A123 Physics Question Thread.

[abd123]

Question 6 from the 2001 VCAA exam.

Full description: On 12 Febuary 2001, A spacecraft named the NEAR Shoemaker landed on Eros, a peanut shaped asteroid between the orbits of Earth and Mars. The 450kg spacecraft (Ans: 4500 Newtons) had a weight of when it landed on the asteroid.

,

i) 4500 Newtons ( (already answered) through

ii) Before landing on Eros, the spacecraft orbited at a radius of 50km from the centre of mass of Eros, with a period of ,

Find the mass of Eros from this information .

[Phy124]

I'm not able to view these pictures and cyberroadie.org won't load for me. Not sure if this is just on my end, but if not, try using http://www.codecogs.com/latex/eqneditor.php

[abd123]

This is off Creelman VCE exam questions, VCAA website it doesn't have the 2001 unit 3 physics exam.

I tried fixing up the latex, but the latex is stuffed up all I see is torn-page icons.

Don't worry about it now, nothings working at the moment.

Thanks mylittlepony.

[Phy124]

The website which supports the forums latex must be down, not to worry.

I think this is your information:

Question 6 from the 2001 VCAA exam.

Full description: On 12 Febuary 2001, A spacecraft named the NEAR Shoemaker landed on Eros, a peanut shaped asteroid between the orbits of Earth and Mars. The 450kg spacecraft (Ans: 4500 Newtons) had a weight of 2.5N when it landed on the asteroid.

450kg x 10 =4.5 x 10³, 2.5N

i) 4500 Newtons 4.5 x 10³ (already answered) through w = m x g, w = 450 x 10

ii) Before landing on Eros, the spacecraft orbited at a radius of 50km from the centre of mass of Eros, with a period of 5.9
 x 10⁴ s.

Find the mass of Eros from this information (G=6.67 x 10-11 Nm²kg^-2)

[Lasercookie]

Googling came up with the VCAA link for the exam. I wonder what other pre-2002 exams are on there.
http://www.vcaa.vic.edu.au/vce/studies/physics/pastexams/physics22001.pdf

Anyway:
I'll latexify this post when it's all working again ----- edit: done
 
Q5. . (which is the answer you got)

Q6. For this question, you use the relationship . Simply solve for M.

Here's the derivation/reasoning of this formula:
In a circular orbit, the centripetal acceleration equals the gravitational field strength at that point.





(these are on the formula sheet given on the exam)

so:










So yeah, to actually answer the question, rearrange for M and sub in your values:










In case you're a bit confused about all those brackets, here is that final bit plugged into Wolfram Alpha M = (4pi^2)((50*10^3)^3)/((6.67*10^-11)((5.9*10^4)^2)
(sorry if I made any other typos throughout as well)

Edit:

This is off Creelman VCE exam questions, VCAA website it doesn't have the 2001 unit 3 physics exam.

Oh yeah, if you're going to be doing the old exam questions (I figure why not), take note that prior to 2004, the units are organised differently e.g. motion was Unit 4.

edit: latexified

[abd123]

Alright this question is some what easy for 'one' part not another.

A car of mass 1600kg is towing a boat up an incline of 25 degrees at a constant speed of 8ms^-1. The total mass of the boat and trailer is equal to 800kg. The total frictional force acting on the car is 300N, while the frictional force acting on the boat is 200N.

a) calculate the magnitude of the reaction force that the road is exterting on the car.

so i gotted it through f_r=mgcos(theta)

f_r=1600 x 9.8 x cos(25)

=1.4 x 10^4N, thats done and dusted.

but however this question is obscure...

b) calculate the tension in the coupling that joins the car to the trailer.

Somewhat the answer is 3.5 x 10^3N.

How is the frictional force is added? I thought I was mean to use the total frictional force of 300N, but instead they used the frictional force thats acting on the boat

can you explain why i have no absolute idea of it and show the solution doing it.

Thanks

[abd123]

Anyways is am I right theres no answer to this question.

Just a work sheet I found in my room.

Is this right?

A 4WD (2,500 kg) is towing a caravan up in an incline of 110. The 4WD and the caravan encounter a constant frictional force of 1,250 N and 800 N respectively. The driving force of 13,250 N provided by the 4WD is enough to accelerate the pair at 1.5 m s-2.

force down the plane

=2500 x 9.8 x sin (11)=4675
F_f=1250N
F_f=800N

force up the plane

f=13,250N

resolving the forces

f=ma=(2500+m) x 1.5
Net force= 13250N - 4675N - 1250N - 800N - m x 9.8 x sin(11)=6525 - m x 9.8 x sin (11)

equate the forces

(2500+m)=6525 - m x 9.8 x sin (11)

m=824kg

Am i doing it right? and is the answers right? can anyone confirm this?