# How High Will the Puck Shoot Up from the Plate?

• Samei
In summary: MvN = mv2 +...+Mv2In summary, the puck will shoot up from the plate high enough that it will never reach the launch point. Because the plate has no inertia, the puck will rapidly match the plate's speed and shoot up to the top.
Samei

## Homework Statement

Problem: [/B]A curved plate of mass M is placed on the horizontal, frictionless plane as shown. Small puck of mass m is placed at the end of the plate and given an initial horizontal velocity v. How high the puck will shoot up from the plate? Regard radius of curvature to be much smaller than the jump height and consider that the puck leaves the plate vertically.

## Homework Equations

Conservation of Energy: KE1 + PE1 - Wf = KE2 + PE2 and Wf = deltaTME = (delta K + delta P)
Also, possibly Angular Speed: w = v/r

## The Attempt at a Solution

[/B]
From the first equation, KE1 + PE1 - Wf = KE2 + PE2
So, ((mv1^2)/2) + (mgh)) - Wf = ((mv2^2)/2) + (mgh))
Now, at the first point, h = 0. Thus, PE1 = 0. Also, since delta K = ((mv2^2)/2) - ((mv1^2)/2) and delta P = mgh - 0:
This extends to ((mv1^2)/2) - (((mv2^2)/2) - (mv1^2)/2) + mgh))) = ((mv2^2)/2) + (mgh))
Simplifying all this makes: (mv1^2)/2) - (mv2^2)/2) + (mv1^2)/2) - mgh = ((mv2^2)/2) + (mgh)) or
2(mv1^2)/2 = 2(mv2^2)/2) + 2(mgh))
Removing 2m, (v1^2)/2 = (v2^2)/2 + 2gh

I need another equation. This is when I turn to the angular speed (?), at least possibly. Momentum is not conserved since friction is present, so I am not sure what equation to make. Any suggestions on my next steps, or if I am on the right track at least? Any help is appreciated!

Note: Do I have to apply Calculus techniques here? Is that why I'm missing? I'm a little rusty on that area.

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It isn't stated, but it seems to me you have to assume there's no friction anywhere, or it becomes insoluble.
You do need to consider the motion and mass of the plate.

haruspex said:
It isn't stated, but it seems to me you have to assume there's no friction anywhere, or it becomes insoluble.
You do need to consider the motion and mass of the plate.

I know this sounds strange, but can I assume there is no friction or can I just denote it as u? From the picture, it seems as if there is some rugged surface involved, but I am with you. I've been racking my brain in coming with a Force for f, but I just can't find one with velocity and mass alone.

Also, can you please elaborate why M is relevant? I can't see how it'll affect the velocity of m.

Ok, I've been tinkering with it and I am guessing since I'm assuming no friction is involved, I can finally apply the conservation of momentum, where angular momentum is the same anywhere on the circle. Am I right on this?

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Samei said:
it seems as if there is some rugged surface
I'd guess that's just staircasing from drawing a freehand curve at low pixel resolution.
Samei said:
why M is relevant?
If the plate has no inertia, as soon as the puck starts to go up the slope the plate will accelerate to match the puck's speed. The puck will never reach the launch point.
Samei said:
angular momentum is the same anywhere on the circle
Circle? The shape of the curve is not given, and is immaterial beyond some basic smoothness requirements.
I don't think angular momentum will be useful here.
Please post/repost your energy and linear momentum equations, taking into account the movement of the plate.

Oh, ok. I'll work on it right now. :)

Sorry for taking so long. I had to do some additional reading to make sure of what I was doing.
Here is what I have:
Assuming no friction, KE1 + PE1 = KE2 + PE2
Again, PE1 starts at an arbitrary h=0, so PE1 is 0.
This becomes KE1 = KE2 + PE2
and ((mv1^2)/2) = ((mv2^2)/2) + mgh
(v1^2)/2 = (v2^2)/2 + gh
I can use this later on to solve for either h or v2.

Now, for momentum. Once again, since the surfaces are frictionless, conservation of momentum can be applied. I'm a bit unsure how to do this, but looking at the problem, this is what I have in mind.
mv1 + Mv1 = mv2 + Mv2
It seems a bit counterintuitive. What did I do wrong here?

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I just realized.
mv1 + Mv1 = mv2 + Mv2 is for collisions.
I'm stuck now.

Samei said:
and ((mv1^2)/2) = ((mv2^2)/2) + mgh
You are still omitting the movement of the plate.
Samei said:
mv1 + Mv1 = mv2 + Mv2
Yes, but be careful about how these velocities relate to your energy equation velocities. In the momentum equation, direction matters.
Might be safest to create separate symbols for the horizontal and vertical components of the puck's initial and final velocities.
Samei said:
mv1 + Mv1 = mv2 + Mv2 is for collisions.
No, it's true whenever there's no external force in the direction of the linear momentum component under consideration.

haruspex said:
You are still omitting the movement of the plate.

Yes, but be careful about how these velocities relate to your energy equation velocities. In the momentum equation, direction matters.
Might be safest to create separate symbols for the horizontal and vertical components of the puck's initial and final velocities.

