Mechanics Help (1 Viewer)

YaMiTe · Jun 14, 2011

Help please

There are solutions but I don't get them

Drongoski · Jun 14, 2011

Q10

$\ddot x = 0$ . . . . . . . . . (1)

$\ddot y = -g$ . . . . . . . . . (2)

$\therefore v = \dot x = \int \ddot x dt = A$ . . . . . (3)

when t=0, $v = \sqrt{4ga} \times cos \alpha$

$\therefore v = \sqrt {4ga} \times cos \alpha$ . . . . . (4)

$\therefore x = x(t) = \int vdt = \int \sqrt{4ga} \times cos \alpha \cdot dt = t \sqrt{4ga} \cdot cos \alpha + B$

When t=0, x=0 ==> B=0

$\therefore x(t) = t \sqrt{4ga} \cdot cos \alpha$ . . . . . (5)

$\dot y = \int \ddot y dt = \int -g dt = -gt + C$

When t=0, $\dot y = \sqrt{4ga }\cdot sin \alpha \implies C = \sqrt {4ga} \cdot sin \alpha$

$\therefore \dot y (t) = -gt + \sqrt{4ga} \cdot sin \alpha$ . . . . . (6)

$\therefore y(t) = \int \dot y dt = \int (above) dt = - \frac {1}{2}gt^2 + t\cdot \sqrt {4ga} sin \cdot \alpha + D$

y(0) = 0 ==> D = 0

$\therefore y(t) = - \frac {1}{2}gt^2 + t \cdot \sqrt{4ga} \cdot sin \alpha$ . . . . . (7)

Now at some time t = T, x(T) = 2a and y(T) = a

$x(T) = T \times \sqrt{4ga} \cdot cos \alpha = 2a \implies T = \frac {2a}{\sqrt {4ga} \cdot cos \alpha}$

Subst this value of T into eqn (7) we get:

$y(T) = - \frac {1}{2}g \left [\frac {2a}{\sqrt {4ga}\cdot cos \alpha} \right ]^2 + \frac {2a}{\sqrt {4ga} \cdot cos \alpha} \times \sqrt {4ga} \cdot sin \alpha = a$

Simplifying we end up with:

$\sqrt {4ga} \cdot sin \alpha \cdot \frac {2a}{\sqrt {4ga}cos \alpha} - \frac {g}{2} \left[ \frac {2a}{\sqrt {4ga}\cdot cos \alpha } \right ]^2 = a$

I'm sick of doing all his LaTeXing. So believe me - this simplifies into:

$2a \cdot tan \alpha - \frac {1}{2} (tan^2 \alpha + 1) = a$

$\therefore tan^2 \alpha -4 tan \alpha + 3 = 0$

$\therefore (tan \alpha - 3)(tan \alpha - 1) = 0$

So $\alpha$ = 72 (71.56 ..) deg and 45 deg (difference: 27 deg)

Got to stop now - inputting in TeX is so tiring for me. Maybe this is still the same as the solution you have and you still need help explaing it. I'm sure some kind soul here will help you.

YaMiTe · Jun 15, 2011

Niceee

You could always write the solutions up on paper and then scan it up.

but anyhow, thanks for explaining the first part

kooliskool · Jun 16, 2011

For 11a, First get the equation:

$x=\frac{V\sin 2\alpha}{g}$
I assume you can get that ya?

Then let the correct position of the target be $R$ metres from the origin, therefore:

$At \; \alpha=45^{\circ},x=R+30, R+30=\frac{V\sin \left ( 2\times 45^{\circ} \right ) }{g}$
$R+30=\frac{V}{g}$
$V=g\left ( R+30 \right )$ ..............[1]

$At \; \alpha=60^{\circ},x=R-60, R-60=\frac{V\sin \left ( 2\times 60^{\circ} \right ) }{g}$
$R-60=\frac{\sqrt{3} V}{2g}$ ..............[2]

Sub [1] into [2]

$2g\left ( R-60 \right )=\sqrt{3} g\left ( R+30 \right )$
$\left ( 2-\sqrt{3} \right )R=120+30\sqrt{3}$
$R=\frac{120+30\sqrt{3}}{2-\sqrt{3}}$

Now to get the $\alpha$ , just substitute what you got into $x$ and V again from before:

$\frac{120+30\sqrt{3}}{2-\sqrt{3}}=\frac{g\left ( \frac{120+30\sqrt{3}}{2-\sqrt{3}}+30\right )\sin {2\alpha}}{g}$

Solving for $\alpha$ is just algebra, so the final answer is:

$\alpha \approx 36^{\circ}24' \; or \; \alpha \approx 53^{\circ}36'$ (to the nearest minute)

Drongoski · Jun 16, 2011

YaMiTe said:
Niceee

You could always write the solutions up on paper and then scan it up.
but anyhow, thanks for explaining the first part

You won't believe it. I don't know how!

kooliskool · Jun 16, 2011

For 11b, you do nearly the same thing, so again first get:

$x=\frac{V\sin 2\alpha }{g}$ ,

Then substitute the correct conditions (and letting $R$ be the correct position), you get:

$R+a=\frac{V\sin 2\alpha }{g}$ ...........[1]
$R-b=\frac{V\sin 2\beta }{g}$ ...........[2]
$R=\frac{V\sin 2\theta }{g}$ ...........[3]

Solving for $V$ first by eliminating R is easier:

Sub [3] into [1]: $\frac{V\sin 2\theta }{g}+a=\frac{V\sin 2\alpha }{g}$

$V=\frac{ga}{\sin 2\alpha-\sin 2\theta}$ ...........[4]

Now Sub [3] and [4] into [2] (since you haven't used that yet), you will eliminate everything and you will get:

$\frac{\frac{ga}{\sin 2\alpha-\sin 2\theta}\sin 2\theta }{g}-b=\frac{\frac{ga}{\sin 2\alpha-\sin 2\theta}\sin 2\beta}{g}$

$\frac{ga}{\sin 2\alpha-\sin 2\theta}\sin 2\theta-gb=\frac{ga}{\sin 2\alpha-\sin 2\theta}\sin 2\beta$

$ga\sin 2\theta-gb\left ( \sin 2\alpha-\sin 2\theta \right )=ga\sin 2\beta$

Cancelling the g and rearranging, you can easily now get:

$\left (a+b \right )\sin 2\theta=a\sin 2\beta+b \sin 2\alpha$

kooliskool · Jun 16, 2011

Hope you understand and find the solutions useful

YaMiTe · Jun 16, 2011

WOW thank you so much.

completely get it!
repped.

btw, are these projectile questions the typical types you'll get in the HSC? (like they generally dont get any harder?)

also, thanks Drongoski!

kooliskool · Jun 16, 2011

Erm, you can't say there's a boundary to questions' difficulty, but all I can say is, the difference between the difficulty in questions you see in 3u and questions in 4u is that in 3u, it's about calculation and derivation of formulas. But in 4u, you need to be good at algebra and manipulation.

Trebla · Jun 17, 2011

Harder projectiles have rarely been assessed in past HSC exams...

cutemouse · Jun 18, 2011

As Trebla said they don't ask harder projectile motion much.

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