Cort's 3/4 Physics Thread

[PB]

PS - PB, Lxnl - did you freaks end up completing all 783 qs? When do you think me and Cort should complete them by?

Cheers

haha I am definitely not smart enough to be a freak, lxnl yes, but not me XD.  How on earth can you find the time to finish all the questions?!?! I think it would have been impossible for me to have finished even half the questions due to other subjects and other time consuming responsibilities. I think I probably did about 1/4 or less of Checkpoints unfortunately
Now, I don't mean to demotivate you when I say this, I am merely telling you what I did :S

I mean, definitely try to finish all the questions if you can, but I honestly think it is quite an ambitious task :PP

My recommendation for you is to simply do all the relevant questions relating to your SACs' topics. Do only minimal questions on the topics that are not tested on your SACs just so you know about them. But once it comes to the end-of-year exam, start to learn the insides and out of the whole course by doing many exams.
That is exactly what I did for Physics and it seemed to work out fine for me
If you can finish all the questions, by all means, go for it. But if you are struggling for time, which most of you probably will be throughout the year - maybe try to give my method a go. It is quite efficient.

The weird thing is that I did at least 70% of Checkpoints questions for Chemistry (a whole heap more than for physics) and I happened to do worse lol ah well.

[Rod]

haha I am definitely not smart enough to be a freak, lxnl yes, but not me XD.  How on earth can you find the time to finish all the questions?!?! I think it would have been impossible for me to have finished even half the questions due to other subjects and other time consuming responsibilities. I think I probably did about 1/4 or less of Checkpoints unfortunately
Now, I don't mean to demotivate you when I say this, I am merely telling you what I did :S

I mean, definitely try to finish all the questions if you can, but I honestly think it is quite an ambitious task :PP

My recommendation for you is to simply do all the relevant questions relating to your SACs' topics. Do only minimal questions on the topics that are not tested on your SACs just so you know about them. But once it comes to the end-of-year exam, start to learn the insides and out of the whole course by doing many exams.
That is exactly what I did for Physics and it seemed to work out fine for me
If you can finish all the questions, by all means, go for it. But if you are struggling for time, which most of you probably will be throughout the year - maybe try to give my method a go. It is quite efficient.

The weird thing is that I did at least 70% of Checkpoints questions for Chemistry (a whole heap more than for physics) and I happened to do worse lol ah well.

Some excellent tips here thank you so much.

And you are a 'freak' PB

[Cort]

In regards to energy, how on earth is work done on the spring the same as energy stored in the spring? I thought that for work done to occur, it conceptually requires some kind of force constant and a sense of displacement? That is, WD = F * x. This is perhaps why WD = F*x = 1/2 * M *v^2, right?

 Or is it because energy stored equation (1/2 * k * delta x^2) does have a force constant (which is the spring constant, k), and there is also a change in displacement as well? Hence, one can assume that:

WD = F * x = 1/2*m*v^2 = 1/2 * k* delta x ^2?

[Cort]

Also, can one also assume that in terms of energy transfers, it's rather obvious by what direction it's travelling to give you an idea of how to set up the equation? That is, if it states that it is travelling horizontally, that means that in terms of energy, it will only be kinetic as work done to some other form; eg. spring p.e. However, the question shows that it has travelled in some form of an angle (like down a hill), it has BOTH Gpe + Ke in the system of Total energy?

I think I'm starting to get the hang of this.

Lastly, is the change in displacement in hooks law/Epe the same as "compression"? Since change in displacement occurs when it is compressed and/or stretched?

[lzxnl]

In regards to energy, how on earth is work done on the spring the same as energy stored in the spring? I thought that for work done to occur, it conceptually requires some kind of force constant and a sense of displacement? That is, WD = F * x. This is perhaps why WD = F*x = 1/2 * M *v^2, right?

 Or is it because energy stored equation (1/2 * k * delta x^2) does have a force constant (which is the spring constant, k), and there is also a change in displacement as well? Hence, one can assume that:

WD = F * x = 1/2*m*v^2 = 1/2 * k* delta x ^2?

