# Energy:Momentum relationship

1. Oct 26, 2008

### avenkat0

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

Car A has a greater mass than car B. Let:

- KA = the kinetic energy of car A
- KB = the kinetic energy of car B
- pA = the magnitude of the momentum of car A
- pB = the magnitude of the momentum of car B
- vA = the speed of car A
- vB = the speed of car B

a) If car A and car B have the same kinetic energy (KA = KB), then

2. Relevant equations
p=mv
Ke= .5(m)(v^2)

3. The attempt at a solution
I tried to set up a situation:
Car A- 10 kg Car B- 5kg Ke=20 J
.5(10)(v^2) = 20 .5(5)(v^2)=20
v= 2 m/s v= 2.828 THEREFORE Vb > Va
Pa=10(2) = 20 Pb=5(2.828) THEREFORE Pb < Pa

But the selection Pb<Pa and Vb>Va is wrong
Is there a flaw in my reasoning

Last edited: Oct 26, 2008
2. Oct 26, 2008

### avenkat0

with Ma=2 and Mb=1 and Ke=10
im getting

Va=3.162
Vb=4.47

MVa=6.92
MVb=4.47

Vb>Va and MVb<MVa

same result as earlier and its wrong

3. Oct 26, 2008

### Hootenanny

Staff Emeritus
I deleted my previous post because you edited your post after I had replied and therefore my previous post didn't make sense.

4. Oct 26, 2008

### avenkat0

sorry about that it was a typo

5. Oct 26, 2008

### borgwal

Your answers look fine. You should, though, try to prove the relations in general, not just for particular values of mass, speed, etc.

6. Oct 26, 2008

### avenkat0

But i tried graphing the m: ke function and it shows that when the mass increases the Velocity decreases
and when the mass increases the momentum increases....
im confuzed haha what am i doing wrong here

pB < pA and vB > vA

was the answer i selected and it came out to be wrong

7. Oct 26, 2008

### borgwal

pB<pA and vB>vA are the correct answers. Who says it's wrong?

8. Oct 26, 2008

### avenkat0

the professor... thanks for your help now that i know that i have someone to confirm it i guess i can email him

9. Oct 26, 2008

### borgwal

good luck :-)