# Homework Help: KE in terms of P Pretty sure Ive got most of it

1. May 29, 2007

### MJC8719

An explosion breaks an object into two pieces, one of which has 1.60 times the mass of the other. If 7100 J were released in the explosion, how much kinetic energy did each piece acquire?

Heavier Piece

Lighter Piece

Express KE in terms of P

I think I have done most of the work here...its the last part of the problem thats tripping me up

Heres my work:

The explosion causes the piece (which I'm assuming explodes while at rest) to break into two pieces. Conservation of momentum states that they must go in opposite directions (you can verify this on your own). So, we have our first equation:

mv1 + 1.6mv2 = 0
mv1 = -1.6mv2
v2 = -v1/1.6

I use v1 for the velocity of the lighter piece, and v2 for the heavier piece. The energy of the explosion is transferred to the pieces and causes it to move, so the sum of the kinetic energies of the two pieces combined must equal 7100 J. Our second equation is:

1/2mv12 + 1/2 (1.6m)v22 = 7100

If we substitute v2 with what we got before with momentum conservation, then we can get the left hand side to be in terms of m and v1 only. We can't find the values of m and v1 directly, but we know that the kinetic energy (of the lighter piece) is 1/2mv12, which we can find after rearranging the equation so that it is in the form of kinetic energy:

1/2mv12 + 1/2(1.6m)(-v1/1.6)2 = 7100 J
1/2m(v12+v12/1.6) = 7100 J
1/2m(1.625v12) = 7100 J
1/2mv12 = 4370 J

Now heres where it starts to get a little tricky:

The final line reads: 1/2 mv1^2 = 4370.

So, we also then know that K = 1/2mv^2 = (1/2)(m^2v^2/m) = p^2/2m

Therefore, p^2/2m = 4370
So, the energy for lighter piece would then equal 4370 - p^2/2m and thus the energy for the heavier piece would just be 1- that.

Is this correct? I am slighty confused by the fact that my answer has to be in terms of P.

Thanks for the help

2. May 29, 2007

### Staff: Mentor

Try this. You know how v_1 and v_2 compare to each other. Use that to compare KE_1 to KE_2. Then use the total KE to find KE_1 and KE_2 explicitly.

3. May 29, 2007

### MJC8719

Ok,

So if i understand you correctly you are saying that we know v2 = -v1/1.6.

We also know that KE1 + KE2 = 7100J

So, KE1 = 1/2mv1^2 and therefore KE2 = 1/2(1.6m)v2^2 but we know that v2 = -v1/1.6 so therefore we can say that

KE2 = (1/2)(1.6m)(-v1/1.6^2)

So we can then have the equation
1/2mv1^2 + (1/2)(1.6m)(-v1/1.6^2) = 7100

Is this correct so far??

If we continue this we would get
1/2 mv1^2 + (0.8m)(v1^2/2.56) = 7100

1/2 mv1^2 + .31250 v1^2 = 7100

p^2/2m + .31250v1^2 = 7100

And this is where i get stuck if i have done the above correctly.... your help is really appreciated

4. May 29, 2007

### Staff: Mentor

So what is KE2 in terms of KE1?

(Forget about expressing KE in terms of P; that's not helpful here.)

5. May 29, 2007

### MJC8719

Hmm,

So if I understand this then if we simplify
KE2 we find that it equals 0.31250mv1^2.
KE1 then equals 0.5mv1^2.

So if we make KE1 ="1" or KE1 - mv1^2 then KE2 = .0625mv1^2 of KE1.

Is this what you mean. If so, then could you give me a hint as to what the next step is?

6. May 29, 2007

### Staff: Mentor

Now express KE2 in terms of KE1. (Just like you wrote v2 in terms of v1: v2 = v1/1.6.)

7. May 29, 2007

### MJC8719

Suddenly, it clicked Doc...Thanks for the help. I was getting hung up on the whole solve for p part

I think the "write K in terms of p" is just a hint:

total momentum before = total momentum after so

0 = M v - m V where M = 1.60 m so V = 1.60 v

Now total energy:

E total = (1/2) M v2 + (1/2) m V2 or

E total = (1/2) M v2 + (1/2) (M/1.60) ( 1.60 v)2

E total = (1/2) M v2 + 1.60 (1/2) M v2 or

E total = ( 1 + 1.60 ) (1/2) M v2 notice that the last bit is just the KE of the larger piece so

E total = 2.60 K of larger piece so

71000 / 2.60 = K or larger piece or

K of larger piece = 27308 J and the smaller piece gets the rest,

71000 - 27308 = 43692 J

Now i Understand it...though i realizied i added an extra zero and made it 71000 instead of 7100.

Thanks for the help

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