How Does the Wedge's Movement Change with the Disk's Motion Upward?

[Rikudo]

Homework Statement

A wedge is placed on a frictionless floor and it is free to move. Its inclination is θ with the horizontal everywhere except at the bottom, where it increases gradually from a vanishingly small value to θ in a very small region (See fig 1). Then, a small disc whose mass is the same as the wedge's is projected towards the wedge with an initial velocity u.

Assuming transit time of the disc on the curved entry to be negligible , find displacement of the wedge when the disc is at the highest point on the wedge!

Assume that every surface is frictionless.

Relevant Equations

Momentum conservation
Newton's law

So, when the mass reached the peak, its horizontal velocity will be the same as the wedge's. Using conservation of momentum :
$$ mu = 2mv$$
$$v = \frac u 2$$
With v is the final velocity for both objects.

Now, what we need is the acceleration of the wedge, which we can find by using Newton's law.
$$a = \frac { g sin\theta cos\theta} {sin² \theta +1}$$ ( I looked this up on a book to make sure that I don't make a mistake)

Lastly, all I need to do is to find the distance that the wedge has moved. I can use this : $$s = \frac {v²-0²} {2a}$$ So, the answer is
$$s = \frac {u² (sin² \theta +1)} {8g sin\theta cos\theta}$$

strangely, according to the book, the answer is wrong.
Why?

 

[Lnewqban]

Why do you think that “when the mass reached the peak, its horizontal velocity will be the same as the wedge's”?

 

[Rikudo]

The wedge's speed always increases, while the horizontal speed of the disc always decreases.

If the horizontal speed of the wedge is larger than the disc's, then the disc will start to slide down. We can use the wedge's frame of reference to prove this. If it is larger, then the disc's velocity relative to the wedge will point to the left, which means it is starting to go down the slope.

 

[erobz]

Just out of curiosity: are we sure the puck has to come to a stop? Couldn’t an interpretion be the it launches off the top of the wedge?

 

[erobz]

Personally, I get a similar answer as you. I applied Newton’s Laws in the non-inertial frame of the wedge yielding three useful equations of motion. You find that $a_w$ does not depend on the velocity of the puck but depends on the slope of the wedge (as you have indicated). So, if the wedge is not initially moving, the kinematic relationship applies to the wedge. I think that is your issue. I believe that factor of “8”should be a “2”, and not $u$ in the denominator but $v_f$.

 

[TSny]

When the disc suddenly changes its direction of motion at the bottom of the wedge, it suddenly loses some horizontal momentum. But horizontal momentum overall must be conserved. Thus, the wedge suddenly gains momentum. So, the wedge does not have zero velocity when it starts its constant acceleration of
$\large \frac{g \sin \theta \cos \theta}{1+\sin^2 \theta}$. The disc essentially gives an impulse to the wedge as it mounts the wedge.

Can you determine the initial velocity of the wedge as it begins its constant-acceleration displacement?

 

[erobz]

When the disc suddenly changes its direction of motion at the bottom of the wedge, it suddenly loses some horizontal momentum. But horizontal momentum overall must be conserved. Thus, the wedge suddenly gains momentum. So, the wedge does not have zero velocity when it starts its constant acceleration of
$\large \frac{g \sin \theta \cos \theta}{1+\sin^2 \theta}$. The disc essentially gives an impulse to the wedge as it mounts the wedge.

Can you determine the initial velocity of the wedge as it begins its constant-acceleration displacement?

I don't know if you are fishing or stating that you can determine it, but:

Assuming transit time of the disc on the curved entry to be negligible

 

[TSny]

I don't know if you are fishing or stating that you can determine it, but:

During the small transit time, the disc exerts a large, impulsive force on the wedge. So, the wedge has a nonzero, initial velocity by the time the disc is "fully on" the wedge (with the disc still essentially at the bottom of the wedge). I derived expressions for the initial velocity of the wedge and the displacement of the wedge when the disc is at max height. It was a little tedious, so I certainly could have made errors.

@Rikudo Can you give us the book's answer for the displacement of the wedge?

 

[erobz]

I thought assuming the transit time to be “negligible” means the impulse during the transition is to be ignored $\Delta t \ll 1$.

 

[TSny]

I thought assuming the transit time to be “negligible” means the impulse is to be ignored

The impulse cannot be ignored. We essentially have a collision between the disc and the wedge as the direction of travel of the disc suddenly changes.

 

[erobz]

While we wait, are you saying the impulse doesn’t depend on entrance geometry?

