# Homework Help: Stopping distance after a collision down a ramp

1. Apr 18, 2014

### NZBRU

Ep = Ek
mgh = Ek
mgh = ½mv²
v = √2gh

As the collision is elastic, m1u1 + m2u2 = m1v1 + m2v2. It is known that m2 = 2m1.

m√2gh = 2m1v2
v2 = (m√2gh)/2m
v2 = (√2gh)/2

Force body diagram of m2:

̂̂̂̂̂̂Fnet = ma
Fnet = Fn + Fg + Ff
Fnet = Ff
m2a = Ff
a = Ff/m2
a = Ff/2m1
a = uN/2m1
a = 0.5*m2g/2m1
a = m1g/2m1
a = g/2 (negative)

v^2=u^2+2ax
0= v2^2 - gx
gx = v2^2
x = (v2^2)/g
x = 2h/9

The answer is 8/9h. My solution differs from the lecturers at this point:

I do not understand what he was done

2. Apr 18, 2014

### rude man

I would not use forces, body diagrams or accelerations. I would use just momentum and energy conservation:

1. compute v of m1 at bottom just before collision
2. equate momenta of m1 and m2 just before and just after collision
3. express energy conservation just before & just after collision for m1 and m2
4. equate k.e. of m2 just before entering the rough patch with the energy dissipated by friction.

4 equations, 4 unknowns: velocity of m1 just before & just after collision, velocity of m2 just after collision, and d.

I would not use forces and accelerations. It's easier to use energies.
1. at bottom before collision, calculate v

3. Apr 18, 2014

### haruspex

That's always true, regardless of elsticity. The elasticity tells you work is conserved too.
You seem to be assuming m1 comes to rest. It won't.

4. Apr 18, 2014

### rude man

It's always true if the surface is frictionless, which it is until after the collision.

5. Apr 18, 2014

### haruspex

Itook the equation as referring to motion immediately before and immediately after the collision. Since there is no sudden impulse normal to the surface, friction has no time to have an effect.

6. Apr 19, 2014

### rude man

Obviously, if so interpreted. But an introductory physics student will not be concerned with momentum conservation in infinitely short time, I don't think. If the surface has friction, momentum is conserved only if externals are also taken into account. If it's frictionless, as is here the case, momentum is conserved within the m1 - m2 system until friction appears.

Or, think of two rubber balls as the masses. The collision starts when they touch and ends when they separate, at least by my definition, which takes finite time which lets friction spoil momentum conservation.

7. Apr 19, 2014

### haruspex

I have no idea what point you are making. Conservation of momentum across the collision must be used to solve the problem. You adverted to this in point 2 of your own post #2, though I'm not entirely happy with the the way you expressed it. (It's the total momentum before and after collision that we equate.). My comment on the OP was pointing out that the application of this principle was not related to elasticity. Elasticity allows us to use conservation of work, which also needs to be used.
As regards applying conservation across an instant of time, that's what we always do with collisions. In fact, it doesn't encounter the friction until a bit later, but that isn't really relevant. Momentum would also be conserved through the collision if the entire ramp were frictional. The difference would be in calculating m1's speed just before collision.

8. Apr 20, 2014

### rude man

Right. But we are trying to justify the conservation of momentum equation as it applies to actual problems. And for those problems, absence of friction is required.

Consider again the two masses m1 and m2. m2 is stationary on the right and sits at the left end of an arbitrarily short stretch of friction. m1 is on the left and moves to the right at constant velocity v1i and collides with m2. Were momentum conserved for the two-mass-only system, the equation we all learn would give m1 v1i = m1 v1f + m2 v2f just after the end of the frictional stretch. But in order for m2 to assume a change in velocity from zero to v2f it must have accelerated to the new velocity, which takes time, which takes distance, which means even that short stretch of friction has had time to slow it down, which means the above momentum equation is invalid.

That equation obviously assumes constant v1i, v1f and v2f or there is no "v1i, v2i or v2f". And for that we need absence of friction - total absence of friction.

