I know these videos are from last year but I am finding them sooo helpful prepping for gamsats in september! Also on torque, it's used heaps in musc-skeletal biomechanics. Not sure if gamsats would actually do a question on that but it would be a super cool real-world application and if my march sitting was anything to go by they do some real random questions haha
Thanks so much for these videos. I’ve realised for the GAMSAT questions what I am struggling with most is the maths due to my mental arithmetic not being hot. I am hoping with practice I can improve it for MARCH.
This is exactly what I had hoped to achieve for people with these so this is great to hear! With trigonometry, I'd say that knowing sin(30)=1/2 and cos(60)=1/2 is useful for applying some shortcuts in certain questions but from what I can see, it isn't necessary to know the trig table. ACER questions often simplify the question by saying things like "let sin(theta) = alpha" and then you just use substitution and the question focuses on the algebra rather than the memorisation of random ratios. If you did want to memorise them, there's this really handy trick if you weren't already aware: m.th-cam.com/video/jI81WXyFrL0/w-d-xo.html The Des stuff does require the use of trig ratios but they're super outdated.
you are actually so amazing, thank you so much for all of your help. i've watched pretty much all of your videos ad you've made me feel more confident. i almost spent my life savings on a gamsat prep course but then you saved me. this will be my first time sitting the gamsat, so thank you for alleviating some of my anxieties. would you say that it is true that the gamsat questions give you all the info you need in the stem of the question to answer it and it is just a matter of reasoning?
Glad to hear you've been able to save your money! It's a difficult question to answer perfectly because everyone has a subjective experience and view on this because of variations in experience and familiarity with different concepts. Most stems are designed such that you can work with the given information, but they may still require some familiarity with the concepts in order to process and apply the information within the expected time frame. For example, if you were given a stem about a type of chemical reaction, chances are you've never heard of it and wouldn't need to have known about it beforehand. BUT, within the stem, there might be words/phrases used that if you had no experience with chemistry might be confusing or difficult to grasp. Probably not a perfect answer to your question, but the best thing to do is learn by doing and if you find you're struggling to grasp what the stem in a practice question is saying, review the core principles behind it and keep moving Best of luck!
Thanks so much for these videos, Jesse. They are proving really helpful in my gamsat study so far. Quick question in regards to torque, is the perpendicular radius perpendicular to the force that is applied? Thanks in advance.
Hey Lorenzo! Yep, that right. I think in the video I say perpendicular radius but obviously you could also think of the force as being perpendicular to the radius too. There are shortcut formulas that calculate it all with an angle included such as FdCos(ø) or FdSin(ø) where F is some force applied at an angle, ø to the object at a distance d from the pivot. The FCos(ø) or FSin(ø) component is just the trig working out the perpendicular component of the force and then multiplying it by 'd' to get the torque. You don't need to know this for GASMAT as trigonometry calculations are often simplified by ACER in the stems anyway, but figured you might come across this and it could also be helpful to see how it reinforces the perpendicular force and radius assumptions.
Another great video! These are proving to be a huge help!! I am taking the GAMSAT in March and I'm wanting to get into Sydney uni in 2023. I might see you there if that's where you decide to go 🙂
Hey Jesse, great video as always! I'm just a bit confused with equations of motion, when would s=ut+1/2at2 be used in comparison to v2=u2+2as? Is it just if we didn't know tfinal, or have I confused this? thanks!
Hi Jesse! I was looking at some “physics work” definitions and I got a bit confused. Definition 1- makes sense The way I think about “work in physics” is simply as “work is done when a force is applied to an object, which displaces or moves that object”… THAT’S IT. (i.e work just centres around displacement/movement). Is this sufficient detail or are there gaps in this definition? Definition 2- confusing “work is done when a force causes an object to move IN THE DIRECTION OF THE FORCE”. The caps lock part is confusing! Because I’ve seen a scenario where you can apply a force, and the object ends up moving/displacing in a “different direction” to the force. Then they use W = F x d x cos(theta) as the formula. See what I mean? The caps lock part doesn’t make sense to me. Can we just think of work as moving an object? By the way, I love your efforts and videos! Keep them up! ❤❤Congratulations for getting into med school! (I have to still watch those videos as I haven’t gotten around to them yet 😂🎉).
