# Block moving towards movable wedge

1. Apr 10, 2015

### Yoonique

1. The problem statement, all variables and given/known data
A small block with mass m is moving towards a stationary slope with velocity v, angle θ and height h, suppose all surfaces are frictionless.
Suppose the slope has mass M and it is able to move freely on the ground, what is the minimum initial speed for the block to be able to reach the top of the slope?

2. Relevant equations
ΣPi = ΣPf
ΔK = -ΔU
3. The attempt at a solution
Case 1: Does the block travel up the wedge at v the moment it meets the wedge? If so, ΣPi in the horizontal axis is mvcosθ if I use the moment of transition of the block going up the wedge as the initial state.

Case 2: However I could also use the initial state when the block have not meet the wedge for my initial momentum, thus ΣPi = mv.

I'm confused because the answer for this question is vmin = √[(2gh)(M+m)/(M+msinθ)]. I will get this answer if I use case 1, and not case 2. But case 2 seems right too, but the answer I will get does not involve θ.

2. Apr 10, 2015

### AlephNumbers

The surfaces may be frictionless, but there is still a normal force exerted on the block from the wedge and vice versa. You do not know the details of the block's collision with the wedge. But you do know that there are no external forces acting in the horizontal. Therefore, you can conclude that momentum is conserved in the horizontal between the moment before the block collides with the wedge, and after the block collides with the wedge. If the block stays in contact with the wedge as it moves up the wedge, then the collision in the horizontal is an inelastic collision

You must consider the initial state to be before the block collides with the wedge, because the velocity that is given is that of the block before it collides with the wedge. When the block collides with the wedge, the velocity begins to change because the y component of the velocity is not conserved. This is due to an external force (gravity).

Last edited: Apr 10, 2015
3. Apr 10, 2015

### AlephNumbers

At least, that is how I understand it. If anything I posted is wrong, I would very much like to be corrected. Basically, Yoonique, I think that what is messing you up is whether or not the block would continue up the ramp at the same velocity as it had before colliding with the ramp. I do not think that it would.

4. Apr 10, 2015

### collinsmark

[I had temporarily removed this post, but I'll put it back now. I'll be sure specify what I'm uncertain about.]

Hello Yoonique,

Wow. What a thought provoking problem, and good insight on your part!

I think you have a very strong case to argue that "Case 2" is correct.

We could analyze the "mini-collision" regarding the first moment the block hits the slope, at the tip of the slope where the slope meets the ground. (Although I don't recommend going into detail for reasons which might become clear later.)
• Clearly, the momentum of the system before this "mini-collision" is simply mv, and is completely in the horizontal direction. Since the surface of the ground is frictionless, no horizontal momentum is imparted to the Earth, and thus the initial, horizontal component must remain the horizontal momentum of the block + slope throughout -- even at the moment immediately after the "mini-collision."
• If immediately after this "mini-collision," the block's horizontal momentum is less than mv [such as mvcos(<something>)], the block must have imparted some horizontal momentum to the slope. It means that immediately after this mini collision, the slope's horizontal velocity is not zero. Furthermore, the angle at which the block moves up, relative to the ground, is not going to be θ (the angle of the slope) since the slope is moving too. This aspect could get complicated.
• The momentum in the vertical direction gets really hairy for this "mini-collision" and even afterwards, since we would ultimately need to take the momentum of the Earth into consideration, or at least model it with external, time varying impulses. This gets hairy.
A different approach is to model the entire problem as one big collision. The initial state is the block is moving horizontally with momentum mv, the the final state is the slope+block are moving together, completely in the horizontal direction, with the block sitting atop the slope. And as usual, momentum is conserved. This way we don't need to muck about with the nasty intermediate steps.

On the surface, this appears to be a completely inelastic collision. If you can make the argument that the all the extra energy of the collision goes into the gravitational potential energy of the block (which, in the end, rests atop the slope), then this should be a pretty easy problem to solve.

[Edit: is this argument valid? Even with frictionless surfaces, can we say for sure for sure that there are no losses to non-conservative forces? That's the crux of the argument.]

(If the argument holds, the implications are that your given answer of vmin = √[(2gh)(M+m)/(M+msinθ)]* is incorrect, and the answer is not even dependent on the angle θ.)

*(Even though your given answer may be wrong, are you sure you typed that in correctly? The way I solved it [using the assumptions in "Case 1"], there was a +/- sign difference in there somewhere, and something else needs to be squared.)

Last edited: Apr 10, 2015
5. Apr 11, 2015

### ehild

I join to AlephNumbers and Collinsmark. During the first short period, when the block makes contact with the wedge, the velocity of both the block and wedge change, because of the impulsive normal forces between them and between the Earth and the wedge. The block gets some impulse, perpendicular to the slope, and the wedge gets a horizontal impulse.
If the block is point-like, those normal forces can account for, see picture.