Work = force by distance only when the force is constant. For a spring, it's not constant.
If we consider stretching the spring by a tiny displacement dx such that the force is effectively constant over the entire displacement, the differential work done dW is given by force*distance = F dx. Now, finding the total work done, we integrate over whatever x values the spring moves over. Now, F = -kx and the change in potential energy is equal to the negative of the work done (work = change in kinetic energy, sum of changes in kinetic and potential energies is zero by conservation of energy), so the potential energy is the integral of kx dx, or just 1/2 kx^2.

As for how kinetic energy is 1/2 mv^2, the easiest way of thinking about this is accelerating an object of mass m from rest to a speed v at a constant rate. Remember v^2 = u^2 + 2ax? Well, if accelerated from rest, u=0, so v^2/2 = ax. For a constant force, work = Fx = max = m(ax) = mv^2/2 which is the familiar formula for energy.

Also: work done on a spring is NOT 1/2*k(change in x)^2. It is 1/2*k(final x^2 - initial x^2).

Also, can one also assume that in terms of energy transfers, it's rather obvious by what direction it's travelling to give you an idea of how to set up the equation? That is, if it states that it is travelling horizontally, that means that in terms of energy, it will only be kinetic as work done to some other form; eg. spring p.e. However, the question shows that it has travelled in some form of an angle (like down a hill), it has BOTH Gpe + Ke in the system of Total energy?

I think I'm starting to get the hang of this.

Lastly, is the change in displacement in hooks law/Epe the same as "compression"? Since change in displacement occurs when it is compressed and/or stretched?

Yes, down the hill the object has both KE and GPE. However the height isn't the distance travelled on the hill; it's just the vertical displacement.

The change in displacement in Hooke's law is just how much the spring is stretched from the point of no force. A compression generally increases the change in displacement, yes.

[Cort]

Work = force by distance only when the force is constant. For a spring, it's not constant.
If we consider stretching the spring by a tiny displacement dx such that the force is effectively constant over the entire displacement, the differential work done dW is given by force*distance = F dx. Now, finding the total work done, we integrate over whatever x values the spring moves over. Now, F = -kx and the change in potential energy is equal to the negative of the work done (work = change in kinetic energy, sum of changes in kinetic and potential energies is zero by conservation of energy), so the potential energy is the integral of kx dx, or just 1/2 kx^2.

Wait, I never knew this. So because Work = Change in Ke, Work also = 0 due to the conversation of energy by Change in Ke + change in Pe? Hence that is why change in pe = negative work (aka change in Ke)? I'm just a tad confused because I need some sort of concrete formula/picture to help me out there. Sorry!

Edit: DON'T SAY THIS IS BECAUSE TOTAL ENERGY = Pe (either Gpe or Epe) + Ke = CONSTANT BECAUSE OF CONVERSION OF ENERGY RIGHT?
ALSO, if that means that it is the negative of work done = the energy would still be 'postive' in the end because you cannot have negative energy?

[lzxnl]

Wait, I never knew this. So because Work = Change in Ke, Work also = 0 due to the conversation of energy by Change in Ke + change in Pe? Hence that is why change in pe = negative work (aka change in Ke)? I'm just a tad confused because I need some sort of concrete formula/picture to help me out there. Sorry!

Edit: DON'T SAY THIS IS BECAUSE TOTAL ENERGY = Pe (either Gpe or Epe) + Ke = CONSTANT BECAUSE OF CONVERSION OF ENERGY RIGHT?
ALSO, if that means that it is the negative of work done = the energy would still be 'postive' in the end because you cannot have negative energy?

Conservation of energy states that so
so

You can't have negative kinetic energy, that is correct. If something has only 2 J of kinetic energy, you cannot possibly do -3 J of work on it. As soon as the object loses all kinetic energy, it's at rest and can only gain kinetic energy, so kinetic energy never becomes negative.

Potential energy is a different story. These can be negative. Gravitational potential energy (in space, not the mgh one on Earth) is always negative, for instance. I don't quite understand your query here though. I've only said that the signs of the changes in energy may be negative.

[Cort]

Conservation of energy states that so
so

You can't have negative kinetic energy, that is correct. If something has only 2 J of kinetic energy, you cannot possibly do -3 J of work on it. As soon as the object loses all kinetic energy, it's at rest and can only gain kinetic energy, so kinetic energy never becomes negative.

