ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Shark 774 on June 09, 2011, 09:57:22 pm

Title: TSSM 09 Question
Post by: Shark 774 on June 09, 2011, 09:57:22 pm

I totally disagree with the answer that TSSM gave to quesion 4, anybody else with me? (I attached their answer too)

Title: Re: TSSM 09 Question
Post by: forumguy on June 09, 2011, 09:59:15 pm

I've always thought it was =mgtan(theta)

SO i got that wrong, i did it this afternoon

Title: Re: TSSM 09 Question
Post by: onur369 on June 09, 2011, 10:03:40 pm

The solution is incorrect.  you should use v=squareroot(radius x gravity x tantheta)

Title: Re: TSSM 09 Question
Post by: Shark 774 on June 09, 2011, 10:06:51 pm

Quote from: onur369 on June 09, 2011, 10:03:40 pm

The solution is incorrect.  you should use v=squareroot(radius x gravity x tantheta)

Bingo, that's exactly what I did. Cheers!

Title: Re: TSSM 09 Question
Post by: LinusX on June 10, 2011, 10:58:48 pm

Awesome
Was getting worried by this.
Does anyone have the actual answer?

Title: Re: TSSM 09 Question
Post by: Vincezor on June 11, 2011, 02:21:58 am

Quote from: LinusX on June 10, 2011, 10:58:48 pm

Awesome
Was getting worried by this.
Does anyone have the actual answer?

 Should be approx 37.42m/s

Title: Re: TSSM 09 Question
Post by: LinusX on June 11, 2011, 08:36:11 am

Another question for this exam (Question 11).

I used v^2 = u^2 + 2ax to work out the final velocity as it started from rest and is only acted upon by
g. When I work it out however I get a different answer to the one they give us in the solutions. How is this possible? Is my method wrong?

In the solutions they use change in gravitational potential = kinetic energy

EDIT: Could someone run through that series of questions? I cant seem to to get question 12 either :(

Title: Re: TSSM 09 Question
Post by: yawho on June 11, 2011, 09:02:00 am

the velocity you obtained using v^2 = u^2 + 2ax is the vertical component

Title: Re: TSSM 09 Question
Post by: LinusX on June 11, 2011, 09:31:48 am

Yes I know its the vertical component but there is no horizontal component for this question...I think I'm missing something here..

Title: Re: TSSM 09 Question
Post by: xZero on June 11, 2011, 11:54:37 am

Would be nice if you post the original question up

Title: Re: TSSM 09 Question
Post by: LinusX on June 11, 2011, 12:59:46 pm

question attached

Title: Re: TSSM 09 Question
Post by: xZero on June 11, 2011, 01:14:49 pm

v^2=u^2+2ax, let a=10 for easy calculation, x=1, u=0
v^2=2*10*1
v^2=20
v=sqrt(20)ms^-1

using energy conservation
1/2mv^2=mg delta h, delta h = 1
1/2mv^2=2*10*1
mv^2=40
v^2=20
v=sqrt(20)ms^-1

unless if i missed something, sqrt(20)=sqrt(20)

Title: Re: TSSM 09 Question
Post by: LinusX on June 11, 2011, 01:37:49 pm

I have no idea what happened there... :-[
Thanks for all the help xZero  :)

Title: Re: TSSM 09 Question
Post by: Shark 774 on June 11, 2011, 03:20:09 pm

Way to steal my thread... :P