PAT 2009 Solutions

Ohhh I see!

for Q22 i used a = (v^2)/r and got A, hard question for 1 mark!!

For Q24:

Let $l, c, v$ be the ship's length, number of crews and cargo capacity respectively.

From the qn, $c=kl^2$ and $v=ml^3$

1. As a quizzer and a roodle taken together have the same length as two pangs, $l_q + l_r = 2l_p \Leftrightarrow l_q = 2l_p - l_r$

2. As the crew of a quizzer is just sufficient to provide crew for two pangs and a roodle, $c_q =2c_p + c_r \Rightarrow k{l_q}^2 = 2k{l_p}^2 + k{l_r}^2 \Rightarrow{l_q}^2 = 2{l_p}^2 + {l_r}^2$

Finding $v_q=m{l_q}^3$,

Using 1.: ${l_q}^3 = (2l_p - l_r)^3 =8{l_p}^3 - 12{l_p}^2l_r + 6l_p{l_r}^2 - {l_r}^3$

Using 1. and 2.: ${l_q}^3 = (2l_p - l_r)(2{l_p}^2 + {l_r}^2)=4{l_p}^3 - 2{l_p}^2l_r + 2l_p{l_r}^2 - {l_r}^3$

Note that constant m is missing in both equations because both sides of the equation have constant m.

Comparing both equations of ${l_q}^3$,

$8{l_p}^3 - 12{l_p}^2l_r + 6l_p{l_r}^2 - {l_r}^3=4{l_p}^3 - 2{l_p}^2l_r + 2l_p{l_r}^2 - {l_r}^3$

$4{l_p}^3 - 10{l_p}^2l_r + 4l_p{l_r}^2=0$

$2l_p(2{l_p}^2 - 5{l_p}l_r + 2{l_r}^2)=0$

$(2l_p - l_r)(l_p - 2l_r) = 0 \Rightarrow 2l_p = l_r, l_p=2l_r$

Recall that a quizzer and a roodle taken together have the same length as two pangs? It is impossible for $2l_p = l_r$, which is the length of 2 pangs is equal to the length of a roodle, then.

Since $l_p=2l_r$, using back the 2nd derived ${l_q}^3$ equation to find the number of pangs and roodles needed to match the cargo capacity of a quizzer,

${l_q}^3 = 4{l_p}^3 - 2{l_p}^2(0.5l_r) + 2(2l_r){l_r}^2 - {l_r}^3 = 4{l_p}^3 - {l_p}^3 + 4{l_r}^3 - {l_r}^3 = 3{l_p}^3 + 3{l_r}^3$

Thus, 3 pangs and 3 roodles are needed.

I feel that this is the most difficult qn in the paper.

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Q 18. (I assume the reason there isn't solution for this one up is that no-one could be bothered? )

Basically all that needs to be done is rearrange the equation for mass then stick in the numbers, do some arithmetic and hey presto:

$t = d\sqrt {\dfrac{m}{2qU} }$

$t^2 = d^2 \times \dfrac{m}{2qU}$

$2t^2qU = d^2m$

$m = \dfrac{2t^2qU}{d^2}$

Time for the numbers, but first:

$30\mu s = 30 \times 10^{-6}s$

$16kV = 1.6 \times 10^4V$

$m = \dfrac{2 \times (30 \times 10^{-6})^2 \times 1.6 \times 10^{-19} \times 1.6 \times 10^4 }{1.5^2}\ kg$

$= \dfrac{2 \times 900 \times 10^{-12} \times 1.6 \times 10^{-19} \times 1.6 \times 10^4 }{2.25}\ kg$

$= \dfrac{2 \times 900 \times 1.6 \times 1.6 \times 10^{-12} \times 10^{-19} \times 10^4 }{\dfrac{9}{4} }\ kg$

$= 2 \times 900 \times \dfrac{4}{9} \times 2.56 \times 10^{-27}\ kg$

$= 2 \times 400 \times 2.56 \times 10^{-27}\ kg$

$= 8 \times 256 \times 10^{-27}\ kg$

$= 2048 \times 10^{-27}\ kg$

$= 2.048 \times 10^{-24}\ kg$

$= 2.0 \times 10^{-24} kg\ \ (2\ sig.\ figs.)$

So D is the correct one.

Q 18. (I assume the reason there isn't solution for this one up is that no-one could be bothered? )

Basically all that needs to be done is rearrange the equation for mass then stick in the numbers, do some arithmetic and hey presto:

$t = d\sqrt {\dfrac{m}{2qU} }$

$t^2 = d^2 \times \dfrac{m}{2qU}$

$2t^2qU = d^2m$

$m = \dfrac{2t^2qU}{d^2}$

Time for the numbers, but first:

$30\mu s = 30 \times 10^{-6}s$

$16kV = 1.6 \times 10^4V$

$m = \dfrac{2 \times (30 \times 10^{-6})^2 \times 1.6 \times 10^{-19} \times 1.6 \times 10^4 }{1.5^2}\ kg$

$= \dfrac{2 \times 900 \times 10^{-12} \times 1.6 \times 10^{-19} \times 1.6 \times 10^4 }{2.25}\ kg$

$= \dfrac{2 \times 900 \times 1.6 \times 1.6 \times 10^{-12} \times 10^{-19} \times 10^4 }{\dfrac{9}{4} }\ kg$

$= 2 \times 900 \times \dfrac{4}{9} \times 2.56 \times 10^{-27}\ kg$

$= 2 \times 400 \times 2.56 \times 10^{-27}\ kg$

$= 8 \times 256 \times 10^{-27}\ kg$

$= 2048 \times 10^{-27}\ kg$

$= 2.048 \times 10^{-24}\ kg$

$= 2.0 \times 10^{-24} kg\ \ (2\ sig.\ figs.)$

So D is the correct one.

Check you formally working it out. I jotted it down with two lines of working:P

+ Mental time to be up on a saturday!

That is some hardcore latexing

I like my workings formal.

I am always up at mental times...



Yeah... it took a while...

, how you anyway?

I'll take a punt at 25...I have never studied circuits like this....

Total resistance:

r=1000ohm resistance.

$R=\dfrac{5r}{8}+125$

$R=\dfrac{5000}{8}+125=750$

Circuits are my weakest area, so I have no idea about 25.

I have however just completed 26. I will type it up in a minute.

...

The work done by an electron moving through a potential difference of 50v :

$1.6 \times 10^{-19} \times 10 \times 5 = 8 \times 10^{-18}$

$W=Fs$

$\dfrac{8 \times 10^{-18}}{0.4 \times 10^{-30}}= a, a=2 \times 10^{13}$

$a= \frac{s}{t^2}$

$t=\sqrt{\frac{0.4}{2 \times 10^{13}}}$

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

$s= \frac{1}{2} \times 10 \times 2 \times 10^{-14}$

$1 \times 10^{-13}m$

I agree with ML's answer. I think you, wcp, have put a 2 where it should be 0.4 in your last line of working

I agree with ML's answer. I think you, wcp, have put a 2 where it should be 0.4 in your last line of working