How to Solve Momentum Problems in Collisions

[SEG9585]

Hey all--

I had a test yestrerday dealing with momentum, energy convservation, collision, etc, and there was one question that was marked incorrect, however I feel that I may have done it correctly. Please check over my work against my teacher's solution to see what you guys think...
anyway here's the question:

EDIT: I made a visual for this (dont criticize i did it in MsPaint)
http://eric.segonline.net/physicsexample.jpg

A 2KG block and an 8-KG block are both attached to an ideal spring (k=200 N/m) and are both initially at rest on a horizontal firctionless surface. A 100-gram ball of clay is thrown at the 2KG block, and the velocity of the clay-block system is 3.9 m/s immediately after contact, and before the spring begins to compress.
What is the final velocity of the 8kg block once the spring compresses and then regains its original length.


My method:
Energy and Momentum are conserved:
.5mv^2=.5mv^2 and mv(f) = mv(i)
combining the 2 gives the equation:
V(1F) = (M1-M2)/(M1+M2) * V(1I)
and
V(2F) = 2M1/(M1+M2) * V(1I)
Note-- this derivation works for elastic collisions, combining both kinetic energy AND momentum conservation.
From this, I find V(2)= 1.62 m/s <--- final velocity of 8kg block
in addition, V(1) = -2.28 m/s <--- final velocity of 2kg block/clay

If you stick these back into the original equations, you will find that momentum is conserved from the original clay/2KG block system, and the initial kinetic energy is the same.


However, my teacher provided this response:

K(i)=V(f)+(K(f)
When spring returns to original length:
V(8KG block) = V(2kg block/clay)

.5(M(clay/2kg block))v^2 = 0 + .5(M(clay/2kgblock) + M(8kgblock))v^2
from this, he finds the final velocity of the 8kg block to be 1.78 m/s. In addition, he claims that when the spring is not compressed, it acts the same way as a rope, meaning velocities are the same.


Which one is more accurate?
Thanks for any input.
Eric

 

[Chi Meson]

I think that you are correct. THe ideal spring and frictionless surface is making this essentially a perfectly elastic collision: simulatneaous conservation of momentum and kinetic energy.

THe statement that "both velocities would be the same when the spring returns to the uncompressed position" is not correct, just as it would be not correct to say that two perfectly elastic balls would be at the same velocity the moment the collison is through. If the two blocks had the same final velocities, the problem would be simple: p_initial = (2.1 + 8) (v_final) which would make the final speed .81 m/s.

Furthermore, when an ideal spring is not compressed nor stretched, it does not act like a rope, it acts like nothing since there would be no restoring force at all: no tension, no compression. It's exactly like nothing is there.

Anyway, I can't figure out what this is supposed to mean: "K(i)=V(f)+(K(f)"
is it a typo? Or is someone seriously trying to add a velocity to a kinetic energy?

I just looked at your teacher's solution again. Even if final velocities are the same, you do not determine this final velocity through conservation of kinetic energy, it's found through conservation of momentum.

I no longer think you are correct. You are definitely correct.


Be nice to your teacher when you give him this information. It's a hard job.

 

[SEG9585]

I'm sorry, what it meant to say was "K(i)=U(f)+K(f)" as in, initial Kinetic energy = final potential energy + final kinetic energy (final PE being 0)

 

[Doc Al]

Maybe I'm missing something, but I think your teacher has been smoking the wacky weed.

Your method makes sense to me. Your teacher, for some unknown reason, assumes that the velocities of the two masses will be equal once the spring expands back to it's original length. Nonsense!

Here are a few ides for showing him the error of his ways:

First, when you bang into this system, it's not going to compress, expand back to it's original length, and mosey along at the original length. The system will oscillate as it's center of mass moves along.

Second, consider conservation of momentum (like you did). The momentum remains the same throughout this collision/compression/expansion. You know the momentum is (2.1)x(3.9)= 8.19. If he claims that the "final" speed of both masses is 1.78, then the momentum must be: (10.1)x(1.78)= 18. I don't think so!

