VCE Physics Question Thread!

[odeaa]

For the last question of the attached. How likely will these sort of motion questions be in the VCAA end of year exams? Struggled with this question since I didn't expect it to get this mathsy in VCE physics, ended up getting it by looking at the worked solution though lmao.

Yeah I've seen a few like that, those graphs are pretty common

[lzxnl]

For the last question of the attached. How likely will these sort of motion questions be in the VCAA end of year exams? Struggled with this question since I didn't expect it to get this mathsy in VCE physics, ended up getting it by looking at the worked solution though lmao.

Questions 9 and 10 are fairly typical year 12 questions.
Question 11 is a little trickier in that you have to equate the areas under two graphs and while that's assessable in VCE physics, I don't recall seeing a question like this in a while. VCE physics motion questions now tend to be fairly simple number crunching questions. A lot has happened in 35 years.

[Muchos Help]

Questions 9 and 10 are fairly typical year 12 questions.
Question 11 is a little trickier in that you have to equate the areas under two graphs and while that's assessable in VCE physics, I don't recall seeing a question like this in a while. VCE physics motion questions now tend to be fairly simple number crunching questions. A lot has happened in 35 years.

Thanks.

I have another tough question I couldn't do, it's the last question of the attached. In the answer it states that the motorcycle travels 100m in every 5secs after t=15 which I don't quite understand how they got that.

Also, they then calculate the time the motorcycle takes to travel the 350m but isn't the car also moving at the same time?

[Adequace]

For the attached on Q6. The answer just said "this ratio must always equal 1", why is that and does this mean anything?

I got 7m/s^2 but they got 7.2m/s^2 for the question prior. Although their answer is closer, I'm not exactly sure how you can guess the correct value.

[wyzard]

For the attached on Q6. The answer just said "this ratio must always equal 1", why is that and does this mean anything?

I got 7m/s^2 but they got 7.2m/s^2 for the question prior. Although their answer is closer, I'm not exactly sure how you can guess the correct value.

The value of 7.2 is gotten by reading the graph more accurate with the use of a ruler to measure the length. FOr the second part, the ratio of 1 means that the gravitational force acting always stays the same.

[Adequace]

The value of 7.2 is gotten by reading the graph more accurate with the use of a ruler to measure the length. FOr the second part, the ratio of 1 means that the gravitational force acting always stays the same.

Thanks.

So, would the gravitational force always equal 1? Would there ever be a situation where it doesn't equal 1?

[wyzard]

Thanks.

So, would the gravitational force always equal 1? Would there ever be a situation where it doesn't equal 1?

Gravitational force don't equal to one, it's the ratio of the forces at different time that equal to one, meaning they're the same.

Of course there are situations when the ratio will not equal to one. When we talk about large distances away from Earth where gravity is no longer uniform, like satellites that orbit the Earth. From there we cannot use anymore, but we'll need another way to calculate gravitational force via Newton's Law of Universal Gravitation which is:

[Adequace]

Gravitational force don't equal to one, it's the ratio of the forces at different time that equal to one, meaning they're the same.

Of course there are situations when the ratio will not equal to one. When we talk about large distances away from Earth where gravity is no longer uniform, like satellites that orbit the Earth. From there we cannot use anymore, but we'll need another way to calculate gravitational force via Newton's Law of Universal Gravitation which is:

Many thanks.

[Maz]

hello...can someone please help me on these...its projectile motion
1. How long will it take a ball thrown at 12ms^1 at an angle of 70 degrees above the horizontal to reach a height of 4m above its launch position?
2. Sketch separate speed versus time graphs for the horizontal and vertical motions of a projectile that it launched directly upwards and lands below the launch position

Thankyou in advance 

[Muchos Help]

hello...can someone please help me on these...its projectile motion
1. How long will it take a ball thrown at 12ms^1 at an angle of 70 degrees above the horizontal to reach a height of 4m above its launch position?
2. Sketch separate speed versus time graphs for the horizontal and vertical motions of a projectile that it launched directly upwards and lands below the launch position

Thankyou in advance 

For 1, you probably just have to use the equations of motion with the variables that the question gives you. (a,x u,v)

2. For the horizontal graph, it would just be a horizontal line at +12m/s since horizontal a=0. I'm not 100% on the vertical graph but I'd say it's a negative slope from +12m/s

[Peanut Butter]

Could someone please help me with the following projectiles question?

Tam throws a ball across a field a horizontal distance of 100m. She throws it at an angle of 30 degrees. It lands at the same height (ignore air resistance). At what speed does the ball leave her hand?

The answer is in the spoiler below

Spoiler

The answer is 34 m/s

Thank you!

[lzxnl]

Could someone please help me with the following projectiles question?

Tam throws a ball across a field a horizontal distance of 100m. She throws it at an angle of 30 degrees. It lands at the same height (ignore air resistance). At what speed does the ball leave her hand?

The answer is in the spoiler below

Spoiler

The answer is 34 m/s

Thank you!

Let's break this down completely. I'm assuming that you recognise this is a constant acceleration in 2D problem (vertical acceleration =g downwards, horizontal acceleration = 0. This part is important; please be sure of this first)

What do we need for a horizontal distance? Well, because the acceleration is zero in that direction (the vertical direction doesn't affect the horizontal direction to a very good approximation; this has been verified by experiment and it follows from the fact that the gravitational force is vertical. I say approximation because the Earth is a rotating sphere and that complicates things but let's disregard that), the distance is just the horizontal velocity * time. The horizontal velocity is related to the speed and the angle, which you have.
So we have 100 = v(x)t = u*cos 30*t

The problem is, this is one equation in two unknowns, so we need another piece of information. This is where we use the vertical motion.

If the ball is thrown across a horizontal field, its total vertical displacement is identically zero. This is also important. In these questions, you need to recognise what words mean what mathematically. That's the art of problem solving: converting given information into useful information.

Anyway, you now have the vertical displacement and the acceleration. Remember how we still don't know the speed at the time? Let's write out another equation that uses them so that we can solve for them. The relevant constant acceleration equation that we want now is the one that involves the initial vertical velocity (to include the speed), the time and the other two known quantities, so our equation becomes 0 = v(y)t - 1/2 at^2.
t isn't zero because we're not interested in the instant the ball is thrown.
So v(y) = 1/2 at -> u sin 30 = 1/2 at, u = at
Plug this into the other equation and solve for the speed.

Of course, you can do this in general and with a bit of algebra, you can find that the general formula for the range for a projectile with no gain in height is R = v^2 sin (2*theta) / g. I'll leave that up to you. Hint: you'll need a double angle trig identity somewhere for this.

[Peanut Butter]

Thank you so much!!

[JI2015]

Hi,

I know someone has already answered the question for you, but you may like to see how I approached this question.

http://jmp.sh/v9IEs96

All the best!

Also, just some advice for physics calculations: When I had trouble trying to start the question, I always wrote down the information the question provided me with and then what information I was trying to find. It makes it much easier to see where you are going and how you can relate each piece of information together to find the answer. If still stuck, my cheat sheet had formulas grouped for each question type (e.g. projectile motion) and I would quickly look and see which one's I would have to use given the specific information provided by the question.

[Peanut Butter]

Thank you!

The formulas at the bottom and the way you set out your solution was really helpful!