 

[erobz]

Actually, It seems like the only way to solve this in terms of $u$, is to have the momentum constraints that @Rikudo has specified. There is no geometry specified anywhere in the problem other than $\theta$. I retract, it seems like the factor of 8 is fine. Anticipating the reveal.

 

[BvU]

according to the book,

might well be $\ \displaystyle{u^2\over 2g\sin\theta}\qquad\qquad -$ or am I too naive ?

[edit] hmmm... too naive... (and too naive for the wrong question - the distance the disc moved along the slope )
Beetter get some sleep - 02:02 here

 

[Rikudo]

@Rikudo Can you give us the book's answer for the displacement of the wedge?

$$\frac {u²} {8gsin\theta} (2\sqrt {1+sin²\theta} - cos\theta)$$

 

[erobz]

Way off! I guess let’s see what @TSny has to offer.

 

[TSny]

$$\frac {u²} {8gsin\theta} (2\sqrt {1+sin²\theta} - cos\theta)$$

That's what I got. For the initial speed of the wedge at the beginning of the constant acceleration I find $$v_0 = \frac{u}{2}\left(1-\frac{\cos \theta}{\sqrt{1+\sin^2\theta}} \right)$$

 

[Rikudo]

The impulse cannot be ignored.

Now I'm confused... When do we can or cannot ignore an impulse?

For example, Let's say that I release a ball from a building, and after that it will hit the ground. At the instant when it collides with the ground, can we ignore the impulse by gravity?why?

 

[jbriggs444]

Now I'm confused... When do we can or cannot ignore an impulse?

For example, Let's say that I release a ball from a building, and after that it will hit the ground. At the instant when it collides with the ground, can we ignore the impulse by gravity?why?

The acceleration of gravity is finite. Decidedly finite. 9.8 m/s^2.

The interval of an impulsive collision is short. Think of it as infinitesimally short. So you have the product of a finite and an infinitesimal. The result is infinitesimal.

In the case of a disc colliding with the base of the ramp we have an infinite force times an infinitesimal duration. That is not well defined. So we have to use other means to decide how much impulse there was. Conservation of energy provides an argument: the disc retains enough of its original speed so that total mechanical energy is conserved. We expect mechanical energy to be conserved since there are no sharp corners anywhere.

 

[Rikudo]

In the case of a disc colliding with the base of the ramp we have an infinite force

How can we know that it is so big? The force that is exerted by the wedge to the disc might be 10 N, 50N, etc.

 

[jbriggs444]

How can we know that it is so big? The force that is exerted by the wedge to the disc might be 10 N, 50N, etc.

The curve is said to be of negligible size. The smaller the curve, the greater the force.

 

[haruspex]

How can we know that it is so big? The force that is exerted by the wedge to the disc might be 10 N, 50N, etc.

When you know that a certain change in momentum has occurred, you know that is the impulse involved. You don't need to worry about the time taken or the magnitude or profile over time of the force. The shorter the time, the greater the average force must be.
This is quite different from the effect of, e.g., gravity. In that case, the force is known, so the shorter the time, the less the impulse, tending to zero in the limit.

 

[erobz]

Did the wedge have a displacement over the transition in this solution?

 

[haruspex]

Did the wedge have a displacement over the transition in this solution?

In the limit, as the transition interval (time and distance) tends to zero, no.

 

[erobz]

Any hints for how to get to this result? I would think the momentum of the puck has to change magnitude of velocity and direction, but how do we know the extent of the change in both these?

 

[TSny]

Here's an outline of a solution where most of the analysis is carried out in the center of mass reference frame (CMF).

(1) In the CMF, what are the speeds of the wedge and the disc before the disc encounters the wedge? Express in terms of $u$, the initial speed of the disc in the lab frame.

Spoiler

Each has speed $u/2$.

(2) What is the system's total energy in the CMF?

Spoiler

$E = \large \frac{mu^2}{4}$

(3) What is the maximum vertical height $y_{max}$ attained by the disc?

Spoiler

$y_{max} = \large \frac{u^2}{4g}$

(4) Let $v_x$ denote the magnitude of the horizontal velocity of the disc (or wedge) in the CMF at some instant while the disc is sliding up the wedge. Let $v_y$ denote the corresponding vertical speed of the disc. Express $v_y$ in terms of $v_x$ and $\theta$.

Spoiler

$v_y = 2 v_x \tan \theta$

(5) Let $v_{x,0}$ and $v_{y,0}$ denote the horizontal and vertical components of the velocity of the disc in the CMF just after the disc starts sliding up the wedge. Use conservation of energy and the result of (4) to find expressions for $v_{x,0}$ and $v_{y,0}$ in terms of $u$ and $\theta$.

