Hello Professor Joe
A link to 2009 HSC exam paper: http://www.boardofstudies.nsw.edu.au/hs ... ex4.html#p  Question 19 b).
I searched the forum for any queries regarding this particular question but I did not find any posts. I am having trouble understanding the solutions in the sample answers. I treated the question as a standard projectile motion question, assuming constant negative acceleration for the particle. I found this using Newton's Second Law, F=ma, which was the constant force down the page experienced by the electron. I got the acceleration as a=1.76 x 10^14 ms^2 downwards, same as the sample answer
2009 HSC Physics Exam Question 19b

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 Joined: Mon Apr 07, 2014 11:44 am
2009 HSC Physics Exam Question 19b
Last edited by Harry_Shan on Mon Apr 07, 2014 12:52 pm, edited 1 time in total.

 Posts: 5
 Joined: Mon Apr 07, 2014 11:44 am
Re: 2009 HSC Physics Exam Question 19b
CONTINUED
I then calculated the vertical velocity of the electron, given by 6.0 x 10^6 x sin60 = 5.196 x 10^6 ms1 up the page, which was the same as the sample answer. Here, the question asks us to calculate when the electron reaches D. I used the equation
y = ut + 1/2 at^2
I set y=0, since the vertical displacement when the electron is at D is the same as where it first starts experiencing the force due to the uniform electrical field. Making appropriate substitutions, I got an answer of t = 5.9 x 10^8 s. However, the sample answers use a different equation of v = u + at and set v=0. This seems incorrect, as this only calculates when the electron is half way to reaching point D (due to assumed symmetry of parabolic trajectory) However, I checked 2 different sources of sample answers, and both use a similar method of v=u + at to find the time.
Could you please explain why the equation v=u + at is used by setting v=0?
Thanks,
I then calculated the vertical velocity of the electron, given by 6.0 x 10^6 x sin60 = 5.196 x 10^6 ms1 up the page, which was the same as the sample answer. Here, the question asks us to calculate when the electron reaches D. I used the equation
y = ut + 1/2 at^2
I set y=0, since the vertical displacement when the electron is at D is the same as where it first starts experiencing the force due to the uniform electrical field. Making appropriate substitutions, I got an answer of t = 5.9 x 10^8 s. However, the sample answers use a different equation of v = u + at and set v=0. This seems incorrect, as this only calculates when the electron is half way to reaching point D (due to assumed symmetry of parabolic trajectory) However, I checked 2 different sources of sample answers, and both use a similar method of v=u + at to find the time.
Could you please explain why the equation v=u + at is used by setting v=0?
Thanks,
Re: 2009 HSC Physics Exam Question 19b
Did the sample answer that solved 0 = v = u + at for t then multiply the time by 2? If so, that would be an easy way of getting to the answer.
But your way is more direct.
Joe
But your way is more direct.
Joe

 Posts: 5
 Joined: Mon Apr 07, 2014 11:44 am
Re: 2009 HSC Physics Exam Question 19b
No, the sample answers didn't multiply by 2.
I have copy pasted from the sample answers:
Will need the perpendicular component of the velocity as the acceleration influences this
component of the velocity only
vh = 6.0 × 106 sin 60°
= 5.196 × 106 m s−1
Determine time using
v = u + at
0 = 5.196 × 106 – 1.759 × 1014 t
5.196 × 106
t =
1.759 × 1014
= 2.9 × 10−8 s
I have copy pasted from the sample answers:
Will need the perpendicular component of the velocity as the acceleration influences this
component of the velocity only
vh = 6.0 × 106 sin 60°
= 5.196 × 106 m s−1
Determine time using
v = u + at
0 = 5.196 × 106 – 1.759 × 1014 t
5.196 × 106
t =
1.759 × 1014
= 2.9 × 10−8 s
Re: 2009 HSC Physics Exam Question 19b
Your answer of 5.9×10^(8) s is correct.
The PDF where they have the Board of Studies sample answers state that the answers may not be complete, or something along those lines.
I think they either meant you "determine time" by doing what they did, and they did not bother multiplying by 2, or they are just wrong (I think the sample answers make mistakes sometimes, sadly).
The PDF where they have the Board of Studies sample answers state that the answers may not be complete, or something along those lines.
I think they either meant you "determine time" by doing what they did, and they did not bother multiplying by 2, or they are just wrong (I think the sample answers make mistakes sometimes, sadly).
Re: 2009 HSC Physics Exam Question 19b
The sample answers (here: http://www.boardofstudies.nsw.edu.au/hs ... ers09.pdf ) say at the start that the sample answers are not intended to be "even complete answers or responses", unfortunately.
Regardless, your original answer is the right one.
I hope that was the one used during actual marking.
Regardless, your original answer is the right one.
I hope that was the one used during actual marking.

 Posts: 5
 Joined: Mon Apr 07, 2014 11:44 am
Re: 2009 HSC Physics Exam Question 19b
a.s.h. wrote:The sample answers (here: http://www.boardofstudies.nsw.edu.au/hs ... ers09.pdf ) say at the start that the sample answers are not intended to be "even complete answers or responses", unfortunately.
Regardless, your original answer is the right one.
I hope that was the one used during actual marking.
I thought this too, and I double checked it with the 'Cambridge Checkpoints Physics 2013' answers for the same question. It had the same answer and incomplete working as BOS, so I just assumed that I was making some sort of careless or conceptual error... Still don't understand why BOS would mislead people by intentionally leaving answers incomplete.
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