# Blocks colliding momentum problem

1. Sep 18, 2011

### brushman

I'm studying for my first exam in Intro to Modern Physics. The question, along with the solution, is attached. Unfortunately I don't understand the solution.

The solution changes to the rest frame of M, which I get. Then, it shows what happens after the collision and it seems like it breaks conservation of momentum (they have Pbefore = -Pafter).

Secondly, when the solution switches back to the original frame, how does the smaller mass suddenly gain velocity? Shouldn't the velocity go down, not up?

Thanks

#### Attached Files:

• ###### hw4.png
File size:
95.9 KB
Views:
84
2. Sep 18, 2011

### danielakkerma

I think you ought to remain in the laboratory reference frame and simply solve using the conservation of momentum and Kinetic energy:
$Mv - 2mv = Mu_1 +mu_2$
$Mv^2+4mv^2 = M{u_1}^2 + m{u_2}^2$

3. Sep 18, 2011

### brushman

Seems kind of messy to solve. Does M >> m tell us anything useful?

4. Sep 19, 2011

### danielakkerma

Something very useful indeed, there will be plenty of reductions there...
Give it a try.
remember that (m/M)^2 is therefore zero, due to this condition, and M - m ~ M..., as is M + m ~ M...
Daniel

5. Sep 20, 2011

### Staff: Mentor

By coincidence, I have been looking at this very problem. I believed I had it solved, but my working doesn't agree with any of the options in your test. :yuck:

See my answer in the second page of this thread: https://www.physicsforums.com/showthread.php?t=529469&page=2" where I have cast the problem as bouncing a ball off the front of an oncoming locomotive.

Last edited by a moderator: Apr 26, 2017
6. Sep 20, 2011

### Staff: Mentor

When you, the observer, change frames while observing some event, the objects being observed don't change -- you do. So your measurements for their velocities will change because your own velocity with respect to those objects has changed. When you switch back to the original frame of reference and you note that the smaller mass has "suddenly gained velocity", it hasn't... you the observer has slowed down with respect to it. Note that we don't bother keeping track of the observer's energy requirements or momentum -- we don't care how he changes frames.

A change of frame of reference can be useful in simplifying the mathematics of collision problems, and it's particularly handy when one object is overwhelmingly more massive than the other as in this case. When you change frames to that of the more massive object, the elastic collision looks just like a ball bouncing off a massive wall. And we know that for that case the incoming speed is equal to the outgoing speed. Simple. The wall is so massive that we assume no change in its momentum. In the original frame of reference the massive object ("wall") overwhelmingly carries the bulk of the momentum, and the small object's change is negligible compared to it.

By the way, the usual change of frame chosen to solve collision problems is that of the center of momentum. That is the frame of reference where the magnitudes of the momenta of the two objects are equal and they are oppositely directed -- net momentum is zero in such a frame. Collisions in such a frame have the extremely useful property whereby the velocities of the objects are simply reflected: change their signs and you're done! (except for the switch back to the original frame of reference).

7. Sep 22, 2011

### Staff: Mentor

EDIT: There was an error in my working for the rebound speed derived in the post referenced below. That has now been corrected.

The formula now agrees with the answer you provided, viz., +4v.

See my answer in the second page of this thread: https://www.physicsforums.com/showthread.php?t=529469&page=2" where I have cast the problem as bouncing a ball off the front of an oncoming locomotive.

Last edited by a moderator: Apr 26, 2017
8. Sep 22, 2011

### danielakkerma

By using the assumptions I provided, you can however readily obtain this answer, through the conservation of momentum & kinetic energy, without sprawling onto other physical problems. That's my point of view, anyhow, that solutions, whenever possible, need to incline to the direct approach...
But whichever way works, and Nascent_Oxygen's path looks very alluring, and certainly devoid of any complicated/excessively burdensome algebra.
Daniel