Olympiad type problem!

A particle, projected with a velocity $u$ is acted upon a force which produces a constant acceleration $a$ in the plane of the motion inclined at a constant angle $\alpha$ with the direction of motion. Obtain the intrinsic equation of the curve described, and show that the particle will be moving in the opposite direction to that of projection at time $\dfrac u{a \cos \alpha} \left( e^{\pi \cot \alpha}-1\right)$ .

How to solve this? Please help.

#Mechanics

Easy Math Editor

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Comments

Simulating this numerically, the anticipated result for the time is easily verified. The (x,y) path doesn't appear to be any ordinary shape.

I should say also that it is easy to write the system of differential equations for this problem. Solving them in closed-form is another matter. Let $v_x$ and $v_y$ be the $x$ and $y$ velocities of the particle:

$\dot{v}_x = \frac{a}{\sqrt{v_x^2 + v_y^2}} (v_x \, \cos \alpha - v_y \, \sin \alpha) \\ \dot{v}_y = \frac{a}{\sqrt{v_x^2 + v_y^2}} (v_x \, \sin \alpha + v_y \, \cos \alpha)$

Steven Chase - 1 year, 11 months ago

I can help with a part of the solution.

Consider the system of differential equations posted by Steven Chase:

Dividing $\dot{v}_y$ and $\dot{v}_x$ , taking $\tan(\alpha) = p$ , and simplifying gives us:

$\frac{dv_y}{dv_x} = \frac{p + \frac{v_y}{v_x}}{1-p\frac{v_y}{v_x}}$

Let $\frac{v_y}{v_x} = z$ . This implies:

$\frac{dv_y}{dv_x} = z + v_x \frac{dz}{dv_x}$

Replacing this in the above differential equation gives us:

$z + v_x \frac{dz}{dv_x} =\frac{p +z}{1-pz}$

This equation can be solved by separating the variables and integrating. Using the conditions that when $v_x=u$ then $z = 0$ gives us the particular solution:

$\tan^{-1}\left(\frac{v_y}{v_x}\right) = \frac{\tan(\alpha)}{2}ln\left(\frac{v_x^2+v_y^2}{u^2}\right)$

From here, we can deduce that when the particle moves opposite to the initial direction, $\tan^{-1}\left(\frac{v_y}{v_x}\right) = \pi$ and $v_y = 0$ and $v_x$ acts along negative x. This step is not mathematically rigorous but is done so by intuition. The speed of the particle at this instant is then:

$v_{xc} = u e^{\pi\cot(\alpha)}$

This is a part of the solution. I am yet to figure out the time computation from here. I'll attempt to do so later and will update this post accordingly.

Karan Chatrath - 1 year, 11 months ago

Thanks nice approach...

Rajyawardhan Singh - 1 year, 11 months ago

A non rigorous way of arriving at the time would be to take the difference between $v_{xc}$ (refer post) and $u$ and divide it by the magnitude of the x component of acceleration at the respective instants (which is $a\cos(\alpha)$ ). If you can figure out a way of proving the required expression for time, do share.

Karan Chatrath - 1 year, 11 months ago

No thoughts are coming how to solve this... Can somebody help me?

Rajyawardhan Singh - 1 year, 12 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$