Hey Jessica, okay so your definition should be fine for 99% of the applications of work. Work is a measure of the energy introduced to a system via a force acting over some distance. The direction thing is referring to the force component that is directed in the same direction as the movement of the object. In most cases, the force applied will be directed in the same direction as the resulting movement of the object. eg. If you push horizontally on the back of a broken down car, it moves forward horizontally. However, it is also possible to apply an oblique force which results in a horizontal or vertical force. In this case, force applied (at an angle) is not in the same direction as the movement. eg. Pushing on the handle of a shopping cart might actually involve pushing down and forward (so down at an angle) but because the floor is horizontal, the shopping cart moves horizontally. In these scenarios, not all of the applied force is doing work on the system so it would be wrong to simply multiply force applied x distance travelled. Instead, we want to calculate the horizontal component of that oblique force which involves using trigonometry (sin or cos of the angle against the vertical or horizontal, respectively. cos of the angle against the horizontal is more commonly used). So the formula W = F x d x cos(theta) is better expressed as W = F x cos(theta) x d where F x cos(theta) represents the force component that is in the same direction as the objects movement. A little tricky to explain in just words (I wish YT comments allowed images) but hopefully this clears things up!
Ohhh so even if we get a force and movement in “different directions”, when we do W = f x cos(theta) x d…. the f x cos(theta) part will try to just get the force and movement in the same direction via trigonometry. So both: W = f x cos(theta) x d W = f x d Will end up with the “same direction” of force and movement in the end.
Hey Jesse, in this video you said not to worry about the formulae because it will be given in the stem, however on your notion page it says that you should memorise formulae such as kinetic, work and gravitational energy, along with torque. And to save another comment the parallel and series effective resistance formulae also fall under ones you should know. Just wondering whether you can confirm whether it's necessary to remember these. Cheers.
Hey Mitchell, I mention that "most" formulas are given in the stem but also say for this crash course not to worry about memorising formulas as I didn't want people to be watching and getting hung up on trying to scribble down and rote learn any formulas they see. Just putting some context to the intentions of the video. The formulas in the cheat sheet that I've placed under "need to know" would be those that aren't most formulas. The number of equations in that list though are minuscule relative to the number of formulas that might pop up in a 'physics based question' and they are largely limited to three or less variables and underpin the foundations of physics concepts needed for basic reasoning. I've put the parallel and series circuit equations under "need to know" because although I don't think it's likely you would use them directly in a question (this would just be routine calculation rather than reasoning), knowing them helps explain the nature and assumptions around the two circuit types eg. Ohm's law, voltage drop, and Kirchoff's law. Better to know them than to not know them, in my opinion :)
@@jesseosbourne Thanks for the response Jesse, as always. I was just working through the practice question booklet and the question on deceleration of blood flow tripped me up, in the worked example they use the v2 = u2 + 2as formula but that wasn't given in the stem. I understand these booklets are quite old and probably reflect the questions as to when the exam was more theory based. Would you recommend memorising the kinematics formulae or do you think with the newer formatted exams they would give you that equation to work with?
@@mitchellgourlay4847 Yeah I'd still say that the kinematics formulas are worth knowing. The trend tends to be that you are given 2 or 3 of the five which would mean a 2 or 3 step process of using simultaneous equations but knowing all 5 means you can shortcut straight to the one formula that will allow you to solve the missing variable in the one step. The five to know are: v=u+at v^2 = u^2 + 2as s = ut + (1/2)at^2 s = vt - (1/2)at^2 s = (u+v)t/2 Keep in mind that these formulas can only be applied over intervals where acceleration is constant
@@jesseosbourne Thank you Jesse, really appreciate your help. Hopefully all this hard work pays off come March. Happy new year brother, I hope all is well as you are preparing for medical school. You're gonna ace it
As the horizontal velocity component of the ball at max height is still greater than zero (as it remains constant throughout the projectile motion), wouldn't this make the kinetic energy more than 0 at max height? Or should we just consider the vertical component when calculating the energy?
