VCE Physics Question Thread!

[Maths Forever]

Hey, I'm just wondering if electricity and motion the only topics that are taught in 1/2 and then expanded on in 3/4?

In units 1 and 2 the topics in my school were:

- Nuclear Radiation
- Electricity
- Light
- Motion
- 2 detailed studies

In units 3 and 4 all schools cover:

- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)

I hope this helps!

[Kel9901]

In units 1 and 2 the topics in my school were:

- Nuclear Radiation
- Electricity
- Light
- Motion
- 2 detailed studies

In units 3 and 4 all schools cover:

- Motion in one and two dimensions
- Electronics and Photonics
- Electromagnetism
- Light and Matter
- 1 detailed study (11 multiple choice questions on the end of year exam)

I hope this helps!

In other words, motion+electricity are very relevant to 3/4, light is only slightly relevant (only part that is relevant is a bit of wave theory like v=fλ, radiation has 0 relevance, and the detailed studies probably aren't relevant

[bobisnotmyname]

hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks

[Maths Forever]

hey guys im having a little trouble with rounding. Well as we know in physics the numbers arn't always exact so its left to do some rounding, however how do we know how much to round by and how many decimal places our answers should have. thanks

The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.

E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.

Now, using the constant acceleration formula, v = u + at:

30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s² ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.

F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.

Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.

[Kel9901]

The answer should have the same number of significant figures (sig figs) as the least accurate piece of information in the question.

E.g. Calculate the net force as a cart of mass 200 kg (3 sig figs) increases its speed from rest to 30 m/s (2 sig figs) in a time of 3.60 seconds (3 sig figs) with constant acceleration.

Now, using the constant acceleration formula, v = u + at:

30 = 0 + a (3.60)
a = 30 / 3.60 = 8.3 m/s² ....the answer should be rounded to 2 significant figures, since 30 m/s was the least accurate piece of data.

F (net) = ma = 3.60 x 200 = 7.2 x 10^2 N ....leave this as 2 significant figures, as we still take into account that 30 m/s was the least accurate piece of data in the question.

Scientific notation is a good method to use if rounding to significant figures. But really, as long as you don't leave your answer to an absurd amount of decimal places, the examiner shouldn't be too harsh. Two or three significant figures are usually fine.

Nope, that's for chem.

In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.

Of course, scientific notation can be used when very large numbers are present

To show what I'm saying:

Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)

Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate

Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth

g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg

[Maths Forever]

Nope, that's for chem.

In physics, you put in all numbers before the decimal point as accurately as possible (ie use unrounded figures), and after the decimal point, do what's reasonable in your opinion.

Of course, scientific notation can be used when very large numbers are present

To show what I'm saying:

Q1: calculate the acceleration of a car travelling around a (circular) roundabout with a radius of 9m at a speed of 15 ms^-1
a=v^2/r=225/9=25 ms^-2 (not 20 or 30)

Q2 a person accelerates from rest to 10ms^-1 in 3 seconds. Calculate its acceleration
u=0, v=10, t=3, a=?
v=u+at
10=3a
a=3.333333333
You should probably say a=3.3 ms^-2, or perhaps 3.33 ms^-2 as they are appropriate

Q3 People on the Earth's surface experience an acceleration due to gravity of 10 ms^-2. The radius of the earth is 6.37*10^6 m. Calculate the mass of the earth

g=GM/R^2
10=1.64*10^-24 M
M=6.08*10^24 kg

Like I said, 2 to 3 significant figures should be fine. But it is important that the answer is not left with more figures than what the information provides. If the answer does not exceed three significant figures (e.g. 450 m), just leave it as it is.

I.e. With your question 2, 3.3 m/s or 3.33 m/s would be appropriate. Examiners are not too strict on that. Like Kel9901 said, this is more of a concern for Chemistry. Don't worry too much for Physics!

[JackSonSmith]

Is work done force x displacement or force x distance?

[Cosec]

Is work done force x displacement or force x distance?

Force x distance.

[lzxnl]

Is work done force x displacement or force x distance?

Technically it's a dot product between the force and displacement vectors; think of it as the force times the component of the displacement in the direction of the force.

[Cosec]

Can someone explain this to me.

Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.

Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff

So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.

Or have i totally just screwed up with the theory?

[Alwin]

Can someone explain this to me.

Say if we have to forces acting
The weight and the normal force.
We can setup a equation like so (from my knowledge) - taking down as negative
-Fw = Fn
Or
Fw=-Fn
Net force = Fn + Fw
And if given a net force we can solve for the normal force acting.

Like wise with friction and the pulling motion of a subject (the horizontal component) - take left as positive
Fd=-Ff
-Fd=Ff

So can some explain why this same approach isnt set out in the Vtextbook video for circular motion?
Have they just not taken into account direction when doing the forces.

Or have i totally just screwed up with the theory?

Hi Cosec Nice to see someone using VT

For your first equation, let's assume that we have a box resting on a table. That way we have a zero net force, and there only two forces that act on the box are the weight force, F_W and F_N.
As forces are vectors, we should put tilde, use arrows or use boldface to show that (see above). Now, if we don't assign a direction as being positive, then yes of course we can write:

Since we know that the net force is zero, we can let the left hand side be zero and move F_N to the other side:


^ which is the set of equations you got.


Now, the alternative method is to take into account the directions first, and then work with just magnitudes (which is what's done throughout the VT vids, as working with vector notation usually confuses the hell out of ppl not doing spesh).
1) Let up be positive for the box example
2) Now, 'convert' the vectors
F_W equals to a force with magnitude of F_W (note no more boldface) downwards. ie: (note how the arrow disappears)
F_N equals to a force with magnitude of F_N upwards. ie:
3) So our equation for the net force becomes, using only magnitudes since we have already taken into account the directions


When this is rearranged, and we sub in zero for the net force, we would get:
... notice the difference in sign because we've already taken into account the direction, whereas in the first example we were still working with vectors!


As for the circular motion in VT vid, let's look at the diagram on the left:
1) Choose on which direction is positive
Take up as positive
2) Look at the forces and add the sign by looking at the direction
Normal force acts up, so it is negative: - N
Weight force acts down, so it is positive: + W
Net force acts down, so it is positive: + F_net
3) Write the equation of motion
F_net = W - N

This is what it would look like if we used vectors

Apologies if this wasn't clear on the video, it's been quite a while haha. Hope it makes sense now

[JackSonSmith]

A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?

[Gentoo]

A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?

s= 16   t=4    u=0    a=?

Sub into s=ut+1/2at^2  to find a.

[lzxnl]

A bicycle accelerates from rest, covering 16m in 4s. The total mass of the bicycle and its rider is 90kg. What is its average acceleration?

The important bit is not to use the right formula, but to know WHY you need to use that formula.
Here you're given the distance, time and initial velocity (0) and you want the acceleration. The only irrelevant piece of information is the final velocity, so you use the one constant acceleration equation that doesn't involve the final velocity. That's why you use x = 1/2 at^2 + ut

[Floatzel98]

Hey guys. I just started gravity and i'm already stuck on this question 3a. What does in terms of the radius on the moon mean. Am i supposed to know the radius of the moon already from my textbook?
The force is 4x weaker at the new radius so the radius is 4x greater? Does this mean its a variation question. Am i supposed to introduce a constant to solve it?

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