Author Topic: HSC Physics Question Thread (Read 1043056 times) Tweet Share

Sanaz · « **Reply #1035 on:** October 22, 2016, 04:26:00 pm »

Hey can anyone help me do this?

RuiAce · « **Reply #1036 on:** October 22, 2016, 04:27:52 pm »

Quote from: student123456 on October 22, 2016, 04:13:14 pm

Hey guys!
Hoping somebody may be able to help me out with solving these problems. I never really know how to handle them. Do you it mathematically? Thanks

A spaceship at a distance r metres from the centre of a star experiences a gravitational force of x newtons. The spaceship moves a distance of r/2 metres towards the star. What is the gravitational force acting on the spaceship in this new location?

$\text{At a distance }r\text{, by Newton's Law of Universal Gravitation we have}\\ x=\frac{Gm_1m_2}{r}$
$\text{Once it travels that distance, it is now }\frac{r}{2}\text{ metres away from the star.}\\ \text{Hence now we have}\\ \begin{align*}F&=\frac{Gm_1m_2}{\left(\frac{r}{2}\right)^2}\\ &= 4 \times \frac{Gm_1m_2}{r^2}\\ &= 4x\end{align*}$

RuiAce · « **Reply #1037 on:** October 22, 2016, 04:40:04 pm »

Quote from: Sanaz on October 22, 2016, 04:26:00 pm

Hey can anyone help me do this?

$\text{For questions like this you should provide the answers to the previous parts.}\\ \text{Here, in part b) we found that }F=4\times 10^{-14}N$
$\text{There are at least two ways of doing this.}\\ \text{Last year, I chose the method Electrical energy }\to \text{ Kinetic energy.}\\ \text{Using }W=Fs\text{, the work done on the electron is}\\ W=4\times 10^{-14} \times 0.02 = 8\times 10^{-16}J$
$\begin{align*}\text{So via }E_k&=\frac{1}{2}mv^2\\ 8\times 10^{-16} &= \frac{1}{2}(9.109\times 10^{-31})v^2\\ v&\approx 4\times 10^7m\, s^{-1}\end{align*}$

Sanaz · « **Reply #1038 on:** October 22, 2016, 04:51:47 pm »

Quote from: RuiAce on October 22, 2016, 04:40:04 pm

$\text{For questions like this you should provide the answers to the previous parts.}\\ \text{Here, in part b) we found that }F=4\times 10^{-14}N$
$\text{There are at least two ways of doing this.}\\ \text{Last year, I chose the method Electrical energy }\to \text{ Kinetic energy.}\\ \text{Using }W=Fs\text{, the work done on the electron is}\\ W=4\times 10^{-14} \times 0.02 = 8\times 10^{-16}J$
$\begin{align*}\text{So via }E_k&=\frac{1}{2}mv^2\\ 8\times 10^{-16} &= \frac{1}{2}(9.109\times 10^{-31})v^2\\ v&\approx 4\times 10^7m\, s^{-1}\end{align*}$

yea thank you <3

Cindy2k16 · « **Reply #1039 on:** October 23, 2016, 12:51:05 pm »

Hi could someone please explain the answers to questions 12 and 18 from the 2014 HSC? The answers are A for q12 and D for q18

student123456 · « **Reply #1040 on:** October 23, 2016, 12:53:48 pm »

Quote from: RuiAce on October 22, 2016, 04:27:52 pm

$\text{At a distance }r\text{, by Newton's Law of Universal Gravitation we have}\\ x=\frac{Gm_1m_2}{r}$
$\text{Once it travels that distance, it is now }\frac{r}{2}\text{ metres away from the star.}\\ \text{Hence now we have}\\ \begin{align*}F&=\frac{Gm_1m_2}{\left(\frac{r}{2}\right)^2}\\ &= 4 \times \frac{Gm_1m_2}{r^2}\\ &= 4x\end{align*}$

Thank you, RuiAce

RuiAce · « **Reply #1041 on:** October 23, 2016, 01:02:24 pm »

Quote from: Cindy2k16 on October 23, 2016, 12:51:05 pm

Hi could someone please explain the answers to questions 12 and 18 from the 2014 HSC? The answers are A for q12 and D for q18

Attack Q18 by a step by step analysis. Incidentally, the right column of the answers is what we find before the left column.

Recall that a cathode ray is an electron. An electron carries NEGATIVE charge. Hence, apply the reverse right-hand rule (or left-hand rule) to determine the magnetic deflection. You will find that the electron is deflected DOWNWARDS by the magnetic field.