Potential energy is a different story. These can be negative. Gravitational potential energy (in space, not the mgh one on Earth) is always negative, for instance. I don't quite understand your query here though. I've only said that the signs of the changes in energy may be negative.

Ah, bloody hell. Lacking logical reasoning sucks.


Another note: I'm having trouble understanding why an object will not go faster if it reduced its mass. Here's the question:

"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."
   Q: Explaining whehter this will be a successful way for Jack to catch up to Jill

I stated:
 - Jack will not catch up to Jill because pushing sideways box means he gets sideways force in the opposite direction.
- Jack will only catch up if he pushed it directly behind him because he gains forward momentum.

My troubles:
I cannot fathom why (as the answer states) "From a conversation of momentum, there will be no change in the forward momentum of the box or tobogan" and "Lightening his toboggan will not cause Jack to gain or lose downhill momentum" (Why so? Is it because he pushed it sideways, and not backwards instead? And how does not lightening his toboggan not allow Jack to gain momentum because of less mass?

[Zealous]

"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."
   Q: Explaining whehter this will be a successful way for Jack to catch up to Jill

For an inclined plane:



There is no mass component in this equation. If he reduces his weight, he will still accelerate at the same rate as before (and Jill will also accelerate at that same rate).

Why? If Jack reduces his weight, it will reduce his weight force (N) due to gravity, leading to a decrease the normal force and also the force pushing him down the slope will also be reduced.



We've got less mass, but we've also got less force (down the hill), so acceleration stays the same.

Hopefully that makes sense.

[Cort]

For an inclined plane:



There is no mass component in this equation. If he reduces his weight, he will still accelerate at the same rate as before (and Jill will also accelerate at that same rate).

Why? If Jack reduces his weight, it will reduce his weight force (N) due to gravity, leading to a decrease the normal force and also the force pushing him down the slope will also be reduced.



We've got less mass, but we've also got less force (down the hill), so acceleration stays the same.

Hopefully that makes sense.

It does. Thank ye

[Cort]

I tend to confuse myself when it comes to solving projectile motions..the thing is, for all types of those projectile motion situations ( such as horizontal launch at a cliff; launch from an angle) I do no know whether my initial velocities (in the vertical component) is correct or not.
  If initial speed (Vo) is given, does that mean initial velocities for the vertical component exist, even if it is launched at a cliff? Because I think that would make some sense. Conversely, if it is projected horizontally from a cliff, and no angle/initial velocity is given, then  Uv = 0?

[Aurelian]

I tend to confuse myself when it comes to solving projectile motions..the thing is, for all types of those projectile motion situations ( such as horizontal launch at a cliff; launch from an angle) I do no know whether my initial velocities (in the vertical component) is correct or not.
  If initial speed (Vo) is given, does that mean initial velocities for the vertical component exist, even if it is launched at a cliff? Because I think that would make some sense. Conversely, if it is projected horizontally from a cliff, and no angle/initial velocity is given, then  Uv = 0?

Just use your common sense and think only about how the object you're dealing with is moving through space. If an object is being launched "horizontally", then that means its initial motion is entirely entirely left or right and that the object is not moving up or down at all to start with. Since the object isn't moving up or down in any way, then clearly that object cannot have an initial (non-zero) vertical velocity; the object is stationary with respect to vertical motion.

An object launched with an initial (non-zero) speed at some angle to the horizontal, however, will be moving up or down as well as left or right. That is, we will be able to represent this angled-speed as some vector combination of a left/right velocity and an up/down velocity. Since the object is initially moving in a vertical direction as well as a horizontal direction, clearly it will have an initial (non-zero) vertical velocity.

Whether a projectile is launched from a cliff or from the ground doesn't change anything concerning the above reasoning; all we need to know about when finding initial velocities is how the object is moving through space at the start of its flight. Surrounding scenery only becomes relevant later on in the problem solving.

Does that help? =)

[Cort]

Just use your common sense

But I have none D:

and think only about how the object you're dealing with is moving through space. If an object is being launched "horizontally", then that means its initial motion is entirely entirely left or right and that the object is not moving up or down at all to start with. Since the object isn't moving up or down in any way, then clearly that object cannot have an initial (non-zero) vertical velocity; the object is stationary with respect to vertical motion.