Hey Patel! Sorry for the wait on a reply until I returned from the break. This is a great question. Technically, the ball does still have a non-zero kinetic energy at the top of the arc due to the horizontal velocity. In the question though, we are only considering the vertical components of motion and therefore only the vertical components' contribution to the energy. This is why it's important to determine the vertical component of its initial velocity in determining kinetic energy rather than the launch speed (at an oblique angle). Hopefully this clears things up :)
Hey Jesse, I'm just wondering why you put g as a negative when you were calculating for t on the earth at 17:32 because I thought gravity would be positive if accelerating downwards? Thanks :) Edit: Nevermind! Just realised you were calculating for when the ball would reach the max height so it would be going against gravity hence the negative!
Any advice for easy resources I can access to apply this knowledge. I say easy because I still feel very lost, I might understand one example, then I'm completely lost of the next one. HELP :(
A good site is libretexts. You can work through specific topics and there's usually some 'exercise' type questions like you'd see in regular textbooks. Be mindful that a lot of physics theory questions can be a little tedious with the calculations because they assume access to a calculator and are generally assessing technical skills a lot more often than conceptual skills as we get in GAMSAT. That being said, it's a good go to. Some other options are just to google random worksheets by searching 'topic name, physics, worksheet/questions, pdf' - this usually returns some good results If you haven't already, check out my notion page (linked in the video description) where I've made a bank of GAMSAT style questions with solutions videos. Although admittedly there may not be many on this specific topic
Hi! Thanks for the videos!! Just wondering why when calculating the max height of the ball trajectory off the raised platform, why is gravity the only form of acceleration accounted for? Are we assuming that the kick/throw has no force?
Hey Rosie, Glad they're helping! It is a little counterintuitive at first, given the ball is still moving upwards on it's upward journey, however there are a few ways we can resolve this. The first thing is that we're applying Newton's second law which is often misquoted as F = ma but it really is NET FORCE = ma. So the acceleration is in the direction of the net force or sum of all the forces combined. 1. Consider the ball in flight as a point and draw it as a force diagram. When in flight the only force acting on the ball is the gravitational force acting downwards. We ignore any air resistance etc in these scenarios. Therefore the net force acting on the ball is equal to the downward gravitational force, due to gravity's acceleration. You'll notice that on the upward phase I use gravity as a negative value because while the ball is going up, gravity is acting to decelerate the ball and slow it down. 2. If you're still wondering about how the ball doesn't have a force and acceleration from the kick, we can look at what happens at the time the ball leaves the foot of the person kicking it (or the hand of the person throwing it) to explain this. While the ball is contact with someone's foot, the foot can be applying a force to the ball and this would accelerate the ball upwards against gravity but as soon as the ball leaves the foot, the foot can no longer apply this force, so the ball is now in 'free fall' ie. only under the effects of gravity. So the kick does still apply a force but only while in contact with the ball. This is what accelerates the ball from zero up to it's initial launch velocity (represented by u in the calculations). Our calculations however are only concerned with the ball from the moment it leaves the foot to the moment it first strikes the ground again. During this phase the foot is no longer providing an acceleration and so the only acceleration is that of gravity. An analogous scenario would be to think of a car accelerating down a straight road. While the accelerator pedal is being pressed, the engine applies a force to accelerate the car forward and speed it up but as soon as you let your foot off the accelerator, the car is now no longer under that acceleration and the only forces still acting on it would be friction at the tyres and some air resistance all pushing in the opposite direction to the cars motion. The car would still roll forward but it would slow down as it is under a deceleration from those resistive forces. Going back to the net force = ma thing, when the car is accelerating, the driving force of the engine is likely much greater than that of the resistive air resistance and frictional forces so the net force of the car is forward and so is it's acceleration. When the