Pw is the impulse between wedge and block, Pg is that between the wedge and ground. The momentum of the block becomes mvb=mvcos(θ) and Pw=mvsin(θ). The momentum of the wedge is MVM=Pwsin(θ)=mvsin2(θ).
The horizontal component of the total momentum is equal to mv, the momentum before collision. But the energy is not conserved during this collision.

If there is a smooth transition between the ground and the wedge, the energy can be conserved, but theta changes during the collision. We do not know what the speeds are when the block reaches the flat part of the wedge.

6. Apr 11, 2015

### Yoonique

There was a typo in the answer, it is supposed to be vmin = √[(2gh)(M+m)/(M+msin2θ)]

7. Apr 11, 2015

### Yoonique

Impulse makes this whole thing so much easier and simple to understand. So I need to consider the ground into the wedge and block system if I want to calculate the linear momentum at the point of collision.
So I can say that the block goes up the wedge at vcosθ at the point of collision.
I assume is the horizontal momentum you are talking about?
mv = mvbcosθ + MVw
So Vw = mvsin2θ/M at the point of collision.
I assume the energy you are referring to is kinetic energy since it is like an inelastic collision since the block gains potential energy after the collision.

8. Apr 11, 2015

### ehild

MVM is the momentum of the wedge, and it is horizontal.

mv = mvbcosθ + MVw
So Vw = mvsin2θ/M at the point of collision.
I talk about the "instantaneous collision", when the block had no time yet to raise on the wedge, and the energy is pure kinetic. After that, the energy is conserved, some part of KE transforms into PE.

9. Apr 11, 2015

### Yoonique

Why is the energy not conserved at the point of instantaneous collision? Isn't the initial kinetic energy = 0.5mv2 and the total kinetic energy at the 'instantaneous' collision = 0.5mvb2 + 0.5MVw2
vb2 + Vw2 = v2cos2θ + vsin2θ = v2 which is an elastic collision at the 'instantaneous' collision where ΔK = 0

10. Apr 11, 2015

### ehild

KE=0.5mvb2+0.5 M Vw2, and
vb=vcos(θ), VM=(m/M)sin2(θ)
KEis not equal to 0.5mv2.

11. Apr 11, 2015

### Yoonique

Isn't VM = mvsin2θ/M
So it does not conserve energy because the change in kinetic energy has not yet changed to potential energy at that moment?

12. Apr 11, 2015

### Staff: Mentor

You can ignore the surface/block connection - a block hitting the wedge horizontally at some point above the ground would give the same result (assuming the wedge does not fall over).

You cannot, the collision will have an inelastic component - otherwise the block would fly into the air and not stay on the ramp (easy to see for very large M but true for all).

Horizontal momentum is conserved (no friction at the surface) and relative momentum parallel to the wedge surface (block as seen by wedge before and after collision) is conserved (no friction at the wedge surface). Those two constraints allow to find the two unknown parameters after the collision, like wedge speed and horizontal block speed.

13. Apr 12, 2015

### haruspex

As ehild noted, that is avoided if there is a smooth transition from horizontal to slope. For point particle, the transition part can be arbitrarily short. To get the given answer (as corrected in post #6), that assumption has to be made. But then, the question should have made it clear, and in the absence of such information I agree it would be appropriate to assume an inelastic element.

14. Apr 12, 2015

### Yoonique

I'm confused. So when is the answer in post 6 be right?

15. Apr 12, 2015

### haruspex

If you assume there is a little curve at the start of the ramp going from horizontal to the angle theta, so that no work is lost in impacts.

16. Apr 12, 2015

### Yoonique

If there is a curve at the start of the ramp (smooth transition) means that the normal force of the wedge on the block is always perpendicular to the velocity of the block thus no work is done on the block. Which also means that the block towards the wedge at u, and move up the wedge at u.

If there is no curve at the start of the ramp (not smooth transition) means that the normal force of the wedge on the block is not perpendicular to the velocity of the block thus there is negative work done on the block, so the block will travel up the wedge at a velocity less than u.
Is my understanding of the situation correct?

17. Apr 12, 2015

### haruspex

If you mean the velocity of the block immediately after the transition in each case, yes.

18. Apr 12, 2015

### Yoonique

Okay thanks. So if the question did not state anything about the transition, which case should I consider?

19. Apr 12, 2015

### haruspex

As I wrote in post #13, from the statement of the question I would have thought there was an inelastic impact, i.e. mfb's view. It is only by seeing the given answer that we are able to infer that the question setter intended a smooth transition.