An object launched with an initial (non-zero) speed at some angle to the horizontal, however, will be moving up or down as well as left or right. That is, we will be able to represent this angled-speed as some vector combination of a left/right velocity and an up/down velocity. Since the object is initially moving in a vertical direction as well as a horizontal direction, clearly it will have an initial (non-zero) vertical velocity.

Whether a projectile is launched from a cliff or from the ground doesn't change anything concerning the above reasoning; all we need to know about when finding initial velocities is how the object is moving through space at the start of its flight. Surrounding scenery only becomes relevant later on in the problem solving.

Does that help? =)

That makes total sense, since I never thought about it that way. I'm still trying to break away my bonds of "sub this crap in and get outta here" mentality..but it's slow progress I guess. Thanks.

[Cort]

This is from Checkpoints:

The ball leaves the tee horizontally and hits the hole. The horizontal distance between the tee and the hole is 155m. The vertical distance is 45m. There is a tailwind that effectively cancels any air resistance.

a) Calculate the speed of the ball off the tee:
I've worked out the angled speed to be ~60ms^-1. Apparently the answer is ~52 -- which is the horizontal speed. Did I read the question wrong? Or am I working it out wrong?

[lzxnl]

Ah, bloody hell. Lacking logical reasoning sucks.


Another note: I'm having trouble understanding why an object will not go faster if it reduced its mass. Here's the question:

"Jack and Jill are racing their toboggans down an icy hill. Jack and Jill are of similar mass are using the same type of tobogann.When Jack is a certain distance from the end of the race they are travelling with the same velocity. Jack is behind Jill and decides that if he is going to win the race he must lighten his tobogann so he pushes a box containing their ice-skating gear off the side of his toboggan."
   Q: Explaining whehter this will be a successful way for Jack to catch up to Jill

I stated:
 - Jack will not catch up to Jill because pushing sideways box means he gets sideways force in the opposite direction.
- Jack will only catch up if he pushed it directly behind him because he gains forward momentum.

My troubles:
I cannot fathom why (as the answer states) "From a conversation of momentum, there will be no change in the forward momentum of the box or tobogan" and "Lightening his toboggan will not cause Jack to gain or lose downhill momentum" (Why so? Is it because he pushed it sideways, and not backwards instead? And how does not lightening his toboggan not allow Jack to gain momentum because of less mass?

One of my students asked me this question today and upon consideration, this is my response.
Lightening the toboggan has no effect on Jack's momentum. Why? Well, the argument is that p=mv, and if we reduce m we must increase v. That argument implicitly uses conservation of momentum on the extended toboggan-box system, and the mass of that system is NOT reduced. Therefore, you can't say that lightening the toboggan will immediately speed it up.

Second issue: when Jack throws the box off the toboggan, the box is still moving down the hill in the same direction as the toboggan. If you consider the toboggan-box group as a single system, they form a system only acted upon by gravity. The force on the box, coming from the toboggan, is an internal force and cannot affect the momentum of the system as a whole.

Also, we can split the system up into the toboggan and the box individually. Although initially we think of the box on top of the toboggan, the momentum of the toboggan is really independent of the box; the toboggan, moving at a fixed speed, has the same momentum regardless of if the box is on top of it. Pushing the box off the tobbogan is then akin to pushing sideways on a wall. As the person is pushing sideways, the downhill motion is unaffected and the toboggan moves at the same rate. You're right, it does have to do with the fact that the box is thrown sideways.

This is from Checkpoints:

The ball leaves the tee horizontally and hits the hole. The horizontal distance between the tee and the hole is 155m. The vertical distance is 45m. There is a tailwind that effectively cancels any air resistance.

a) Calculate the speed of the ball off the tee:
I've worked out the angled speed to be ~60ms^-1. Apparently the answer is ~52 -- which is the horizontal speed. Did I read the question wrong? Or am I working it out wrong?

So...initial vertical speed is zero. Vertically, we have 45 = 1/2 at^2 = 5t^2. t=3 (zero initial vertical velocity)
Horizontally, as the horizontal acceleration is zero, we have 155m = speed * time = speed * 3s. Speed is around 51.67 m/s.