Third, he thinks an expanded spring is like a string? Yikes! I think he's confusing this with the fact that when the spring fully compresses the two masses move as one, with the same velocity--momentarily!

 

[Chi Meson]

sorry, I forgot to mention the obvious proof of his wrongdoing:

The final momentum, according to his method, is 18 kgm/s. That is more than twice the initial momentum of 8.19 kgm/s. THis is so blatently and obviously wrong.

I will reserve passing judgment, but how long has this guy been a physics teacher? This is so wrong that I would question everything else on your test. Wanna run any other questions by this forum?

 

[SEG9585]

Yeah, I couldn't believe his response either--
He is a very good teacher -- regarded as one of the best in the school, an MIT graduate. The class is AP Physics by the way... The thing is, I also have the physics mind (Aerospace engineering is my intended major), and I look at things logically from different perspectives, which maybe he didnt see when he gave this solution? Anything else I get wrong on tests, usually comes from stupid mistakes that I agree I messed up on.

One thing I presented to him, was that with his given final velocity, I used that V to solve for the original momentum in the system (which you just presented to me also), and showed him that the momentum was not conserved. In addition, he added that the collision is not elastic, and some KE was lost in the spring. Being an "ideal spring", I couldn't agree with this. Even after this response, I didnt understand why he would compare final and inital energy to begin with if he thought this was true.
Thanks for the response, hopefully I can get my 5 points back hehe.

 

[Doc Al]

Originally posted by Chi Meson
Be nice to your teacher when you give him this information. It's a hard job.

Apparently, for some it's too hard!

 

[Doc Al]

I think it finally dawned on me what your teacher was thinking and the source of his error. It is certainly true that (post collision) energy is conserved:

KE_initial = KE_final + PE_final

And it is also true that the spring will stretch until both mass are moving at the same speed (momentarily). His mistake is in thinking that this is the original unstretched length of the spring. Not so - the spring stretches. The final PE is not zero!

Perhaps he meant to ask: what is the speed of the masses when the spring expands to its maximum extension. In which case it's a simple conservation of momentum problem.

But you correctly solved the question he did ask.

 

[SEG9585]

I think, he misinterpreted the question in that case-- because he didnt write it lol. It was from a 1994 AP Exam, but either way, he may be right about the velocities when the spring is fully compressed, rather than completely uncompressed

 

[Chi Meson]

Originally posted by SEG9585
Yeah, I couldn't believe his response either--
He is a very good teacher -- regarded as one of the best in the school, an MIT graduate. The class is AP Physics by the way... The thing is, I also have the physics mind (Aerospace engineering is my intended major), and I look at things logically from different perspectives, which maybe he didnt see when he gave this solution? Anything else I get wrong on tests, usually comes from stupid mistakes that I agree I messed up on.

One thing I presented to him, was that with his given final velocity, I used that V to solve for the original momentum in the system (which you just presented to me also), and showed him that the momentum was not conserved. In addition, he added that the collision is not elastic, and some KE was lost in the spring. Being an "ideal spring", I couldn't agree with this. Even after this response, I didnt understand why he would compare final and inital energy to begin with if he thought this was true.
Thanks for the response, hopefully I can get my 5 points back hehe.

OK. If he is new to teaching, there is a common case of "forgetting the easy things." Still I find it hard to believe that anyone faced with a violation of the conservation of momnetum would not back down and admit his error.

THe whole idea of an "idal spring" is one that has no mass and loses no mechanical energy. BUt regardless, you can't violate the conservation of momentum.

Print out a selection of these responses. YOu are correct. YOu deserve the five points. He needs to relearn this chapter.

 

[SEG9585]

Hey all--
Thanks for your responses,
I showed him the responses in the thread today, and he saw my point about the entire issue. I think there was a miscommunication in my response-- he thought that I meant "the collision with the clay and block" is elastic, rather than my intended "block-clay and 8kn block system" is elastic.

He's not exactly a new teacher...this was the first mistake that I've found... otherwise, he really does a great job teaching. So, as far as I know, I should be getting my points back hehe.

Thanks again!
Eric