No, it's true whenever there's no external force in the direction of the linear momentum component under consideration.
Ok, let me work this out again.
KE1 (of m) + KE1 (of M) + PE1 (of m) = KE2 (of m) + KE2 (of M) + PE2 (of m)
KE1 (of M) does not have initial velocity applied to it, so it zero. Initial height arbitrarily is zero so PE1 is also zero. KE2 (of m) will be at zero velocity at maximum height so this will also be zero.

KE1 (of m) = KE2 (of M) + PE2 (of m)

For momentum, m(v1,x) + M(v1,x) = m(v2,y) + M(v2,x)
where x is the horizontal component and y is the vertical component. Now, I'm curious is (v2,y) is zero here and would that mean my goal is to solve for (v2,x)?

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I forgot to add: Is v1 for m different from v1 of M? I assumed it for the energy equations since no velocity is applied there but I am not sure if that is valid.

Samei said:
KE1 (of M) does not have initial velocity applied to it,
Right, so why the M v1x term in the momentum?
Samei said:
m(v1,x) + M(v1,x) = m(v2,y) + M(v2,x)
Momentum is a vector. In the vertical direction there is a normal force from the ground, which varies, so you can't use momentum conservation in that direction. Momentum is conserved in the horizontal direction since there's no friction, but v2y doesn't feature in that.
What will be the relationship between the horizontal velocities of puck and plate at the launch point?

Thanks for sticking with me on this problem by the way! I'm still working on it, and after discussing this with other people, I found out that if I were to neglect friction, then the bigger plate (M) would not move. Friction is the one that is pushing M to be on the same direction as m. With that in mind, I can say that some KE is being converted from the original m to M. This is described by my energy equations as:
KE of m = KE of M + PE of m
It's basically what I wrote before, but now I have a firm grasp on what's happening. It's a Eureka moment! :)
Or am I completely misunderstanding the laws of physics yet again?
haruspex said:
What will be the relationship between the horizontal velocities of puck and plate at the launch point?

I can use the conservation of energy to form this relationship, at least possibly? I know that the velocity of the puck will be greater than the plate, but smaller than initial. That statement may not be directly quantifiable, but it may help.

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Samei said:
if I were to neglect friction, then the bigger plate (M) would not move
Wrong. The top of the plate is not horizontal. As soon as the puck starts to go up a slope the normal force is not vertical. This will slow the puck and accelerate the plate.
Samei said:
haruspex said:
What will be the relationship between the horizontal velocities of puck and plate at the launch point?
I can use the conservation of energy to form this relationship, at least possibly?
No, but you can use geometry. How steep does the slope appear to be at the launch point?

Have another try at the equations in post #8, this time using separate symbols for the horizontal and vertical components of the puck's velocity.

haruspex said:
No, but you can use geometry. How steep does the slope appear to be at the launch point?

Have another try at the equations in post #8, this time using separate symbols for the horizontal and vertical components of the puck's velocity.

I think I have an idea. There is a reason why the questions suggested that the plate moved vertically.

Here are my updated equations:
Energy: KE1 (of m) = KE2 (of M) + PE2 (of m)
Momentum:
mv1(cos(0)) + Mv1(cos(0)) = mv2(cos(90)) + Mv2(cos(0))
mv1 + Mv1 = Mv2

If all is good, I can work this out.

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Samei said:
I think I have an idea. There is a reason why the questions suggested that the plate moved vertically.

Here are my updated equations:
Energy: KE1 (of m) = KE2 (of M) + PE2 (of m)
Momentum:
mv1(cos(0)) + Mv1(cos(0)) = mv2(cos(90)) + Mv2(cos(0))
mv1 + Mv1 = Mv2

If all is good, I can work this out.
No, sorry, both those equations are wrong.
They both assume the puck ends with no horizontal velocity. What makes you think that?

haruspex said:
No, sorry, both those equations are wrong.
They both assume the puck ends with no horizontal velocity. What makes you think that?

I guess what I really mean was that final vertical velocity would be zero. But how would horizontal velocity play a part in this? I am not exactly sure where to add it for which I can then solve.

Are my equations only lacking?

Samei said:
what I really mean was that final vertical velocity would be zero.
When at its maximum height, yes, but first we have to find the vertical component of the launch speed (i.e. the speed at the time it leaves the plate). Concentrate on that for now.
Your nomenclature for the variables is confusing. Please use these variable names, or something similar:
v = initial (horizontal) speed of puck (given)
vx=horizontal speed of puck at launch time
vy=vertical speed of puck at launch time
w=speed of plate at launch time
Samei said:
how would horizontal velocity play a part in this?
The horizontal velocity that the puck retains at launch is taking up some of the KE, so reduces the vertical launch speed.
haruspex said:
How steep does the slope appear to be at the launch time?
When you have answered that correctly you should try to answer this question:
haruspex said:
What will be the relationship between the horizontal velocities of puck and plate at the launch time?

haruspex said:

From the information given and the picture, the slope is as steep as it can be. That means it is about 90 degrees, as it shoots upward. Also, like you said, only the horizontal momentum is conserved. So this is the relationship between the velocities of the puck and plate at launch time: mv = Mw + mvx

Rewriting the energy equation, at launch:
KE1, Puck = KE2, Puck - Horizontal + KE2, Puck - Vertical + KEPlate (Horizontal) + PEPuck
I would have three unknowns here, namely h, w, vx, and vy.

