Morin 3.7 -- Block sliding sideways on an inclined plane

[masteralien]

TL;DR Summary: In Morin 3.7 sliding sideways on a plane I used a completely different method than he did and got the correct answer is my method right

The problem statement is as follows
I split up the friction force into x and y components derived a diff eq for v_y in terms of v_x then took the limit as v_x goes to 0

My solution was this

$$m\ddot{x} = -\mu mg\cos(\theta) \cos(\varphi)$$
$$\\m\ddot{y}= -mg\sin(\theta)-\mu mg\cos(\theta)\sin(\varphi)$$
$$\\\mu = \tan(\theta)$$
$$\\v_x=\dot{x}$$
$$\\v_y=\dot{y}$$
where $$\varphi$$ is the angle the velocity vector makes
$$\\\dot{v}_x=-g\sin(\theta) \frac{v_x}{\sqrt{v_x^2 + v_y^2}}$$
$$\\\dot{v}_y=-g\sin(\theta)-g\sin(\theta) \frac{v_y}{\sqrt{v_x^2 + v_y^2}}$$

At long times v_x’ will be 0 so we can set final v_x = 0 as setting v_x’=0 means v_x will be 0

I then divided to get

$$\frac{dv_y}{dv_x}=\frac{v_y}{v_x} + \sqrt{1+\left(\frac{v_y}{v_x}\right)^2}$$

I substituted v_y/v_x = u and with some algebra setting the lower bound of v_x as V and upper as v_x to take the limit later and for v_y I took the upper bound as v_y and lower as 0 get


$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$

$$\frac{v_x^4}{V^2}-\frac{2v_y v_x^2}{V}+v_y^2=v_x^2+v_y^2$$

Then get
$$\frac{v_x^2}{V^2}-\frac{2v_y}{V}=1$$

Finally taking v_x goes to 0

$$v_y = -\frac{V}{2}$$

With the final speed being
$$V_f=\frac{V}{2}$$
I want to know if this solution is correct as this gives the correct answer

This is the book’s solution

 

[haruspex]

I substituted v_y/v_x = u and with some algebra setting the lower bound of v_x as V and upper as v_x to take the limit later and for v_y I took the upper bound as v_y and lower as 0 get

$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$

Unfortunately, that does not give the right answer if you plug in $v_x=0$.
Please post the omitted algebra.

 

[masteralien]

Its right below the algebra

 

[masteralien]

Unfortunately, that does not give the right answer if you plug in $v_x=0$.
Please post the omitted algebra.

The steps are right below. I even checked with a graphing calculator that as v_x approaches 0 v_y approaches -V/2

 

[haruspex]

The steps are right below.

You wrote

and with some algebra

Can you post that?

I even checked with a graphing calculator that as v_x approaches 0 v_y approaches -V/2

But if you substitute $v_x=0$ in the equation I quoted you get $v_y=0$, no?

 

[masteralien]

You wrote

Can you post that?

But if you substitute $v_x=0$ in the equation I quoted you get $v_y=0$, no?

No my steps are right below I subtracted v_y and squared both sides then divided by v_x^2 then took the limit. You need to take the limit as it approaches as this for long times its a bit like terminal velocity.

 

[haruspex]

No my steps are right below

I am asking you to show how you got from

$$\frac{dv_y}{dv_x}=\frac{v_y}{v_x} + \sqrt{1+\left(\frac{v_y}{v_x}\right)^2}$$

to

$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$

I do not see that anywhere.

I subtracted v_y and squared both sides

Squaring introduces extra solutions. Specifically here, you are now solving
$$\frac{v_x^2}{V}=\pm\sqrt{v_x^2+v_y^2}+v_y$$
If we substitute $v_x=0$ in the extra solution that introduces:
$$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$
we get 0=0, so this spurious solution is the one that, through taking the limit, arrives at $v_y=V/2$.

$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$ cannot be right because $v_x=0$ implies $v_y=0$, whether by taking limits or simply plugging in. It seems likely that it should have been $$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$.
You can avoid introducing extra solutions by dividing that through by $v_y$, applying the binomial expansion and taking limits.

 

[masteralien]

Also v_x would never truly reach 0 just get very close and its best to express the solution a bit differently. Also I have plugged in my equation the first into desmos to check and as you decrease v_x it goes to -V/2 but never set v_x=0
$$v_y=v_x\sinh(\ln(\frac{v_x}{V}))$$
Taking x to 0
=$$V x\sinh(\ln(x))$$
=$$V x/2 (x-1/x)= V /2 (x^2-1)$$
=$$-V/2$$ as x goes to 0

 

[masteralien]

I am asking you to show how you got from

to

I do not see that anywhere.


Squaring introduces extra solutions. Specifically here, you are now solving
$$\frac{v_x^2}{V}=\pm\sqrt{v_x^2+v_y^2}+v_y$$
If we substitute $v_x=0$ in the extra solution that introduces:
$$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$
we get 0=0, so this spurious solution is the one that, through taking the limit, arrives at $v_y=V/2$.

$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$ cannot be right because $v_x=0$ implies $v_y=0$, whether by taking limits or simply plugging in. It seems likely that it should have been $$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$.
You can avoid introducing extra solutions by dividing that through by $v_x$, applying the binomial expansion and taking limits.

Also is my solution physically sound like is the setup logical physically

 

[masteralien]

I am asking you to show how you got from

to

I do not see that anywhere.


Squaring introduces extra solutions. Specifically here, you are now solving
$$\frac{v_x^2}{V}=\pm\sqrt{v_x^2+v_y^2}+v_y$$
If we substitute $v_x=0$ in the extra solution that introduces:
$$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$
we get 0=0, so this spurious solution is the one that, through taking the limit, arrives at $v_y=V/2$.

$$\frac{v_x^2}{V}=\sqrt{v_x^2+v_y^2}+v_y$$ cannot be right because $v_x=0$ implies $v_y=0$, whether by taking limits or simply plugging in. It seems likely that it should have been $$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$.
You can avoid introducing extra solutions by dividing that through by $v_x$, applying the binomial expansion and taking limits.

You would have to solve for v_y

 

[haruspex]

You would have to solve for v_y

I meant divide by $v_y$, of course.
$$\frac{v_x^2}{V}=-\sqrt{v_x^2+v_y^2}+v_y$$
$$\frac{v_x^2}{Vv_y}=-\sqrt{1+(\frac{v_x}{v_y})^2}+1$$.
$$\approx-\frac 12(\frac{v_x}{v_y})^2$$
$$v_y=-V/2$$
Starting with $$\frac{v_x^2}{V}=+\sqrt{v_x^2+v_y^2}+v_y$$ does not lead to that.

 

[Gavran]

If $\frac {dv_y}{dv_x} = \frac {v_x}{V}$ is true, from here it can be get that $v_y = \frac {v_x^2}{2V} - \frac {V}{2}$. For $v_x=0$ $v_y$ will be $-\frac {V}{2}$.