driving force is removed (no accelerator) then the car is coasting along and the only forces are the resistive forces making up the net force and so the acceleration is now entirely due to them a = net force/mass of car) and so it will begin to slow down until it stops. So that you'v got a broader understanding of how to apply this reasoning in other scenarios and questions, generally speaking, a force and therefore an acceleration can only be applied to an object if it is in surface contact with something. Field forces such as gravitational, electric, and magnetic are exceptions to this. But in terms of kinematics and motion, if the object is in contact, force can be applied, but as soon as it leaves the surface, that force can no longer be applied. A simple example is our sense of weight actually comes from the normal force of the ground pushing up against our gravitational force. Jump in the air and you feel 'weightless' because the normal force disappears because you aren't in contact with the ground. Hopefully this has helped clear a few things up :)
Hi! I'm just doing practice questions from a book and its giving me different formulas for max height and time it takes for the ball to reach max height. The formulas involve the sin(angle of trajectory).. How come the formulas you use dont involve any trigonometry and these do? I needless to say prefer the formulas you use in the video but i get different answers using each
Hey Rosie, yep so when working with projectiles you always need to work with the vertical component of velocity to determine the max height and the time interval to get there. If you're just given the initial launch speed of the object (at an angle to the ground) then you would need to use trigonometry to solve the vertical component of this initial velocity. ie. Vertical initial velocity = launch velocity x sin(theta) where theta is the angle against the horizontal. The reason why I don't use this in the video is because knowing exact values of sine, cosine, and tan is not required for GAMSAT (at least from what I can see) so instead I've given the vertical velocity as 30m/s so that we can just dive into the rest of the calculations and bypass the trig. If you were to use 30m/s as just the angled launch velocity, then apply sine to it, you'd get a smaller value and therefore an incorrect answer. Technical trigonometry is actually not really required, and I would argue even knowing SOHCAHTOA etc is not entirely relevant (but is helpful) as the stem will always bypass the need to actually use it by just saying let sin(theta) = alpha or something similar. A good example of this is the ACER sample questions at around Q44 with the physics question about a man's spine. You'll see in the notes they give the horizontal and vertical components of a force as simplified expressions and bypass the trig.
Time stamp 20:23 is the distance travelled to max height 3 times as much as the equation can be simplified and 6 can be divided by 2, or is my math just shit?
Hey Tay! Mathematically there is a 6/2 to give 3 but the reason why I leave this out and keep the 2 separate is because I want to draw out the factor change in the distance In simpler terms, since the distance on Earth is -u^2/2g (the green writing), the 2 is actually part of the earth distance travelled. This is then multiplied by 6 for the moon calculation If we were to simplify the 6/2 to 3 we would have distance on moon as 3 x -u^2/g but -u^2/g without the 2 is not the distance travelled on Earth So your maths is good but we actually want to keep the expression for distance on earth , with the 2 included, isolated from the x6 You could also use some algebra and earlier, say let d = -u^2/2g, that way we can see that distance on the moon = 6 x d as this would effectively hide the 2 and prevent us from simplifying it to 3 Hopefully this clears things up :)
I know these videos are from last year but I am finding them sooo helpful prepping for gamsats in september! Also on torque, it's used heaps in musc-skeletal biomechanics. Not sure if gamsats would actually do a question on that but it would be a super cool real-world application and if my march sitting was anything to go by they do some real random questions haha
Jesse is by far the best tutor I've ever come across. Thanks so much for these awesome videos.
You're very welcome, Nick. I hope the study has been going well :)
Loving these crash courses! You’re an excellent teacher😊
Thanks so much! Glad they’re helping, Sophia!
Really appreciate your videos Jesse, so helpful for even for those with a science background! 💐
Thanks so much for these videos. I’ve realised for the GAMSAT questions what I am struggling with most is the maths due to my mental arithmetic not being hot. I am hoping with practice I can improve it for MARCH.
It's definitely a skill to work on but can be done if you regularly practice regularly. Good luck!