Spoiler

$\large v_{x,0} = \frac{u}{2} \frac{\cos \theta}{\sqrt{1+\sin^2 \theta}} \,\,\,\,\,\, v_{y, 0} = u \frac{\sin \theta}{\sqrt{1+\sin^2 \theta}}$

(6)The time interval $\Delta t$ that it takes the disc to reach maximum height is just $\Delta t = \large \frac{y_{max}}{\langle v_y \rangle}$, where $\langle v_y \rangle$ is the average vertical velocity of the disc. Find $\Delta t$ in terms of $u$, $g$, and $\theta$.

Spoiler

$\Delta t= \large \frac{u}{2g} \frac {\sqrt{1+\sin^2 \theta}}{\sin \theta}$

(7) The displacement $\Delta x_{cm}$ of the wedge in the CMF while the disc slides up the wedge is $\Delta x_{cm} = \langle v_{x} \rangle \Delta t$. Find $\Delta x_{cm}$ in terms of $u, g$ and $\theta$.

Spoiler

$\Delta x_{cm} = \left(-\frac{v_{x,0}}{2} \right) \Delta t = \large -\frac{u^2 \cot \theta}{8g }$

(8) Find the displacement $\Delta x_{lab}$ of the wedge in the lab frame.

Spoiler

$\Delta x_{lab} = \Delta x_{cm} + \frac{u}{2} \Delta t = \large \frac{u^2 \left ( 2 \sqrt{1+\sin^2 \theta} - \cos \theta \right )}{8g \sin \theta}$

 

[kuruman]

The bottom line shown in the spoiler, post #25, $$\Delta x_{lab} = \frac{u^2 \left ( 2 \sqrt{1+\sin^2 \theta} - \cos \theta \right )}{8g \sin \theta}$$ bothers me because of its behavior in the limiting cases.

When $\theta \rightarrow 0,$ the expression correctly predicts zero displacement. No incline means no transfer of energy and momentum. Newton's first law rules.

However, when $\theta=\dfrac{\pi}{2},$ the expression predicts $$\Delta x_{lab} = \frac{u^2 \left ( 2 \sqrt{2} \right )}{8g }$$ and this bothers me. When the incline becomes a vertical wall, the collision is completed instantly. The disk does not slide up the incline and there ought to be zero displacement of the wedge from the point of the collision.

Am I misinterpreting something?

 

[erobz]

The bottom line shown in the spoiler, post #25, $$\Delta x_{lab} = \frac{u^2 \left ( 2 \sqrt{1+\sin^2 \theta} - \cos \theta \right )}{8g \sin \theta}$$ bothers me because of its behavior in the limiting cases.

When $\theta \rightarrow 0,$ the expression correctly predicts zero displacement. No incline means no transfer of energy and momentum. Newton's first law rules.

However, when $\theta=\dfrac{\pi}{2},$ the expression predicts $$\Delta x_{lab} = \frac{u^2 \left ( 2 \sqrt{2} \right )}{8g }$$ and this bothers me. When the incline becomes a vertical wall, the collision is completed instantly. The disk does not slide up the incline and there ought to be zero displacement of the wedge from the point of the collision.

Am I misinterpreting something?

I don't know if you are misinterpreting anything, but why does it become instantaneous when the wall becomes vertical? If the geometry of the transition is irrelevant, then it would seem that it could change its momentum to vertically up the wall over some finite interval of time so long as the transition was smooth - say a quarter circle? It's quite likely that I mis gauge a subtle argument here, as I did in the first place about the mass having negligible transit time over the transition.

 

[haruspex]

When the incline becomes a vertical wall, the collision is completed instantly. The disk does not slide up the incline and there ought to be zero displacement of the wedge from the point of the collision.

In all cases short of vertical, we are assuming a smooth transition, with no loss of mechanical energy. The limit of such a sequence should be the same, specifically, the left over KE after the wedge and disc reach the same horizontal speed must be a vertical motion of the disc. so we still have to consider the time it takes to reach its maximum height.
If the disc moves initially at speed u then they both get a horizontal speed of u/2. The left over KE is $\frac 14mu^2$, so the vertical speed of the disc is $\frac u{\sqrt 2}$. The disc will reach its highest point after $\frac u{g\sqrt 2}$, at which time the wedge will have moved $\frac {u^2}{2g\sqrt 2}$.