The Proof is 'not-so' Trivial - Physics Edition

To be honest every single problem on this thread has been a homework question so far... actually worse... a 'bookwork' question.

I think you'd struggle to google the answer to this one unless you knew where to look.

How can you say with certainty that all the questions thus far a homework question? None of the ones I've posted were book work or homework. I don't get questions to prove for homework.

I've had all those problems as homework...

Problem 19 : What is the energy-momentum relation for a space-like particle? For example, the hypothetical particle known as the tachyon.

What unusual properties can you deduce about the allowed energies of these particles?

(I posted this question in a different thread, but realised it would be well suited here.)

Given the like, is anyone actually trying to solve this problem/attempted it? If not, I will post a solution for people to browse should they please.

Problem 20**

A particle ejected into a fluid at $t=0$ experiences a drag represented by $a_d=-\text{k}v$ where $\text{k}$ is a constant and $v$ is the velocity of the projectile. If the initial speed is $v_o$ show that the y-component of the velocity is

$\dfrac{v_y}{\text{k}} + g = \dfrac{\text{A}e^{-\text{k}t}}{\text{k}}$

Problem 17**/***

Suppose that a block is on a inclined plane with a weight, $W$, and force,$P$, acting horizontally

Show that if motion occurs up the incline

$P > W[\dfrac{\tan\theta + \mu_s}{1-\mu_s\tan\theta}]$

where $\mu_s$ is the co-effiecient of static friction

My maths isn't great... is this:



basically the right idea? If it's more complicated than that then I'll likely run around in circles for days.

EDIT: when you say for drag, a=-kv, is that supposed to be an acceleration or a force? Just asking because my answer currently has m's in it that I have no idea how to get rid of, and if the total drag was supposed to be d=-mkv, then the m's would all have cancelled in the second line or so. Otherwise, I've just done something stupid and wrong.

a is for acceleration. That'll make things clearer for you, though you can treat the k as the mass which is also a constant but yeah, you get the picture.

The $a_d=-kv$ represents the drag force as a function of velocity, no? Thus lerjj should end up with a second order D.E

Assume a hollow uniform cylinder of a set mass.

And Yea, com is highest in either full or empty bottle. I want the equation for combined com, then the minimum point of this equation.

Posted from TSR Mobile

The $a_d=-kv$ represents the drag force as a function of velocity, no? Thus lerjj should end up with a second order D.E

It's $F_{drag} = -mkv$ which ends up making this not that bad of an equation. I do currently have slightly the wrong answer though.

Its really easy form here, though. Basic GCSE maths.

I must be missing something, because there's no way I can see to get from
$v_y -\dfrac{g}{k} = Ae^{-k/t}$
to
$\dfrac{v_y}{k} + g = \dfrac{Ae^{-k/t}}{k}$

To get the g term right, you need to multiply by -k everywhere, and that doesn't give the correct answer for any of the other terms. I assume that I just made a mistake somewhere.

$v_y -\dfrac{g}{k} = Ae^{-k/t}$

Factorising k we get, your negative sign is wrong though

$k(\dfrac{v_y}{k} - g) = Ae^{-k/t}$

$(\dfrac{v_y}{k} - g) = \dfrac{Ae^{-k/t}}{k}$

No, factoring k you get

$k(\dfrac{v_y}{k}-\dfrac{g}{k^2}) = Ae^{-k/t}$
$\dfrac{v_y}{k} - \dfrac{g}{k^2}=\dfrac{ Ae^{-k/t}}{k}$

Which is different. I'm actually inclined to think the -ve sign is correct, as moving the term to the other side shows that the y velocity is greatest when you have a larger acceleration, which makes sense.

Here's the solution.

Taking into account the effect of gravity.

$a_p = -\text{k}v_p - g\mathbf{j}[br][br]a_x\mathbf{i} + a_y\mathbf{j} = -\text{k}(v_x \mathbf{i}+ v_y\mathbf{j}) - g\mathbf{j}$

Equating the j terms



solve integral to get (starting from initial conditions so c = 0)

$v_y + \dfrac{g}{k} = Ae^{-kt}$