STAV 2010 problem

[m@tty]

Really? I thought area under graph

(taking the direction of the braking force as positive).

How do the answers calculate 13.6 ??

[kenhung123]

Area=1/2(4+8)*6000 then they equate to mv. Maybe they are wrong?

[m@tty]

Seems they used the wrong values for the side lengths of the trapezium. They used 4 and 8, while, to me, it appears they should have used 8 and 12...

[asterio]

For this question I found the area under the graph to be 30x1000x2=60000
Then I equated that value to 1/2mv^2 to find v. Is that wrong?

I'd do it this way

f * (change in time) = I = m * (change in v)

60000=2200(change in v)

I did this, and it is somehow wrong.

the answer is 30 000/2200 = 13.6m/s

[lachymm]

Yeah they used the wrong numbers, there answer is wrong.

[kenhung123]

OK this question is rather strange because its not the typical "would X increase when Y intensity increases" or something like that which always involve some explanation saying the Vout needs to maintain so if this decrease then this needs to increase...Is there anyway I can work out how to solve this question?

[m@tty]

With less light there will be more resistance. Therefore at lower levels of light the voltage drop across the LDR will be higher. As they have set the system up to activate when the voltage is below a certain level you must place the switch in position X, as there will be a lower V here when there is lower light intensity.

Very weird question...

[kakar0t]

It was a great question IMO, really makes you think hard and uses all your knowledge of voltage divider circuits.

[kenhung123]

Hmm...so in this case the Vout fluctuates and does not maintain because there is not fixed resistor?

[jasopan]

Does WORK have direction?
A question with a Force VS Distance Graph, makes you equate Work to Change in Kinetic Energy to solve for mass
I got
2.25*10^(5) = 1/2 (m)v^2 - 1/2mu^2
and if you solve for M you get a negative mass.
So do we have to state work is the BRAKING force and thus negative?

EDIT:
Is it important to put the 'net' sign?
eg. NET Eki = NET Ekf when doing those elastic/inelastic collision problems?

EDIT2: Does impulse have direction? a braking force vs time question;
I = m change in V , where v final =0 and we're calculating v initial.
This gives a negative direction, but is this because it is the BRAKING force thus backward force?
Should we right
-I = mv
or when we get a final answer of say -20m/s we write 20m/s forward?

[jasopan]

anyone?

[Greggler]

i dont belive work done has direction
not sure about impulse. dont usually get asked about direction though, more about 'size of force of A on B' which doesn't really matter since they're equal and opposite anyway

And for inelastic/elastic ones i just treat them as proofs and have:
 'Initial Ke' as one equation and 'final Ke' as the other