Author Topic: TSSM Exam 1 2009 (Read 3276 times) Tweet Share

Hooligan · « **on:** October 27, 2009, 02:53:30 pm »

Some problems I had... (yet again.

):

Question 2c.

I don't get why $z_{1} \in T$ such that $\[z_{1}w \in T \] $ $\implies$ $z_{1}=z_{2}w$
I get that after arriving at the above equation, that when you multiple by $i$ (since w= $i$ ); that you are rotating $\frac{\pi }{2}$ anticlockwise (right?)

Then after that.... I just get confused where they get the answers $\[z_{1} = 2, -2, 2i, -2i\] $

Question 4.
Shouldn't the first normal equation be $y = 2$ not $x = 2$ ?

Question 7.
I had problems finding the range..... (if you're reading TrueTears, I did it the way we discussed yesterday, but I got a range of $[0,\infty)$ 'cause, like you said, range of cartesian = range of $y(t)$ and so from there, we know $y(t) = 3t$ , and since this is a straight line and the only restriction on it is $t \geq 0$ , the range of $y(t)$ and hence the range of the cartesian is $[0,\infty)$ ?)

Question 8.
I'm confused how they did it, but this is how I started it, and then got stuck proving it:

First, a=g, t=0.4s, s=8, u=?, v=0.... using $\[s = ut + \frac{1}{2}at^{2}\] $ I get u = $20 - \frac{g}{5} m/s^{2}$ or $\frac{100-g}{5} m/s^{2}$
Then, u= $20 - \frac{g}{5} m/s^{2}$ , v=0, s=?, a=g, and solving for s I get s= $- \frac{(100-g)^{2}}{50g} m$ compared to the answer which is... s= $\frac{(100+g)^{2}}{50g} m$

Is this the right way of going about it? $:-\$

Thanks in advance.

TrueTears · « **Reply #1 on:** October 27, 2009, 03:05:08 pm »

2c.

Multiplying a number of i means rotation 90 degs anticlockwise. Thus if you rotate the 2 parallel lines and you rotate them 90 deg anticlockwise one will pass through (-2,0) and (0,2) and the other one will pass through (0,-2) and (2,0)

Now it says $z_1 \times i \in T$

T was all the complex numbers ON the two lines before rotation, now if you rotate it what are the numbers now that were still ON the two lines before rotation? That's right, those 4 points namely, (-2,0) and (0,2) and (0,-2) and (2,0)

So your answer would be $z$ = -2, 2i, -2i, 2

If you still are unsure about this question I'll sketch it up for you

4.

The first normal equation is undefined meaning you have something like $\frac{dy}{dx} = \frac{x+y}{0}$ . This can only happen if it's a vertical line, since rise on run is 0. (There is no 'run' for vertical lines). Thus it is x = 1 as the equation.

7.

Yes you are right for this question

. The answers are 100% wrong, I remember checking with kamil and Mao in my thread, the range is NOT [0,30]

If you wanna have a read: http://vcenotes.com/forum/index.php/topic,9192.msg184255.html#msg184255

TrueTears · « **Reply #2 on:** October 27, 2009, 03:12:27 pm »

8. (This is my way, I didn't do it the answers way)

The motion of the object until it is 8 m above the ground.

$u = 0$ $a = g$ and $t = T$

$v = u + at$

Thus $v = gT$

$x = ut + \frac{1}{2}at^2$

$x = \frac{1}{2}gT^2$

Now for when the ball is travelling from 8 m above the ground until it touches the gruond

$u = gT$ $a = g$ $x = 8$ $t = 0.4$

$x = ut + \frac{1}{2}at^2$

$8 = gT + \frac{1}{2}g(0.4)^2$

Work out $T$ from there.

Now that you worked out $T$ , sub back into $x = \frac{1}{2}gT^2$ to work out x.

Now you know how far the ball travelled from the top of the tower until when it is 8 m above ground. But the question asks for the height of tower so your final answer would be $x + 8$ . A bit of rearranging will get the form they want.

GerrySly · « **Reply #3 on:** October 27, 2009, 03:23:21 pm »

I worked 8 out differently as well

a=g, t=0.4, s=8

$\begin{align*} s&=\frac{1}{2}(u+v)\cdot t\\ &=\frac{1}{2}(v-at+v)\cdot t\\ 8&=\frac{1}{2}(2v-g\frac{2}{5})\frac{2}{5}\\ &=\frac{2v}{5}-\frac{2g}{25}\\ v&=\frac{5(8+\frac{2g}{25})}{2}\\ &=\frac{100+g}{5} \end{align*}$

a=g, u=0, v= $\frac{100+g}{5}$ , s=h

$\begin{align*} v^2&=u^2+2as\\ \left ( \frac{100+g}{5} \right )^2 &=2gh\\ h&=\frac{(100+g)^2}{25(2g)}\\ &=\frac{(100+g)^2}{50g} \end{align*}$

Leahvce09 · « **Reply #4 on:** October 27, 2009, 06:23:40 pm »

for question 7:
how can the range be [o,∞]??
the cartesian equation end up being y=(30/∏)arccos(x/2), which only has a range of [0,30]. Yes, you don't need to restrict the range due to y=3t, however, the arccos graph cannot go to infinity...

