VCE Physics Question Thread!

[JI2015]

Draw your forces on the mass on the diagram. There is a tension force vertically upwards acting on the mass and a vertically downwards force of gravity acting on the mass. Since the mass is at rest (pay attention to the wording), it means the net force at this point is zero. So, the magnitude of the tension in the rope is equal to the magnitude of the weight force which in this case is 40N. Also, specify that tension is in the upwards direction.

[knightrider]

Draw your forces on the mass on the diagram. There is a tension force vertically upwards acting on the mass and a vertically downwards force of gravity acting on the mass. Since the mass is at rest (pay attention to the wording), it means the net force at this point is zero. So, the magnitude of the tension in the rope is equal to the magnitude of the weight force which in this case is 40N. Also, specify that tension is in the upwards direction.

Thanks JI2015 

just wondering is my working out below correct in relation to this situation?

let up be positive

[JI2015]

Treating upwards as postive:

F(net) = F(centripetal) = Tension - Weight = 0

Tension - Weight = 0

therefore, Tension = Weight

Remember to refer to tension rather than F(N) as it is more specific. Also with the net force equation, make sure that there is a negative before the Fg/ Weight as it is acting in the opposite direction to the tension force.

What I also want to point out is that you can treat the equation with the variables as vectors or scalars. I prefer scalars (you account for direction with negative signs). Your net force equation is fine if you keep vectors in mind, but since you specified up as positive then treating them as scalars is the way to go. This also means that when referring to the forces in the equations, you are only concerned with their magnitudes when solving. If you do this correctly for all force equations, you will only get positive values for magnitudes of forces when solving as you have accounted for the direction (whether positive or negative) already in the equation.

If you would like me to show you both methods (vector and scalar) and how the solutions present themselves when you solve, Iet me know and I'll prepare something tomorrow.

[knightrider]

Treating upwards as postive:

F(net) = F(centripetal) = Tension - Weight = 0

Tension - Weight = 0

therefore, Tension = Weight

Remember to refer to tension rather than F(N) as it is more specific. Also with the net force equation, make sure that there is a negative before the Fg/ Weight as it is acting in the opposite direction to the tension force.

What I also want to point out is that you can treat the equation with the variables as vectors or scalars. I prefer scalars (you account for direction with negative signs). Your net force equation is fine if you keep vectors in mind, but since you specified up as positive then treating them as scalars is the way to go. This also means that when referring to the forces in the equations, you are only concerned with their magnitudes when solving. If you do this correctly for all force equations, you will only get positive values for magnitudes of forces when solving as you have accounted for the direction (whether positive or negative) already in the equation.

If you would like me to show you both methods (vector and scalar) and how the solutions present themselves when you solve, Iet me know and I'll prepare something tomorrow.

Thanks so much JI2015  Really clarified things 

It would be great if you could  show me both methods (vector and scalar) and how the solutions present themselves when you solve.

Thanks so much for your help   

[JI2015]

Look for a post later today (since it is past midnight) about this as I want you to understand fully.

Some quick points:

When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.

Good Night!

And also well done with your commitment to physics as working hard and understanding this subject is really important.

[knightrider]

Look for a post later today (since it is past midnight) about this as I want you to understand fully.

Some quick points:

When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.

Good Night!

And also well done with your commitment to physics as working hard and understanding this subject is really important.

Nice tips thanks again !!

Also it is with the help of such wonderful people like you and the many others on this site , that my skills and understanding  in physics improves .!!

Good night 

[lzxnl]

Look for a post later today (since it is past midnight) about this as I want you to understand fully.

Some quick points:

When simply adding vectors for your net force, you might get a negative if you solve which indicates opposite direction to the direction you are taking as positive. When treating as scalars, the sign and direction is already accounted for. Another benefit of the scalar equation is that you are proactively thinking about directions which aids in your understanding.

Good Night!

And also well done with your commitment to physics as working hard and understanding this subject is really important.

Technically, those aren't scalars. Scalars are actually quantities that are invariant (don't change) under coordinate transformations like rotation or reflection. So, rotating an object doesn't change its mass or any distances so those are scalars. Dealing with vector quantities in 1D and treating their direction as a scalar quantity is strictly not correct. Your 'scalar' is not invariant under rotation as rotating your axis 180 degrees changes the sign. So technically you're still dealing with a vector, just representing it as a real number.

[JI2015]

Hi knightrider,

As promised I have prepared a more thorough explanation for you. Hope this clears up any hesitation.

http://jmp.sh/bnRFytP

[JI2015]

More of a 1/2 question, but I'll ask anyway since it's relevant to 3/4. When adding or subtracting numbers, do we get our answer to the least amount of sig figures after the decimal point? For example, is 6.9536 (4 sig figs after decimal point) + 3772.650 (3 sig figs after d.p) equal to 3779.604 (3 sig figs after d.p)?

When multiplying or dividing, is our answer to least amount of sig figures (regardless of decimal place)? For example, is 63.84 (4 sig.figs) / 2.5010 (5 sig figs) = 25.53 (4 sig figs)

Your reasoning is spot on!

Adding or Subtracting: Decimal places of answer is to the lower amount of decimal places that one of the terms has.
Multiplying or Dividing: Answer has the same significant figures as the number with the lower amount of significant figures of the two used in the calculation.

Usually people only have trouble with the Addition/Subtraction but you seem to understand the rules well!

[knightrider]

Hi knightrider,

As promised I have prepared a more thorough explanation for you. Hope this clears up any hesitation.

http://jmp.sh/bnRFytP

Wow awesome explanation  Thanks so much JI2015 

[Adequace]

For projectile motion are we allowed to use pre-derived formulas? My teacher encourages us to use a pre-derived formula for time which is convenient. I've also heard that assessors don't mind it.

[JI2015]

For projectile motion are we allowed to use pre-derived formulas? My teacher encourages us to use a pre-derived formula for time which is convenient. I've also heard that assessors don't mind it.

You can use the 'shortcut' formulas but just make sure you know how to use them properly (i.e.when the object lands at the same height that it starts at). Maybe even learning how they are derived may benefit your understanding.

Also, just to add, I used some of these formulas in practice exams and also the actual exam and had no problems, so yeah assessors do not mind at all as long as your answer is correct.

[Syndicate]

For projectile motion are we allowed to use pre-derived formulas? My teacher encourages us to use a pre-derived formula for time which is convenient. I've also heard that assessors don't mind it.

if the assessors don't mind it, then why not?

[lzxnl]

if the assessors don't mind it, then why not?

Simply because it's not good for your understanding in general. It's sooo easy for students to forget where a formula is applicable. For instance, range = v^2 sin 2 theta / g only for landing at the height of launch.

[Adequace]

Thanks all for the insight.