Physics is such a huge topic, but these small videos focusing on what is important specifically for the gamy is so helpful!! Thanks so much :D
Also Jesse, just a question, do you think it would be wise to memorise the trigonometry table for sin, cos , tan?
This is exactly what I had hoped to achieve for people with these so this is great to hear! With trigonometry, I'd say that knowing sin(30)=1/2 and cos(60)=1/2 is useful for applying some shortcuts in certain questions but from what I can see, it isn't necessary to know the trig table. ACER questions often simplify the question by saying things like "let sin(theta) = alpha" and then you just use substitution and the question focuses on the algebra rather than the memorisation of random ratios.
If you did want to memorise them, there's this really handy trick if you weren't already aware: m.th-cam.com/video/jI81WXyFrL0/w-d-xo.html
The Des stuff does require the use of trig ratios but they're super outdated.
@@jesseosbourne Perfect this helps a bunch!
you are actually so amazing, thank you so much for all of your help. i've watched pretty much all of your videos ad you've made me feel more confident. i almost spent my life savings on a gamsat prep course but then you saved me. this will be my first time sitting the gamsat, so thank you for alleviating some of my anxieties.
would you say that it is true that the gamsat questions give you all the info you need in the stem of the question to answer it and it is just a matter of reasoning?
Glad to hear you've been able to save your money! It's a difficult question to answer perfectly because everyone has a subjective experience and view on this because of variations in experience and familiarity with different concepts. Most stems are designed such that you can work with the given information, but they may still require some familiarity with the concepts in order to process and apply the information within the expected time frame. For example, if you were given a stem about a type of chemical reaction, chances are you've never heard of it and wouldn't need to have known about it beforehand. BUT, within the stem, there might be words/phrases used that if you had no experience with chemistry might be confusing or difficult to grasp.
Probably not a perfect answer to your question, but the best thing to do is learn by doing and if you find you're struggling to grasp what the stem in a practice question is saying, review the core principles behind it and keep moving
Best of luck!
Thank you so much Jesse for your incredible videos.🙏
Must be nice to just know all this stuff, this is like the third time I’ve come back to these vids and I have a science background haha
Trust me, if I hadn't been a high school tutor this whole time this'd look like another language to me by now 😂
so helpful, once again! Thank you Jesse!!
Awesome, no worries, Nina! :)
you're the best Jesse! thank you!
You're very welcome! :)
Thanks so much for these videos, Jesse. They are proving really helpful in my gamsat study so far. Quick question in regards to torque, is the perpendicular radius perpendicular to the force that is applied? Thanks in advance.
Hey Lorenzo! Yep, that right. I think in the video I say perpendicular radius but obviously you could also think of the force as being perpendicular to the radius too.
There are shortcut formulas that calculate it all with an angle included such as FdCos(ø) or FdSin(ø) where F is some force applied at an angle, ø to the object at a distance d from the pivot. The FCos(ø) or FSin(ø) component is just the trig working out the perpendicular component of the force and then multiplying it by 'd' to get the torque. You don't need to know this for GASMAT as trigonometry calculations are often simplified by ACER in the stems anyway, but figured you might come across this and it could also be helpful to see how it reinforces the perpendicular force and radius assumptions.
@@jesseosbourne Thanks Jesse. That definitely helps. All the best for your gamsat.
You're awesome at explaining things :)
Thanks Ruby!
Great video! Are you writing the notes on an ipad and casting it to your monitor?
Hey Tom, I actually use a Wacom intuos sketchpad that I connect to the computer
Another great video! These are proving to be a huge help!! I am taking the GAMSAT in March and I'm wanting to get into Sydney uni in 2023. I might see you there if that's where you decide to go 🙂
Hopefully all works out well for the both of us, then! Awesome to hear that the videos are helping. Best of luck in the prep for March! :)
Hey Jesse, great video as always! I'm just a bit confused with equations of motion, when would s=ut+1/2at2 be used in comparison to v2=u2+2as? Is it just if we didn't know tfinal, or have I confused this? thanks!
Hi Jesse!