Hooligan · « **Reply #5 on:** October 27, 2009, 06:44:11 pm »

Quote from: TrueTears on October 27, 2009, 03:05:08 pm

2c.

Multiplying a number of i means rotation 90 degs anticlockwise. Thus if you rotate the 2 parallel lines and you rotate them 90 deg anticlockwise one will pass through (-2,0) and (0,2) and the other one will pass through (0,-2) and (2,0)

Now it says $z_1 \times i \in T$

T was all the complex numbers ON the two lines before rotation, now if you rotate it what are the numbers now that were still ON the two lines before rotation? That's right, those 4 points namely, (-2,0) and (0,2) and (0,-2) and (2,0)

So your answer would be $z$ = -2, 2i, -2i, 2

If you still are unsure about this question I'll sketch it up for you

Oh, I get it now... soo, it was just pure luck we got the same points (after we rotated) as there were before we rotated? right? Otherwise, there could be question where you have to eliminate answers, where it does not do this.

Quote from: TrueTears on October 27, 2009, 03:05:08 pm

4.

The first normal equation is undefined meaning you have something like $\frac{dy}{dx} = \frac{x+y}{0}$ . This can only happen if it's a vertical line, since rise on run is 0. (There is no 'run' for vertical lines). Thus it is x = 1 as the equation.

If it is undefined, then, it would have to be x = 2, however, if it is 0, then can it be y=2... ?
This is how I did it:
$\frac{\mathrm{dy} }{\mathrm{d} x} = \frac{4x-2y}{2y+2x}$
At $(1,2)$ , $\frac{\mathrm{dy} }{\mathrm{d} x}$ = $\frac{4(1)-2(2)}{2(2)+ 2(1)} = 0$
Then, using: $y - y_{1} = m(x - x_{1})$ :
$y - 2 = 0(x - 1)$
$\implies y=2$ ?

Quote from: TrueTears on October 27, 2009, 03:05:08 pm

7.

Yes you are right for this question . The answers are 100% wrong, I remember checking with kamil and Mao in my thread, the range is NOT [0,30]

If you wanna have a read: http://vcenotes.com/forum/index.php/topic,9192.msg184255.html#msg184255

Oh yay! Thanks TrueTears!! Your mathematical ways are unstoppable!! haha! take that TSSM!!

Quote from: TrueTears on October 27, 2009, 03:12:27 pm

8. (This is my way, I didn't do it the answers way)

The motion of the object until it is 8 m above the ground.

$u = 0$ $a = g$ and $t = T$

$v = u + at$

Thus $v = gT$

$x = ut + \frac{1}{2}at^2$

$x = \frac{1}{2}gT^2$

Now for when the ball is travelling from 8 m above the ground until it touches the gruond

$u = gT$ $a = g$ $x = 8$ $t = 0.4$

$x = ut + \frac{1}{2}at^2$

$8 = gT + \frac{1}{2}g(0.4)^2$

Work out $T$ from there.

Now that you worked out $T$ , sub back into $x = \frac{1}{2}gT^2$ to work out x.

Now you know how far the ball travelled from the top of the tower until when it is 8 m above ground. But the question asks for the height of tower so your final answer would be $x + 8$ . A bit of rearranging will get the form they want.

Whoa... thanks TT!!

TrueTears · « **Reply #6 on:** October 27, 2009, 06:44:22 pm »

Quote from: Leahvce09 on October 27, 2009, 06:23:40 pm

for question 7:
how can the range be [o,∞]??
the cartesian equation end up being y=(30/∏)arccos(x/2), which only has a range of [0,30]. Yes, you don't need to restrict the range due to y=3t, however, the arccos graph cannot go to infinity...

The arccos graph can go to infinity.

We merely restrict it to be a function but the path that the particle follows doesn't have to be a function.

The arccos graph is simply the cos graph but on the y axis. We merely restrict the range and domain to make it a function.

TrueTears · « **Reply #7 on:** October 27, 2009, 06:47:27 pm »

Quote from: Hooligan

Oh, I get it now... soo, it was just pure luck we got the same points (after we rotated) as there were before we rotated? right? Otherwise, there could be question where you have to eliminate answers, where it does not do this.

That's right

Quote from: Hooligan

If it is undefined, then, it would have to be x = 2, however, if it is 0, then can it be y=2... ?
This is how I did it:
$\frac{\mathrm{dy} }{\mathrm{d} x} = \frac{4x-2y}{2y+2x}\] At \[(1,2)\], \[\frac{\mathrm{dy} }{\mathrm{d} x}\] = \[\frac{4(1)-2(2)}{2(2)+ 2(1)} = 0\] Then, using: \[y - y_{1} = m(x - x_{1})\]: \[y - 2 = 0(x - 1)\] \[\implies y=2$ ?

Your working is fine but the question asks for the equation of the normal whereas you have found the equation of the tangent. The normal to the tangent equation of y = 2 is just the vertical line x = 1.

And it's x = 1 because that's the vertical line that passes through the point (1,2). If you had x = 2 it'd pass through another coordinate.

EDIT: I fail at quoting >_>

Hooligan · « **Reply #8 on:** October 27, 2009, 06:50:10 pm »

Quote from: GerrySly on October 27, 2009, 03:23:21 pm

I worked 8 out differently as well

a=g, t=0.4, s=8

$\begin{align*} s&=\frac{1}{2}(u+v)\cdot t\\ &=\frac{1}{2}(v-at+v)\cdot t\\ 8&=\frac{1}{2}(2v-g\frac{2}{5})\frac{2}{5}\\ &=\frac{2v}{5}-\frac{2g}{25}\\ v&=\frac{5(8+\frac{2g}{25})}{2}\\ &=\frac{100+g}{5} \end{align*}$

a=g, u=0, v= $\frac{100+g}{5}$ , s=h

$\begin{align*} v^2&=u^2+2as\\ \left ( \frac{100+g}{5} \right )^2 &=2gh\\ h&=\frac{(100+g)^2}{25(2g)}\\ &=\frac{(100+g)^2}{50g} \end{align*}$

You did it practically the same way I did, except your way, somehow you got $\frac{100+g}{5}$ whereas I got $\frac{100-g}{5}$ ...? I used a different formula, but they're all constant acceleration formulas, so shouldn't we arrive at the same answer?

For the second bit, how come you made v= $\frac{100+g}{5}$ ? Doesn't that mean that when it hits the ground it still has velocity, since you have s = h?

Hooligan · « **Reply #9 on:** October 27, 2009, 06:53:03 pm »

Quote from: TrueTears on October 27, 2009, 06:47:27 pm

Quote from: Hooligan
If it is undefined, then, it would have to be x = 2, however, if it is 0, then can it be y=2... ?
This is how I did it:
$\frac{\mathrm{dy} }{\mathrm{d} x} = \frac{4x-2y}{2y+2x}\] At \[(1,2)\], \[\frac{\mathrm{dy} }{\mathrm{d} x}\] = \[\frac{4(1)-2(2)}{2(2)+ 2(1)} = 0\] Then, using: \[y - y_{1} = m(x - x_{1})\]: \[y - 2 = 0(x - 1)\] \[\implies y=2$ ?
Your working is fine but the question asks for the equation of the normal whereas you have found the equation of the tangent. The normal to the tangent equation of y = 2 is just the vertical line x = 1.

And it's x = 1 because that's the vertical line that passes through the point (1,2). If you had x = 2 it'd pass through another coordinate.

Ohhhh... right... ummm... okay, so once I find out the gradient of the normal is undefined, how to do go about finding the line x = ... ? is there a formula? I don't see the logic behind it now....

EDIT: I GOT IT NOW!!!!!! NEVERMIND! haha! (I re-read your post 3 times until it clicked! thank you!!)

TrueTears · « **Reply #10 on:** October 27, 2009, 06:57:19 pm »

There's not really a formula you just have to kinda visualize it:

So say the graph looks something like this at (1,2) [Doesn't matter at all what it looks like, as long as you know there's a stationary point at (1,2) since the tangent has a gradient of 0]

Now since it asks for the equation of the normal, the ONLY vertical line that passes through the (1,2) is x = 1. Thus the equation of normal is x = 1

EDIT: haha posted this before I saw your EDIT, I'll leave it up here anyways xD

thethingexe · « **Reply #11 on:** October 27, 2009, 10:02:43 pm »

Quote from: Hooligan on October 27, 2009, 06:50:10 pm

Quote from: GerrySly on October 27, 2009, 03:23:21 pm
I worked 8 out differently as well

a=g, t=0.4, s=8

$\begin{align*} s&=\frac{1}{2}(u+v)\cdot t\\ &=\frac{1}{2}(v-at+v)\cdot t\\ 8&=\frac{1}{2}(2v-g\frac{2}{5})\frac{2}{5}\\ &=\frac{2v}{5}-\frac{2g}{25}\\ v&=\frac{5(8+\frac{2g}{25})}{2}\\ &=\frac{100+g}{5} \end{align*}$

a=g, u=0, v= $\frac{100+g}{5}$ , s=h

$\begin{align*} v^2&=u^2+2as\\ \left ( \frac{100+g}{5} \right )^2 &=2gh\\ h&=\frac{(100+g)^2}{25(2g)}\\ &=\frac{(100+g)^2}{50g} \end{align*}$

You did it practically the same way I did, except your way, somehow you got $\frac{100+g}{5}$ whereas I got $\frac{100-g}{5}$ ...? I used a different formula, but they're all constant acceleration formulas, so shouldn't we arrive at the same answer?

For the second bit, how come you made v= $\frac{100+g}{5}$ ? Doesn't that mean that when it hits the ground it still has velocity, since you have s = h?

v is the speed at which it hits the ground, it doesn't hit the ground with 0m/s.

the equation he used basically, which isn't on the formula sheet is.

s=vt-(1/2)at^2

because if you think about it, if it was dropped from h and h>8

at 8m from ground, the u won't really help solve the problem.

you just need to find the speed it hits the ground, v.

so you use

8=v(0.4)-(1/2)g(0.4)^2
you will get
v= (g+100)/5

in your initial working out you use the s=ut+(1/2)at^2 formula.

it's hard to explain, but i think you just can't think of the object stopping instantly when it hits the ground.

Hooligan · « **Reply #12 on:** October 27, 2009, 10:54:41 pm »

Quote from: TrueTears on October 27, 2009, 06:57:19 pm

EDIT: haha posted this before I saw your EDIT, I'll leave it up here anyways xD

awwww... *feels guilty* should have made a complete new post stating I got it. *feels extremely guilty* SORRY TT!

EDIT: Argh! Can't send you 'sorry karma'! Have to wait 12 hours! >.< (I can send Karma now! yay! )

Hooligan · « **Reply #13 on:** October 27, 2009, 11:12:58 pm »

Quote from: thethingexe on October 27, 2009, 10:02:43 pm

v is the speed at which it hits the ground, it doesn't hit the ground with 0m/s.

the equation he used basically, which isn't on the formula sheet is.

s=vt-(1/2)at^2

because if you think about it, if it was dropped from h and h>8

at 8m from ground, the u won't really help solve the problem.

you just need to find the speed it hits the ground, v.

so you use

8=v(0.4)-(1/2)g(0.4)^2
you will get
v= (g+100)/5

in your initial working out you use the s=ut+(1/2)at^2 formula.

it's hard to explain, but i think you just can't think of the object stopping instantly when it hits the ground.

Hmmm... I sorta get what you mean, but I 've come across textbook questions, where since the object hits the ground, you make v=0... eg. (from the top of my head) A truck is travelling with a constant acceleration of 2 and an initial velocity of 3m/s, when it hits its breaks. What time does it take for the truck to come to a stop? (answer: 3/2 secs) - you have to make v=0 to solve this problem, don't you?

However, I am still stuck as why when I used just the $\[s = ut + \frac{1}{2}at^{2}\] $ , but I get the negative, whereas GerrySly, used a constant acceleration in another constant acceleration formula, and that managed to him/her a positive??

why? I'm totally confused.

TrueTears · « **Reply #14 on:** October 27, 2009, 11:36:33 pm »

Quote from: Hooligan on October 27, 2009, 10:54:41 pm

Quote from: TrueTears on October 27, 2009, 06:57:19 pm
EDIT: haha posted this before I saw your EDIT, I'll leave it up here anyways xD

awwww... *feels guilty* should have made a complete new post stating I got it. *feels extremely guilty* SORRY TT!

EDIT: Argh! Can't send you 'sorry karma'! Have to wait 12 hours! >.< (I can send Karma now! yay! )

haha it's fine, glad to help xD

With regards to your other question, yes normally when something stops with v = 0 the question should specify something like this: "...comes to rest" or else it is really ambiguous.

Quote from: Hooligan

# First, a=g, t=0.4s, s=8, u=?, v=0.... using $s = ut + \frac{1}{2}at^{2}$ I get $u = 20 - \frac{g}{5} m/s^{2}$ or $\frac{100-g}{5} m/s^{2}$

Your way is fine

$8 = u(0.4) + (0.5)(g)(0.4)^2$

$u = \frac{100-g}{5}$

$a = g$ $u = 0$ $v = \frac{100-g}{5}$

$v^2 = u^2 + 2ax$

$\left(\frac{100-g}{5}\right)^2 = 2gx$

$x = \frac{(g-100)^2}{50g}$

then $\frac{(g-100)^2}{50g} + 8 = \frac{(g+100)^2}{50g}$

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