Momentum with friction is conserved only in the sense that the center of mass of the universe has not moved. Since we are practical scientists let us approximate and say that the c.m. of the Earth has not moved.

9. Apr 20, 2014

### haruspex

If the frictional stretch is arbitrarily short then its effect on motion is arbitrarily small, i.e. zero in the limit.
Suppose m1 and m2 sit on a ramp at angle θ, kinetic friction μ > tan θ. m1 is given an initial speed u down the ramp. After travelling distance d it collides elastically with m2. Find where each mass comes to rest. How would you solve this without using conservation of momentum (or equivalent) across the instant of impact?

10. Apr 20, 2014

### rude man

Well, I didn't mean zero friction run. Any finite friction run would destroy the momentum equation for before and after the friction run.

That's all I ever meant to point out, and I don't think you disagree with that.

Momentum conservation as you mean it is true only in the conceptual form of an impulse force F = Δpδ(t) at the moment of impact t=0. The velocities, while needed to solve the problem, are conceptual only since infinite accelerations and Dirac delta functions are not realizable. They are never actually seen.

I think we're beginning to argue orthogonally here. Yes, momentum conservation is realized conceptually in the instant of collision. No, friction destroys the m1-m2 momentum conservation for even the smallest friction stretches starting at the point of impact and extending a finite distance after.

Last edited: Apr 20, 2014
11. Jun 23, 2017

### neilparker62

Collision is elastic so transfer of momentum Δp to block 2 given by the equation 2μΔv where Δv = √(2gh) and μ = reduced mass (m x 2m)/(m + 2m) = 2m/3. Thus:

Δp = 4m/3 x √(2gh) and dividing by block 2's mass (2m) we obtain it's velocity = 2/3 √(2gh).

After that just apply work energy F x d = Δ(KE) where F is the frictional force.

12. Jun 23, 2017

### sophiecentaur

I have come to this thread late but I don't think everyone can have been reading the actual question fully. It seems to me that there are three clearly separate parts to the Question. First, you have to calculate the velocity of M1, after falling down the slope. There is then an impact between the two masses and you can determine their velocities easily because the impact takes place on a frictionless surface. M2 then travels onto the friction surface (on its own) and will stop when its KE has been dissipated against the friction.
Each of those steps is very straightforward and, for an elastic collision, the time taken doesn't matter (as long as the collision doesn't extend as far as the beginning of the friction surface and it must be assumed that it doesn't because there is no information about the details of the collision). I can't see what there is to argue about there except that the actual arithmetic needs to be correct.

Incidentally, M1 will be moving left after the impact and it will go back up the slope and come down again, also to lose its KE on the friction surface. (Not included in the Question)

13. Jun 23, 2017

### haruspex

I think everyone understood that, but the OP's difficulty was only in the second phase.
I never understood what rude man's issue was either. As you say, the collision is still on the frictionless part, and it wouldn't matter if it were not. What would be a problem is if the smaller mass were to transit a non-infinitesimal stretch of the frictional part before the collision were complete.

14. Jun 23, 2017

### sophiecentaur

I could believe it but I haven't proved it for myself - except for the general argument that it's all linear. But the retarding forces would be different for both masses and the work done on each would be different. Whether or not that would make any difference, I'd have do work it out. I am not sure what you mean by a stretch of the frictional part. For an elastic collision, the spacing between the CMs could be changing in an unspecified way and different for each mass and that would be without any need for the surface to stretch at all. Too late for that and I have to take a dog for a walk.
Edit: I re-read what you wrote. You mean a section of when you say 'stretch'

15. Jun 23, 2017

### haruspex

Yes, sorry. My natural style involves "elegant variation", i.e. deliberately using a range of words to avoid repetition, but it's not ideal for technical discussions.

16. Jun 27, 2017

### neilparker62

Quite right:

Step 1: Change PE to KE: v = √(2gh)
Step 2: Transfer of momentum: Δp = 2μΔv (elastic collision)
Step 3: Work Energy: Fd = Δ(KE)

And if you complete step 3 after the other 2 as per previous post, you should get the required solution.