Then, if the electron is deflected downwards, we need the electric field to counteract it by deflecting it UPWARDS.
If we want the electron to go up, it must be that the POSITIVE plate is at the top, whereas the negative plate is at the bottom.

Finally, recall that electric field lines go from + to -. Hence, they go from the top to the bottom, i.e. downwards

Hence D.

An explanation on Q12 is found in post #872

jamonwindeyer · « **Reply #1042 on:** October 23, 2016, 01:22:17 pm »

Quote from: Cindy2k16 on October 23, 2016, 12:51:05 pm

Hi could someone please explain the answers to questions 12 and 18 from the 2014 HSC? The answers are A for q12 and D for q18

Welcome to the forums Cindy!

Cindy2k16 · « **Reply #1043 on:** October 23, 2016, 05:28:09 pm »

Quote from: jamonwindeyer on October 23, 2016, 01:22:17 pm

Welcome to the forums Cindy!

Thanks

I've been lurking for a while now but finally signed up today!

Cindy2k16 · « **Reply #1044 on:** October 23, 2016, 05:32:59 pm »

Hi another question- the 2014 HSC sample answer for q22 said that if the angle of re-entry for a spacecraft is too steep, then the "spacecraft will decelerate too quickly" I'm a bit confused by this. I thought that if a spacecraft enters at too steep an angle, then it wont be able to slow down fast enough - and so deceleration is too little?

jakesilove · « **Reply #1045 on:** October 23, 2016, 05:50:41 pm »

Quote from: Cindy2k16 on October 23, 2016, 05:32:59 pm

Hi another question- the 2014 HSC sample answer for q22 said that if the angle of re-entry for a spacecraft is too steep, then the "spacecraft will decelerate too quickly" I'm a bit confused by this. I thought that if a spacecraft enters at too steep an angle, then it wont be able to slow down fast enough - and so deceleration is too little?

The answer given by BOSTES is definitely a bit confusing. It's not so much that the ship won't have time to decelerate quickly enough; it's more than the deceleration caused by interaction between the atmosphere and the ship will be so great that huge amounts of heat will be released, and the G-force experienced by the astronauts will be deadly. The more quickly the ship goes straight downwards, the more it has to deal with the effects of friction. This will heat the ship up, causing it the burn up in the atmosphere. You can really write whatever you want in this section (ie. G-forces, heat, etc. etc.) as long as what you're saying makes sense. Don't always trust the BOSTES answers, because they often DON'T make sense.

ml125 · « **Reply #1046 on:** October 23, 2016, 05:56:43 pm »

Quote from: Cindy2k16 on October 23, 2016, 05:32:59 pm

Hi another question- the 2014 HSC sample answer for q22 said that if the angle of re-entry for a spacecraft is too steep, then the "spacecraft will decelerate too quickly" I'm a bit confused by this. I thought that if a spacecraft enters at too steep an angle, then it wont be able to slow down fast enough - and so deceleration is too little?

"Decelerate" in this sense, refers to a downwards acceleration. So the magnitude of the acceleration is increasing, but the direction of the movement is downwards - the value for acceleration is becoming increasingly negative and is therefore classified as a deceleration.

RuiAce · « **Reply #1047 on:** October 23, 2016, 06:05:43 pm »

Quote from: ml125 on October 23, 2016, 05:56:43 pm

"Decelerate" in this sense, refers to a downwards acceleration. So the magnitude of the acceleration is increasing, but the direction of the movement is downwards - the value for acceleration is becoming increasingly negative and is therefore classified as a deceleration.

I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...

ml125 · « **Reply #1048 on:** October 23, 2016, 06:09:06 pm »

Quote from: RuiAce on October 23, 2016, 06:05:43 pm

I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...

This is how my physics teacher explained this specific question to us - not 100% sure though ._.

jakesilove · « **Reply #1049 on:** October 23, 2016, 06:10:15 pm »

Quote from: RuiAce on October 23, 2016, 06:05:43 pm

I'm not too sure if accelerating downwards and deceleration mean the same thing... I think deceleration refers to only the magnitude...

'Deceleration' is a stupid term that means nothing in Physics for fairly obvious reasons, however generally it does only refer to the magnitude. Ie. 'slowing down' as opposed to 'becoming less positive'. Let's ignore the term altogether though, if possible

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