I was looking at some “physics work” definitions and I got a bit confused.
Definition 1- makes sense
The way I think about “work in physics” is simply as “work is done when a force is applied to an object, which displaces or moves that object”… THAT’S IT. (i.e work just centres around displacement/movement). Is this sufficient detail or are there gaps in this definition?
Definition 2- confusing
“work is done when a force causes an object to move IN THE DIRECTION OF THE FORCE”.
The caps lock part is confusing! Because I’ve seen a scenario where you can apply a force, and the object ends up moving/displacing in a “different direction” to the force. Then they use W = F x d x cos(theta) as the formula. See what I mean? The caps lock part doesn’t make sense to me. Can we just think of work as moving an object?
By the way, I love your efforts and videos! Keep them up! ❤❤Congratulations for getting into med school! (I have to still watch those videos as I haven’t gotten around to them yet 😂🎉).
Hey Jessica, okay so your definition should be fine for 99% of the applications of work. Work is a measure of the energy introduced to a system via a force acting over some distance.
The direction thing is referring to the force component that is directed in the same direction as the movement of the object. In most cases, the force applied will be directed in the same direction as the resulting movement of the object. eg. If you push horizontally on the back of a broken down car, it moves forward horizontally.
However, it is also possible to apply an oblique force which results in a horizontal or vertical force. In this case, force applied (at an angle) is not in the same direction as the movement. eg. Pushing on the handle of a shopping cart might actually involve pushing down and forward (so down at an angle) but because the floor is horizontal, the shopping cart moves horizontally.
In these scenarios, not all of the applied force is doing work on the system so it would be wrong to simply multiply force applied x distance travelled. Instead, we want to calculate the horizontal component of that oblique force which involves using trigonometry (sin or cos of the angle against the vertical or horizontal, respectively. cos of the angle against the horizontal is more commonly used). So the formula W = F x d x cos(theta) is better expressed as W = F x cos(theta) x d where F x cos(theta) represents the force component that is in the same direction as the objects movement.
A little tricky to explain in just words (I wish YT comments allowed images) but hopefully this clears things up!
Ohhh so even if we get a force and movement in “different directions”, when we do W = f x cos(theta) x d…. the f x cos(theta) part will try to just get the force and movement in the same direction via trigonometry.
So both:
W = f x cos(theta) x d
W = f x d
Will end up with the “same direction” of force and movement in the end.
Hey Jesse, in this video you said not to worry about the formulae because it will be given in the stem, however on your notion page it says that you should memorise formulae such as kinetic, work and gravitational energy, along with torque. And to save another comment the parallel and series effective resistance formulae also fall under ones you should know. Just wondering whether you can confirm whether it's necessary to remember these.
Cheers.
Hey Mitchell, I mention that "most" formulas are given in the stem but also say for this crash course not to worry about memorising formulas as I didn't want people to be watching and getting hung up on trying to scribble down and rote learn any formulas they see. Just putting some context to the intentions of the video.
The formulas in the cheat sheet that I've placed under "need to know" would be those that aren't most formulas. The number of equations in that list though are minuscule relative to the number of formulas that might pop up in a 'physics based question' and they are largely limited to three or less variables and underpin the foundations of physics concepts needed for basic reasoning.
I've put the parallel and series circuit equations under "need to know" because although I don't think it's likely you would use them directly in a question (this would just be routine calculation rather than reasoning), knowing them helps explain the nature and assumptions around the two circuit types eg. Ohm's law, voltage drop, and Kirchoff's law. Better to know them than to not know them, in my opinion :)
@@jesseosbourne Thanks for the response Jesse, as always. I was just working through the practice question booklet and the question on deceleration of blood flow tripped me up, in the worked example they use the v2 = u2 + 2as formula but that wasn't given in the stem. I understand these booklets are quite old and probably reflect the questions as to when the exam was more theory based. Would you recommend memorising the kinematics formulae or do you think with the newer formatted exams they would give you that equation to work with?
