VCE Physics Question Thread!

[Muchos Help]

I'm wondering why v^2=u^2+2ax doesn't work, just to clarify..

[Syndicate]

I'm wondering why v^2=u^2+2ax doesn't work, just to clarify..

As I stated, the formula will get you the velocity of the projectile, while you question clearly states that, it wants the distance.

Therefore, it is relevant to use the formula in this situation

[Muchos Help]

As I stated, the formula will get you the velocity of the projectile, while you question clearly states that, it wants the distance.

Therefore, it is relevant to use the formula in this situation

Normally you could just rearrange the first formula for x, but since a=0 you can't. I just need some who knows the theory on why it isn't possible.

Thanks though.

[lzxnl]

OK. I see your problem now.

If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error

However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.

To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)

[Syndicate]

OK. I see your problem now.

If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error

However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.

To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)

never knew cheat sheets were allowed in VCE exams

[Muchos Help]

OK. I see your problem now.

If a = 0, and you try and solve for x, what's going to happen?
v^2 - u^2 = 2ax, x = (v^2 - u^2)/2a
But a = 0 -> divide by zero error

However, if x = ut + 1/2 at^2 = ut, then you don't have this divide by zero error.

To put it simply, if acceleration = 0, it's constant velocity and speed, so you just use distance = speed * time. Knowing what formula to use and why is important for VCE physics, given that you don't even need to know the formulas from memory (*cough cheat sheet)

Ah okay, thanks.

[lzxnl]

never knew cheat sheets were allowed in VCE exams

They shouldn't be allowed. But ask any maths or physics student. It's silly how there's a formula sheet in physics AND you get to bring in 2 A4 double-sided. At least, that was way back in 2013.

[Syndicate]

They shouldn't be allowed. But ask any maths or physics student. It's silly how there's a formula sheet in physics AND you get to bring in 2 A4 double-sided. At least, that was way back in 2013.

is the formula sheet like the one on a maths exam?

EXAMPLE:
thermodynamics
q = ml
q = mc*change in temperature

velocity
v(final) = v(initial) + at

??

[lzxnl]

is the formula sheet like the one on a maths exam?

EXAMPLE:
thermodynamics
q = ml
q = mc*change in temperature

velocity
v(final) = v(initial) + at

??

Difference is, in maths there's no limit on how much stuff you bring as long as it's bound. Otherwise...it's the same really.

[Adequace]

Quick question when you're using F=-kx, I see multiple/all of the time that the questions' worked solution just uses this formula as F=kx, so when the spring constant will always be positive. Are you allowed to do this?

Both F and x is also positive so I'm not sure how you could cancel out the negative of k?

[JI2015]

Quick question when you're using F=-kx, I see multiple/all of the time that the questions' worked solution just uses this formula as F=kx, so when the spring constant will always be positive. Are you allowed to do this?

Both F and x is also positive so I'm not sure how you could cancel out the negative of k?

A quick online search:
https://answers.yahoo.com/question/index?qid=20101231173352AAfckwX

I used F=kx, it doesn't really matter which you use as the magnitude of the force is the same. Negative just indicates direction which you can find out from a diagram so it's not the most useful to have a negative in the equation, especially if you haven't set a positive direction. Bottom line, draw a diagram and label the direction the force and use F=kx to find the magnitude.

[lzxnl]

Quick question when you're using F=-kx, I see multiple/all of the time that the questions' worked solution just uses this formula as F=kx, so when the spring constant will always be positive. Are you allowed to do this?

Both F and x is also positive so I'm not sure how you could cancel out the negative of k?

It matters when the sign of the force matters. Otherwise, as mentioned, you can just relate the force to magnitudes.

When does it matter? When you're actually solving for the equation of motion or the energies.
The potential energy is given by -int F dx, so the negative sign for the force gives rise to U = 1/2 kx^2. Without the negative sign, the potential energy would be -1/2 kx^2
Similarly, if you're trying to solve for the shape of the motion, you would have F = -kx -> ma = -kx
a = -kx/m = d^2x/dt^2
The general solution to this differential equation is a sinusoid, x = A cos(sqrt(k/m)t + constant)

Without the negative sign, the solutions would be sums of exponentials and there'd be no oscillation (try it yourself if you want xD)

[Adequace]

Thanks all

[Maz]

hey- just a few questions related to tension? please...i don't really understand this...
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou

[lzxnl]

hey- just a few questions related to tension? please...i don't really understand this...
what does it mean by 'vertical of tensions is equal to the gravitational force acting on the object'? its in my book but i don't understand it
2. is a word problem... A 0.20kg objects is whirled in a vertical circle on the end of the string of length 0.6m. the speed of the object is at a constant 2ms^'1. what is the tension of the string at the top and bottom of the table.
my book isn't really explaining tension much and I'm a little confused. id appreciate any help
Thankyou

Tension is the force on the string pulling on the object. So, if it's a vertical circle, at the bottom of the circle the tension is pointing up, and at the top it's pointing down.
To do this question, you would treat the tension force like a normal reaction force and just solve net force = mv^2/r
Then assign the direction based on the direction requirement for circular motion