 1, 2005

### B-80

well even if you do conservation... then you'll still have Va' and Vb' as unknowns how can you solve with 2 unknowns

Last edited: Dec 1, 2005
8. Dec 1, 2005

### Staff: Mentor

You'll actually have only one unknown, since conservation of momentum will allow to write Vb' in terms of Va'. Then you can use the given total KE to solve for the KE of each piece.

Why don't you start by writing out the conservation of momentum and see what it tells you. (Hint: What's the total momentum?)

9. Dec 1, 2005

### B-80

it just seems like it can be simpler, i mean once i look a it i can tell that masss a is 1.5 time mass b, so work should be as well, but i dont know what to do to 7500 to do this

10. Dec 1, 2005

### mybsaccownt

if you want to take an algebraic approach, which it seems like you do

the amount of energy that the first chunk took plus the energy that the second one took (same energy of the first chunk*1.5) = 7500 right?

so, let's call the energy they took x

the small chunk's energy x + 1.5x = 7500

so solve for x, then once you have x, ...there's just one small step left

11. Dec 1, 2005

### mybsaccownt

but why would you want to do it that way? you don't really learn anything about physics that way

same for a problem like...

a car of mass 1500kg with a wheelbase of 3m and center of mass at 1.2m from the front

i solved the problem using torques, then i had a lot of time left over and i checked it algebraically by finding that the mass towards the front was 60% and the rear 40%, taking 60 and 40 % of mg gave the same answer as solving with torque and vertical forces

but so what? it didn't teach me anything, it just helped me check my answer

i suggest you solve your problem using physics, then if you like you can check it algebraically (by then it wouldn't be necessary though...so i don't see the point)

whatever, rant rant rant lots of words

lol

12. Dec 1, 2005

### Staff: Mentor

That the energy of one is 1.5 times the energy of the other is certainly true, but I'd prefer that B-80 didn't just assume this to be true but instead proved it using physics. (How do you know that it's true? What law of physics is invoked to prove it?)

13. Dec 1, 2005

### mybsaccownt

are you asking me how i know it's true?

or is that directed at the original poster?

14. Dec 1, 2005

### emptymaximum

isn't conservation of energy make it be true? and T = Integral[momentum,dv]? i guess it is momentum then.

you don't need a relation with the speed because there's a relation with the kinetic energy (i know those two are themselves related. see above.). i would just worry about that. that's one of the reasons people have so much trouble with this stuff.

15. Dec 1, 2005

### mybsaccownt

it's true because

KE1 + KE2 = KEtotal

KE1 = mv^2/2

KE2 = 1.5mv^2/2

so KE2 = 1.5KE1

KE1+1.5KE1 = KEtotal

16. Dec 1, 2005

### emptymaximum

that's what i said. 10 posts ago.

17. Dec 1, 2005

### mybsaccownt

no, mr. liar, you didn't say it

you started down that path, but you didn't keep walking

i know that's because you wanted the OP to solve it himself, but then in your 2nd last post i think you might have confused him/her more, so i clarified it

18. Dec 1, 2005

### emptymaximum

no, my second to last post was directed to the guy pushing momentum.
no need for namecalling. i'm on your side on this.

19. Dec 2, 2005

### mybsaccownt

ah i see, i thought you were implying my posts had been useless

i wasn't really serious anyway, who addresses someone as mr.liar? lol

anyways, sorry if i offended you

Last edited: Dec 2, 2005
20. Dec 2, 2005

### Staff: Mentor

Of course.

In my opinion, showing that this is true is the key step of the problem; the rest is arithmetic.

21. Dec 2, 2005

### Staff: Mentor

I guess you mean me--the guy "pushing momentum". :rofl: I suppose it's just a mere coincidence that the the title of this thread has the word "momentum" in it.

The only physics in this problem is showing that KE1 = 1.5 KE2; that requires conservation of momentum.