@@mitchellgourlay4847 Yeah I'd still say that the kinematics formulas are worth knowing. The trend tends to be that you are given 2 or 3 of the five which would mean a 2 or 3 step process of using simultaneous equations but knowing all 5 means you can shortcut straight to the one formula that will allow you to solve the missing variable in the one step.
The five to know are:
v=u+at
v^2 = u^2 + 2as
s = ut + (1/2)at^2
s = vt - (1/2)at^2
s = (u+v)t/2
Keep in mind that these formulas can only be applied over intervals where acceleration is constant
@@jesseosbourne Thank you Jesse, really appreciate your help. Hopefully all this hard work pays off come March. Happy new year brother, I hope all is well as you are preparing for medical school. You're gonna ace it
As the horizontal velocity component of the ball at max height is still greater than zero (as it remains constant throughout the projectile motion), wouldn't this make the kinetic energy more than 0 at max height? Or should we just consider the vertical component when calculating the energy?
Hey Patel! Sorry for the wait on a reply until I returned from the break. This is a great question. Technically, the ball does still have a non-zero kinetic energy at the top of the arc due to the horizontal velocity. In the question though, we are only considering the vertical components of motion and therefore only the vertical components' contribution to the energy. This is why it's important to determine the vertical component of its initial velocity in determining kinetic energy rather than the launch speed (at an oblique angle). Hopefully this clears things up :)
Hey Jesse, I'm just wondering why you put g as a negative when you were calculating for t on the earth at 17:32 because I thought gravity would be positive if accelerating downwards? Thanks :)
Edit: Nevermind! Just realised you were calculating for when the ball would reach the max height so it would be going against gravity hence the negative!
Haha yep you got it!
Any advice for easy resources I can access to apply this knowledge. I say easy because I still feel very lost, I might understand one example, then I'm completely lost of the next one. HELP :(
A good site is libretexts. You can work through specific topics and there's usually some 'exercise' type questions like you'd see in regular textbooks. Be mindful that a lot of physics theory questions can be a little tedious with the calculations because they assume access to a calculator and are generally assessing technical skills a lot more often than conceptual skills as we get in GAMSAT. That being said, it's a good go to.
Some other options are just to google random worksheets by searching 'topic name, physics, worksheet/questions, pdf' - this usually returns some good results
If you haven't already, check out my notion page (linked in the video description) where I've made a bank of GAMSAT style questions with solutions videos. Although admittedly there may not be many on this specific topic
I really hope you get into med school 😊
Hi! Thanks for the videos!! Just wondering why when calculating the max height of the ball trajectory off the raised platform, why is gravity the only form of acceleration accounted for? Are we assuming that the kick/throw has no force?
Hey Rosie, Glad they're helping!
It is a little counterintuitive at first, given the ball is still moving upwards on it's upward journey, however there are a few ways we can resolve this.
The first thing is that we're applying Newton's second law which is often misquoted as F = ma but it really is NET FORCE = ma. So the acceleration is in the direction of the net force or sum of all the forces combined.
1. Consider the ball in flight as a point and draw it as a force diagram. When in flight the only force acting on the ball is the gravitational force acting downwards. We ignore any air resistance etc in these scenarios. Therefore the net force acting on the ball is equal to the downward gravitational force, due to gravity's acceleration. You'll notice that on the upward phase I use gravity as a negative value because while the ball is going up, gravity is acting to decelerate the ball and slow it down.
2. If you're still wondering about how the ball doesn't have a force and acceleration from the kick, we can look at what happens at the time the ball leaves the foot of the person kicking it (or the hand of the person throwing it) to explain this. While the ball is contact with someone's foot, the foot can be applying a force to the ball and this would accelerate the ball upwards against gravity but as soon as the ball leaves the foot, the foot can no longer apply this force, so the ball is now in 'free fall' ie. only under the effects of gravity.
So the kick does still apply a force but only while in contact with the ball. This is what accelerates the ball from zero up to it's initial launch velocity (represented by u in the calculations). Our calculations however are only concerned with the ball from the moment it leaves the foot to the moment it first strikes the ground again. During this phase the foot is no longer providing an acceleration and so the only acceleration is that of gravity.
