Mr. Study's 3 and 4 Question.

[Mr. Study]

Hey, thanks for that guys. Question 1 looks so much easier in comparison to two. xD

Just need some clarification with the horizontal/vertical concept.

Say if I chucked a ball horizontally at 20 ms^-1.

Can I consider the horizontal movement to be a straight line? And therefore use the five equations of motion? With inital velocity being the 20 ms^-1.

But once I consider the vertical movement, I cannot use the given velocity?

If my question doesn't make sense, I'll try and re-word it.

Thanks.

[trinh]

Firstly, just remember:
- To split the velocity of the projectile into horizontal and vertical components for most calculations (I didn't know this at first and wasted so much time ><)
- The horizontal component of velocity remains constant (assuming no air resistance)
- The only force acting on the ball during its flight is gravity (again, assuming no air resistance); don't think that there's a 'pushing' or 'driving' force behind the ball causing to move throughout its flight

So if you chucked a ball horizontally at 20 ms^-1:

- The initial velocity would be 20 ms^-1
- Having split the velocity of the ball into horizontal and vertical components, I guess you can kinda consider the horizontal movement to be in a straight line, in that the horizontal velocity component remains constant (assuming no air resistance). Furthermore, since the horizontal velocity component remains constant, you probably won't need to use the equations of motion. It will probably involve the use of , or more specifically, (which is a form of an equation of motion xD)
- As you probably know, the initial vertical velocity would be 0. The magnitude of the vertical velocity would increase due to acceleration caused by gravity. So you'd probably have to use the five equations of motion here

- The 'complete' velocity of the ball would be given by the vector sum of the horizontal and vertical components, so you would use pythagoras to find it

Hope I could help

EDIT:

Can I consider the horizontal movement to be a straight line?

Yes you can, since you have 'picked out' only the horizontal movement (or velocity) for analysis. Because horizontal movement, by definition, is in a straight line.

[Mr. Study]

^. Thanks for that trinh.

New question: A child pulls a 4.0kg toy card along a horizontal path with a rope, so that the rope makes an angle of 30 degrees with the horizontal. The tension in the rope is 12N.

I got A) but B) and C) are a bit hard.

b) What is the component of tension in the direction of motion?
I do not understand what the question wants. :S Could someone explain it and I'll post up what I MAY think is the answer.

c) What is the magnitude of the normal reaction?
Looking at the previous question I asked, I'm not too sure if I should be using the equation . I dont think I would use it as the child isn't pushing down the rope. xD.

Thanks.

EDIT: Would someone be able to tell me what it is I'm doing wrong?
Question 3: Dodgem car, mass 200kg, driving south, hits a barrier at 5.0 ms-1. It rebounds at 2.0 ms-1. It was in contact with the barrier for 0.20s. Find the average acceleration, during it's interaction with the barrier.

This is my working out:
 
                 
                 



However, the answer is . Would someone be able to explain what I'm assuming is wrong? (If the question is too vague, I'll be happy to re-write it). Thanks again.

[Phy124]

Firstly, I'll just sort out your latex.

When using latex you can't use the "sub" or "sup" wraps you have to use _{Insert what you want "subbed" here} or ^{Insert what you want "supped" here}. Also to do the theta symbol it is \theta

Now onto the questions...

Here is a picture I illustrated showing what is happening in the question. You may want to look at it after attempting the question, but up to you.

b) I believe the component of tension in the direction of motion is the horizontal component of tension

Working out here, for when you have attempted by yourself (I think that's what you wanted to do?)

c) Forces down = Forces up (F_net vertically = 0)

Let T_V = the vertical component of tension

Working out here.

3. Hmm, your answer looks right to me, so maybe the given answer is incorrect or I'm just not thinking properly.

[Mr. Study]

Hey, Thanks for that Phy124. I had to cheat a little and look at the diagram first before attempting the question again. . However, it made all my other questions easier!

Last question and I will be done with Forces/Tension/Motion. :O
An old light globe hangs by a wire from the roof of a train. What angle does the glove make with the vertical when the train is accelerating at

I can find angle, if given more values, but I have NEVER done one with only acceleration. Would someone be able to explain what I should do but not, yet, give the answer?

Thanks.

[Lasercookie]

Hmm, well the vertical component of acceleration would be 10 m/s^2.

You should be good to go with that hint.

[Mr. Study]

Haha! I got it now! Thanks.

EDIT: New topic: Electricity. Question 1: Three 10 ohm and one 20 ohm are joined in a series, across a 10.0V supply. What is the potential difference across each?

Would someone be able to explain how I should approach this question.

Thanks.

NVM, Worked it out!

[yawho]

Hmm, well the vertical component of acceleration would be 10 m/s^2.

You should be good to go with that hint.

Has the light globe got acceleration vertically?

[Lasercookie]

Hmm, well the vertical component of acceleration would be 10 m/s^2.

You should be good to go with that hint.

Has the light globe got acceleration vertically?

Yes, because gravity is acting upon it - which is why the vertical component (emphasis on the word component) is 10 m/s^2. You could have also used 9.8 m/s^2 or whatever other value of g you prefer but 10 is nice (and is also the value of g you're given on the exams).

It's being held up by a wire, which is why it doesn't just drop to the ground (we're not given any details about the wire, other than the fact that it's there).

[yawho]

So the light globe has acceleration in the vertical direction?

[Lasercookie]

So the light globe has acceleration in the vertical direction?

Yes.

[yawho]

So the light globe has acceleration in the vertical direction?

Yes.

So Newton's second law is not applicable to the light globe in the vertical direction?

[Lasercookie]

So the light globe has acceleration in the vertical direction?

Yes.

So Newton's second law is not applicable to the light globe in the vertical direction?

Why would it not be applicable? I'm kind of confused about what you're getting at, do you want to explain what you're thinking and I'll be able to figure out what exactly is confusing you/explain better.

[yawho]

Vertically the net force on the light globe is zero, don't you think the acceleration is also zero in the vertical direction. I learned this in year 11.

[Lasercookie]

Vertically the net force on the light globe is zero, don't you think the acceleration is also zero in the vertical direction. I learned this in year 11.

Well my approach above was to just look at the ratio between the relevant acceleration. I was kind of stuck in my thinking with that approach, so I did say something incorrect up there (sorry about that) - but yes, you're right the net vertical force is zero and yes there is zero net vertical acceleration - as the force of the wire holding up the globe opposes the force of gravity. If you want to look at it in terms of forces though:

You have the horizontal force of the wire given by:
You have the vertical force given by: - this force here is opposed by the upwards force exerted by the wire, giving the net vertical force of zero and net vertical acceleration of 0.

So then we find that theta is given by:


I immediately started off by just thinking about this ratio between the relevant accelerations - pretty much skipping this cancelling out of the masses step and going straight to

So solving for theta: