Momentum & Kinetic Energy MC *quick*

mhz

1. The problem statement, all variables and given/known data
Two objects, X and Y, are held at rest on a horizontal frictionless surface and a spring is
compressed between them. The mass of X is 2/5 times the mass of Y. Immediately after the
spring is released, X has a kinetic energy of 50 J and Y has a kinetic energy of:
A. 20 J
B. 8 J
C. 310 J
D. 125 J
E. 50 J

2. Relevant equations
p = p'
E_kinetic = 1/2 mv^2

3. The attempt at a solution
Alright, if I use p = p' to set up an explosion/recoil equation where we have two variables v1' and v2' then I can use the kinetic energy equation to find v1' (50 = 1/2 (2/5) v1^2) and then plug this value into the p = p' equation, finding v2'. I then use v2' to find it's kinetic energy which should be the answer to the question and I get 20 J.

Apparently this is incorrect. How?

Please provide proof as to how I could possibly be wrong. Thanks :)

LawrenceC

Taking your answer of 20J, work the problem backwards to see if you come up with 50J for the KE of mass x.

.5M1V1^2 + .5M2V2^2 = 70

From momentum you can relate V1 and V2. Solve for KE1 which is quantity .5M1V1^2. If you get 50, your answer must be correct.

Steely Dan

Kinetic energy of the two blocks is not conserved, because of the compressed spring. Only conservation of momentum is legitimate to use to solve this problem.

Doc Al

I'd say that your answer is correct. Why do you think it's wrong?

mhz

This PDF file has the question on it, #77. It says the answer is D, 125 J.

http://myfizika.ucoz.com/_ld/0/43_Test_Bank_9.pdf

@ Steely Dan: However it says immediately after the spring is released, the kinetic energy is 50 J so wouldn't the spring energy be entirely transferred at this point? I don't think that's relevant. If it is, how could it affect the answer?

@ Lawrence: All the Math works out correctly forward and backward.

Doc Al

I'd say that that answer is wrong. After all, X is the lighter and thus faster mass so it must have greater KE.

It looks like they mixed up X and Y. (Try solving it assuming: The mass of Y is 2/5 times the mass of X.)

castro94

i would say that they mean that beacuse of the frictionless surface both gain the same velocity. if they both have the same velocity but different mass , the one with the higher mass will have more kinectic energy.
and the answer 125 sounds then logical. if the velocity is a constant then the differense in kinectic energy is direclty proportional to the mass.
if the kinectic energy of X with mass 2/5 is 50 j , then 1/5 of the mass is 25 j, 25 j times 5 is 125 giving the kinectic energy of Y with "5/5" mass

Doc Al

That would contradict momentum conservation.

LawrenceC

The above is wrong. The spring pushes on each with the same force because the ma of one is equal and opposite the ma of the other. Therefore the heavier mass will not attain as high a velocity due to a lower acceleration rate.

castro94

Im just saying that this is the only way the answer makes sence.
besides the way i understand it the conservation rule of mommentum applies when two objects collide , but here there is a third object , the spring , that is resposible for the eenergy provided.
( i just resently got aquinted with the momentum , so i may have understood it wrong , im just saying that it makes sence and seems to be right according to the answer given )

Steely Dan

Well, the assumption you must make here to use momentum conservation is that the spring comes to rest after the blocks are released (or that the momentum of both sides of the spring is equal and opposite, or that the momentum is negligible). If you do that, then since the spring was also at rest to begin with, you can use the approach of equal and opposite momenta to work out the final energy of block Y.

Edit: Actually, now that I've thought about it more, I guess the only valid assumption, if you want to get an answer, is that the momentum of the spring is negligible.

castro94

well , i think that this cannot be the case here. If you think about the objects as two fallen objects in vacuum ( as the surface is frictionless ) , the force acting on them is the same , gravitational pull ( anloguos to the energy given from the spring ) . The way you put it, it seems to me that you mean that two fallen objects will gain speed at different levels in vaccum , which is not the case. following from this order of ideas the two objects will accelerate at the same rate

Steely Dan

It is possible for two objects to have different forces on them and to still accelerate at the same rate. This is the essence of the equivalence principle, after all. The object with more mass will have a larger gravitational force on it, but in its motion this is exactly compensated by the lesser propensity to be accelerated.

castro94

I agree , i just used the simplified version of newtons law of motion. but from this it will still follow that both balls will gain the same velocity

mhz

That's what I was thinking happened, I noticed that* too.

Doc Al

But your answer violates the laws of physics! So what does it matter that it gives you 'the answer'?

Doc Al

You didn't use any version of Newton's laws of motion. You just assumed that the velocities would be the same.