An analogous scenario would be to think of a car accelerating down a straight road. While the accelerator pedal is being pressed, the engine applies a force to accelerate the car forward and speed it up but as soon as you let your foot off the accelerator, the car is now no longer under that acceleration and the only forces still acting on it would be friction at the tyres and some air resistance all pushing in the opposite direction to the cars motion. The car would still roll forward but it would slow down as it is under a deceleration from those resistive forces.
Going back to the net force = ma thing, when the car is accelerating, the driving force of the engine is likely much greater than that of the resistive air resistance and frictional forces so the net force of the car is forward and so is it's acceleration. When the driving force is removed (no accelerator) then the car is coasting along and the only forces are the resistive forces making up the net force and so the acceleration is now entirely due to them a = net force/mass of car) and so it will begin to slow down until it stops.
So that you'v got a broader understanding of how to apply this reasoning in other scenarios and questions, generally speaking, a force and therefore an acceleration can only be applied to an object if it is in surface contact with something. Field forces such as gravitational, electric, and magnetic are exceptions to this. But in terms of kinematics and motion, if the object is in contact, force can be applied, but as soon as it leaves the surface, that force can no longer be applied. A simple example is our sense of weight actually comes from the normal force of the ground pushing up against our gravitational force. Jump in the air and you feel 'weightless' because the normal force disappears because you aren't in contact with the ground.
Hopefully this has helped clear a few things up :)
@@jesseosbourne Thank you so much!! this was so helpful and its much clearer now :)
Hi! I'm just doing practice questions from a book and its giving me different formulas for max height and time it takes for the ball to reach max height. The formulas involve the sin(angle of trajectory).. How come the formulas you use dont involve any trigonometry and these do? I needless to say prefer the formulas you use in the video but i get different answers using each
Hey Rosie, yep so when working with projectiles you always need to work with the vertical component of velocity to determine the max height and the time interval to get there. If you're just given the initial launch speed of the object (at an angle to the ground) then you would need to use trigonometry to solve the vertical component of this initial velocity. ie. Vertical initial velocity = launch velocity x sin(theta) where theta is the angle against the horizontal.
The reason why I don't use this in the video is because knowing exact values of sine, cosine, and tan is not required for GAMSAT (at least from what I can see) so instead I've given the vertical velocity as 30m/s so that we can just dive into the rest of the calculations and bypass the trig.
If you were to use 30m/s as just the angled launch velocity, then apply sine to it, you'd get a smaller value and therefore an incorrect answer.
Technical trigonometry is actually not really required, and I would argue even knowing SOHCAHTOA etc is not entirely relevant (but is helpful) as the stem will always bypass the need to actually use it by just saying let sin(theta) = alpha or something similar. A good example of this is the ACER sample questions at around Q44 with the physics question about a man's spine. You'll see in the notes they give the horizontal and vertical components of a force as simplified expressions and bypass the trig.
@@jesseosbourne thank you so so much!! that makes much more sense :)
Time stamp 20:23 is the distance travelled to max height 3 times as much as the equation can be simplified and 6 can be divided by 2, or is my math just shit?
Hey Tay! Mathematically there is a 6/2 to give 3 but the reason why I leave this out and keep the 2 separate is because I want to draw out the factor change in the distance
In simpler terms, since the distance on Earth is -u^2/2g (the green writing), the 2 is actually part of the earth distance travelled. This is then multiplied by 6 for the moon calculation
If we were to simplify the 6/2 to 3 we would have distance on moon as 3 x -u^2/g but -u^2/g without the 2 is not the distance travelled on Earth
So your maths is good but we actually want to keep the expression for distance on earth , with the 2 included, isolated from the x6
You could also use some algebra and earlier, say let d = -u^2/2g, that way we can see that distance on the moon = 6 x d as this would effectively hide the 2 and prevent us from simplifying it to 3
Hopefully this clears things up :)