Physics: Discussion, Questions & Potential Solutions

[Richard Feynman 101]

6 b

I think Freddie means 9.8 instead of 0. Due to the difficulty of such a question (2 marks, 1 mark correct response (0) and exp) and maybe 2 marks (0 marks, incorrect answer but I would give 1 mark if your reasoning is correct but on your answer).

Also, not sure if 6c is correct, did it ask for the ∆x for max speed (if so, then ∆x = 0.25 m).

[MrTeacherMan]

Let's review the spring question (Q6).

The question states that "[t]he ball and the spring come to rest when they are at a distance of 0.50 m below the uncompressed position of the spring."

The wording of "come to rest" is slightly ambiguous. It may mean they completely come to rest, or momentarily come to rest.

However using the most obvious definition of "at rest", we can assume this means when velocity is equal to zero, which would then mean when the ball and spring are first momentarily at rest (at the lowest point of spring compression).

Using this definition, then part a. can be done using a conservation of energy approach and the correct answer can be obtained (in fact the question can only be answered if you assume this definition).

Part b. asks for you to find the acceleration at maximum speed. Many people assumed that the maximum speed occurs just before the ball hits the spring. This is not true. It is true that the maximum acceleration occurs just before the ball hits the spring, however as the ball begins to compress the spring, the acceleration will decrease, during which time the velocity is still increasing, just at a slower rate.

Thus the maximum speed will then occur as acceleration decreases to zero.

For part c., you are asked to find the compression at maximum speed. From b., the acceleration at this point will be zero, and hence net force will be zero. Thus mg = kx, and solving for x (using k = 392 N/m given in part a.), x = 0.05 m.

[G-Fr3sh]

Square wave; zero for the loop outside and inside sections, positive for loop entering, negative for loop exiting.

It asked for induced current, therefore zero for outside, negative gradient for travelling inside and positive gradient for travelling outside.

I think both would be accepted

[oyba]

Have the solutions been released yet?

[wanigara]

Solutions Q7 - Projectiles

(a) d=v*t = 1.2 m
(b) S= ut+ 1/2at^2 = 0- 5 x0.4^2= -0.8 m or 0.8m
(c) There are two main methods (i) Conservation of energy (ii) find vertical velocity just before hit the ground and find the resultant speed (Pythag) Ans: 5ms^-1

[wanigara]

Q8 Motion
(a) Acceleration of the system 8.0 ms-2 , Using Fnet = ma where Fnet = 40 N and m= (4+1) kg
Then isolate B. Only horizontal foce on B is F on B by A = 1.0 * 8 = 8.0 N
(b) The magnitude of  Fon A by B is same as Fon B by A (Newtons 3rd law). But in opposite direction. Answer: 8.0 N to the Left

[wanigara]

Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno

Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.

[G-Fr3sh]

Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno

Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.

1. I used GMm/r^2, the question did not explictly state to use the curve and got 4.75*10^3 N

2. You didn't need to calculate area under curve

You can use Change in mgh, however keep in mind, the value of g will change with height.
-> Change in mgh= Eg(f)-Eg(initial)
= m((13*1*10^8)-(3*2*10^8))
=1.05*10^12
It lead me to 1.1*10^12

[aussiboi]

TWM send solutions its been past your organised time

[KiNSKi01]

1. I used GMm/r^2, the question did not explictly state to use the curve and got 4.75*10^3 N

Yeah I ultimately did the same as you but I was unsure whether we were meant to use the graph cos you get two slightly answers depending on the method you use 

They will accept both methods right? since the question didn't specifically ask us to use the graph??

[Richard Feynman 101]

Q9. Gravity
(a) Gravitational Force = mg, where g = 3.0 Nkg-1 (from graph) so Fg = 4.50x10^3 N
(b) Magnitude of change in Gravitational potantial energy = area under the graph from 2.0x10^8 to 1.0x10^8 x mass of Juno

Count the square = 14
Value of one square = 0.5x10^8 x1 = 5.0x10^7 J/kg
Change in GPE = 1500x14x5x10^7 = 1.05x10^12 J or 1.1x10^12 J
(c) Use Kepler's equation and re-arrange it
T =sqrt( 4[PI]^2 R^3/GM) = 3.06x10^5 sec or 3.1x10^5 sec.

Just with the counting of the squares VCAA will have to accept a range for instance, I got 14 sqaures, however, 13 sqaures is perfectly reasonsable. Still don't know why integration is out of the scope but oh well.

[KiNSKi01]

Yeah I counted 13 too (its closer to 13 if you do integration right?)

[Richard Feynman 101]

Yeah I counted 13 too (its closer to 13 if you do integration right?)

Yes. Perfectly fine.

[aussiboi]

can someone send multiple choice answers through pls its been a day

[vox nihili]

can someone send multiple choice answers through pls its been a day

Why don't you put up your answers?