Physics Question Help (1 Viewer)

[Khan.Paki]

Hey guys

Can someone help out with part B in this question?

Cheers

 

[astroman]

 

[Mr_Kap]

Hey guys

Can someone help out with part B in this question?

Cheers

View attachment 32492

what year is this from?

 

[Mr_Kap]

Why is the force positive when you use it here? Do you always use absolute value of force in F = ma?
So the negative direction (force downwards) doesn't matter in this case when finding acceleration

 

[rand_althor]

what year is this from?

2009 HSC Question 19.

 

[InteGrand]

Why is the force positive when you use it here? Do you always use absolute value of force in F = ma?
So the direction doesn't matter in this case when finding acceleration

He is using the downwards direction as the positive direction. You can define any direction you like to be the positive direction, so normally you pick the one that makes things easiest; in this case, easiest to make downwards the positive vertical direction, so we don't need to worry about minus signs.

 

[Mr_Kap]

He is using the downwards direction as the positive direction.

So is there a time when you wouldn't want to define the force as positive?

 

[InteGrand]

So is there a time when you wouldn't want to define the force as positive?

If there is only one force in a simple case like this, it may be best to define whatever the direction of the force is (in this case, down) as positive. In more complex problems, you may have multiple forces, and some could be in opposite directions to each other, so in that case, whichever you define positive direction, you would have some forces pointing in the negative direction. In this case, it would probably be best to define the positive direction to be the most intuitive direction (e.g. "upwards" as in pointing towards the sky, or something).

Anyway, I realised was actually just using positive F in order to calculate the magnitude of the acceleration (magnitudes are positive always), and then he put a minus sign in from of it when doing the v = u + at part (as his u is positive, so he is using the upwards direction as positive, since u is the vertical component of initial velocity, which is pointing upwards).

 

[Khan.Paki]

I am getting confused in one thing. The equation finds horizontal component of velocity right? So why do these use v.sin(theta)? Shouldn't it be v.cos(theta) as that's for horizontal? Furthermore, when finding the time, how come you don't times it by 2? Through the HSC calculations, it seems its just finding the time it takes to reach maximum height?

 

[Drsoccerball]

I am getting confused in one thing. The equation finds horizontal component of velocity right? So why do these use v.sin(theta)? Shouldn't it be v.cos(theta) as that's for horizontal? Furthermore, when finding the time, how come you don't times it by 2? Through the HSC calculations, it seems its just finding the time it takes to reach maximum height?

You do multiply by 2. The answer above needs to be multiplied by 2 assuming there's no resistance... We find the time it takes to get to its max height which is when vertical velocity=0 we don't care about horizontal for this question.

 

[Khan.Paki]

but the final answer says its 2.9 x10-8. Shouldn't it be 5.9 x10-8?

 

[Drsoccerball]

but the final answer says its 2.9 x10-8. Shouldn't it be 5.9 x10-8?

Yes like i said astro forgot to double it.

 

[Khan.Paki]

But thats not from astroman. that answer is from the BOS HSC sample answers. im getting confused here.

 

[InteGrand]

But thats not from astroman. that answer is from the BOS HSC sample answers. im getting confused here.

Yeah I remember the Board of Studies had a mistake for this Q in their sample answers (they didn't double for some reason). In the fine print at the start of the sample answers, they do say that the sample answers may not even be 'complete'. haha