By direction of components, is it not sufficient that I include the angle (i.e., sin(90) or cos(0))?

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Samei said:
That means it is about 90 degrees, as it shoots upward.
Right, but directly upward?
Samei said:
mv = Mw + mvx
Yes, that's the horizontal momentum equation, but it's not the relationship between w and vx that you can deduce from the vertical slope at the launch point.
What will the relative horizontal velocity be?
Samei said:
I would have three unknowns here, namely h, w, vx, and vy.
Not sure how you are defining h here.
The question says to treat the plate's radius of curvature as much smaller than the jump height. (Alternatively, it could have asked for jump height relative to its initial height, not its height at launch.) Either way, this means we can ignore the gain in PE from initial position to launch point. We just need to find vy from conservation of KE. We can then find the jump height from that.

haruspex said:
Right, but directly upward?
No. I didn't realize this before, but some of the horizontal velocity will let it move further.

haruspex said:
What will the relative horizontal velocity be?
I am still reluctant, but is the velocity of the plate equal to that of the puck? I've also been using geometry to explain some things, but I don't think it has lead me anywhere. Say, I know that the puck will be perpendicular to the plate at launch.

Energy equation, with minor revision: KEinitial = KEpuck at launch - horizontal + KEpuck at launch - vertical + KEplate at launch, considering that PE gained is negligible.

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Samei said:
No. I didn't realize this before, but some of the horizontal velocity will let it move further.

I am still reluctant, but is the velocity of the plate equal to that of the puck? I've also been using geometry to explain some things, but I don't think it has lead me anywhere. Say, I know that the puck will be perpendicular to the plate at launch.

Energy equation, with minor revision: KEinitial = KEpuck at launch - horizontal + KEpuck at launch - vertical + KEplate at launch, considering that PE gained is negligible.
Yes, that all looks good. Just bring in the horizontal momentum equation and you have enough to solve it.

haruspex said:
Yes, that all looks good. Just bring in the horizontal momentum equation and you have enough to solve it.
Ok, I will try it yet again. I can almost see the light at the end of the plate!

Again, thanks for all the help.

haruspex said:
Yes, that all looks good. Just bring in the horizontal momentum equation and you have enough to solve it.

Just an update, I have equated for it and it all seems to fall into place. This turned out to be more complicated than I initially thought it would be, but I guess it was because I simplified the situation.

Samei said:
consider that the puck leaves the plate vertically.
Is it clear that this means vertically in relation to the plate or is it vertically in relation to the table?

insightful said:
Is it clear that this means vertically in relation to the plate or is it vertically in relation to the table?

haruspex said:
I referenced "The problem statement." Sorry, I don't understand your question.

insightful said:
I referenced "The problem statement." Sorry, I don't understand your question.
You also need to consider the picture. The only way that the puck could end up moving vertically relative to the table is if the plate curled back at the edge, which it clearly does not.

haruspex said:
You also need to consider the picture. The only way that the puck could end up moving vertically relative to the table is if the plate curled back at the edge, which it clearly does not.
I did consider the (very crude) picture. I would ask the original source for clarification; just my opinion.

insightful said:
I did consider the (very crude) picture. I would ask the original source for clarification; just my opinion.

I'm not quite sure myself, but the teacher does want the students to assume as they will. Looking at the picture, it does not look like it curls.

Just to be safe, I was told to write an observation whenever possible.

## 1. What is the Work-Energy Theorem?

The Work-Energy Theorem is a fundamental principle in physics that states that the work done on an object is equal to the change in its kinetic energy. In other words, the work done on an object will result in a change in its speed or direction, depending on the direction of the force.

## 2. How is the Work-Energy Theorem used in problem solving?

The Work-Energy Theorem is used to solve problems involving the motion of objects by relating the work done on an object to its change in kinetic energy. This allows us to calculate the speed, displacement, or other properties of an object based on the amount of work done on it.

## 3. What is the formula for the Work-Energy Theorem?

The formula for the Work-Energy Theorem is W = ΔK, where W is the work done on an object, and ΔK is the change in its kinetic energy. This can also be written as W = 1/2mv2, where m is the mass of the object and v is its velocity.

## 4. How is the Work-Energy Theorem related to other laws of physics?

The Work-Energy Theorem is closely related to the laws of conservation of energy and momentum. It is also related to Newton's laws of motion, as the work done on an object is equal to the force applied to it multiplied by the distance it travels in the direction of the force.

## 5. Can the Work-Energy Theorem be applied to all types of motion?

Yes, the Work-Energy Theorem can be applied to all types of motion, including linear, rotational, and oscillatory motion. It can also be applied to both macroscopic and microscopic scales, making it a versatile tool in solving problems in various fields of physics.

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