• YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page

Trial paper problem (1 Viewer)

mathsbrain

Member
Joined
Jul 16, 2012
Messages
161
Gender
Male
HSC
N/A
Hi all,

I am studying for my trials and i found these two questions to be unbelievably hard, can someone please lend a helping hand.

In the kings 2007 trial:

Q5a) I get a "plus" between the RHS terms rather than "minus". Can someone please explain with solutions?

Q6a) (i) i had no idea....even though its only 1 mark.

In the Ruse Trial(note i have uploaded the trial solution as well):

part c) (ii) in their solution, if you read their 2nd line of working for c(ii), they substituted v=1/2v only in one of the v's, but sin and cos both are in terms of v, shouldnt they be substituted as 1/2v as well?
 

Attachments

rolpsy

Member
Joined
Apr 9, 2011
Messages
94
Gender
Male
HSC
2012
okay here's 5(a)
Resolving forces along the bank can be tricky. If all else fails, resolve horizontally and vertically, then solve simultaneously.

 

Attachments

mathsbrain

Member
Joined
Jul 16, 2012
Messages
161
Gender
Male
HSC
N/A
okay here's 5(a)
Resolving forces along the bank can be tricky. If all else fails, resolve horizontally and vertically, then solve simultaneously.

Hi THanks for your help, but here is my working, can you please find out what is wrong?
 

Attachments

mathsbrain

Member
Joined
Jul 16, 2012
Messages
161
Gender
Male
HSC
N/A
okay here's 5(a)
Resolving forces along the bank can be tricky. If all else fails, resolve horizontally and vertically, then solve simultaneously.

Hi Frank,

I think in your solution you've assumed the car is falling downwards along the line BA, hence you claim resultant force is (mv^2/r) *cos(theta), but i think the resultant force along BA should be 0 as the car should not be sliding up or down
 

rolpsy

Member
Joined
Apr 9, 2011
Messages
94
Gender
Male
HSC
2012
remember that mv2/r is the resultant force (horizontally); the other forces must add to give this.
the same principle applies when you resolve this resultant force into its components.

to convince yourself, resolve horizontally and vertically, then solve simultaneously (times vertical by sin, horizontal by cos, then subtract)


and lol my name isn't frank...:p that's the first word that popped into my head
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
6 a (i) is fairly easy.
Just use ratio intercept thm since AQ is parallel to SD which gives (PS)/(QD)=(PS)/(AS), but
you are given (PD)/(QD)=(PS)/(QS), equating the two gives AS=QS.

Ill look at reast now
 

math man

Member
Joined
Sep 19, 2009
Messages
503
Location
Sydney
Gender
Male
HSC
N/A
the question is stupid to use v and not v_0 or some other distinguishable variable for optimal speed in (i), which is what confused you
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top