Hi Jesse! For the stretcher problem, how come we say 17/20 of the weight pushes towards B and 3/20 of the weight pushes toward A? The 17/20 of the weight is above point A, vice versa. Could you please clarify this?
Hey Acelya! Yep so at first it does seem a little counter intuitive that the side that has the larger length has the smaller force but this is because we consider the entire weight of the object (the patient and the bedding in this case) as a point with its force acting at the point of centre of mass. So in this case, because the centre of mass sits 1.7m from support A, it is 17/20ths of the way towards Support B meaning that 17/20ths of the weight force is acting through support B. To better understand this principle, think about the entire weight force acting directly over support A. All the weight force would be acting through support A. But if we began to shift this weight force along the length of the bed, the relative amount of force through A would start to decrease and instead would begin to increase over at support B as the force gets closer. At the halfway point, the force would be split 50-50 through A and B and then as it gets closer to B, the force would continue to increase at B and decrease at A, despite the fact that the length to point A is much larger now. As a simple summary though, we can just apply the force by proportion of the length on the alternate side to the support. Hopefully this clears things up! :)
Hi @@jesseosbourne, I am still a little confused about why there is more force on the leg that is closer to the centre of mass. I think my confusion lies with the idea that if the bed is stationary than the forces have to be in equilibrium, so I don't understand how having more force on one leg than the other keeps the bed stationary. Thank you!
Hey Elysia! I can see what you're saying. I'll come back to the equilibrium bit in a second. First, we can get a different understanding of the weight through each leg pf the stretcher by instead thinking about placing each leg on a set of scales. If we had an empty bed and placed it on the scales, we should expect half the weight to press down on one leg and half on the other. If though, we were to have a patient stand in the exact middle of the bed, then the weight force through each leg would increase due to their weight. Although, the weight shown on each scale would still be the same as it is equally balanced. Now, what if the patient stepped a little closer to the right side of the bed. Would the scales show different values? We'd expect that the right leg scale would now take on a greater value (greater proportion of the total weight of the bed and patient). If the patient keeps shuffling closer to the right side, that proportion would increase. Does that make sense? Now, in terms of equilibrium, we can still achieve equilibrium in this state if we also consider the normal force applied by the ground pushing back up against the legs of the stretcher. If we had 1000N of force, but 800N was pressing down through the right leg and only 200 down through the left leg then the normal force applied as a reaction force by the ground (upwards) would be 800N and 200N respectively. So although the weight through each leg is not equal, they are are being cancelled out or counterbalanced by the reaction normal force of the ground pushing upwards, therefore the bed won't move. If the ground were not there and it were floating above the ground, then it would not be in equilibrium and it would instead be spinning as a result of the unbalanced forces on either end because the ground is not there to provide a normal force. Hopefully that clears a few things up for you :)
Hi Jesse. Just want to thank you for creating these tutorials! I'm a biochem grad sitting the GAMSAT for the first time this year in The UK. I can't afford to/don't want to sit it twice and was so extremely intimidated by the thought of tackling physics questions. Really appreciate your time and effort in talking us through these concepts and their applications. Thank you!
"Jesse working hard to educate and inspire future med students" "Doggo in the background also wanting to contribute and demonstrates an example of force and friction by scratching up the carpet". All being said you are a true life saver and miracle worker Jesse! Thank you!
Jesse I must begin by saying that you the most wholesome and real person in this gamsat space. And these services you provide are simply invaluable in a time when company prey on vulnerable students with exorbitant prices. I would say that you explain things with such alacrity and your resources are so good. I can see that you truly want to equalise the field and have a great heart. You will be a phenomenal doctor mate amongst many other things
Thanks so much for your comment! I’m Really glad that people are able to take so much from these videos and avoid falling into the trap of paying for courses
Great video Jesse, Thanks. Filling up gaps in my knowledge not through rote learning concepts, instead with Jesse's videos where he makes you understand the concepts.
Thank you so much for this video! It was the first time I FINALLY understood these concepts! Coming from a non-science background has been super overwhelming but this video has made me feel as though I can overcome the GAMSAT - so THANK YOU!!!!
Hi Jesse! Thanks for your guidance! I really struggle with question types alike to the final one on this video. How do you practice developing this type of skill set? I understand as I watch you, but feel this type of thinking is what is needed to unlock the gamsat. How can I work on this type of logic/rationale?
Hi Jesse!! Randomly read a comment on reddit that your crash courses were amazing, but turns out that it's MORE than amazing!! Thank you sooo much for all the valuable resources you are providing!! Really appreciate it. I am planning to finish all three crash courses and move on to more of your other video playlists as well. Wish I could know you earlier but well, glad to know you now :) Good luck on the March exam coming soon!
Hey Jenn! Super happy to hear that the videos are already making an impact. Yep, I'd say that working through the crash courses as a starting point is a good plan if you are in the early stages of your study. Will you be sitting in September?
Yes! and the March one this week too haha😂 Not prepared much tbh, but think it’s better to take the exam than not to. Gaining one more real-exam experience!
I have been watching your videos on section 3 crash courses and solving bunch of questions related to each topic. You are doing us a huge favour. Thank you so much.
hey Jesse! thank you for the videos and the extra explanations in the comments section, makes so much sense! Just wondering if you have ever done math olympia or things like that? you are soooo smart!!!
Nope, never done any kind of competition stuff. It sounds very cliche but I've never liked the idea of education/knowledge becoming a competition, I see it as a communal resource. I did always love a good 'times tables race' in primary school though haha
Hi again Jesse, physics is somewhat alien to me, particularly with vectors. In your first example we wanted to find an unknown side, so we could use pythagorus theorem, for the second example I'm not quite sure what we were trying to find. You mentioned at the end about two tennis balls colliding, so perhaps it was the impact of two vectors colliding together? and for the 3rd example we wanted to know the net direction and movement of the box. Is there a set of skills that would be helpful to know in all the vector kind of questions? e.g. pythagorus theorum, SOHCAHTOA etc. Physics and math concepts seem to overlap quite a bit, and I'm a bit rusty on math concepts too haha
Yep that's right! So I'm effectively explaining how to add two vectors together and using the tennis ball example as an analogy to give a more conceptual understanding of what is happening. Adding vectors in physics involves breaking them down into horizontal and vertical components first and then summing these together to get the net vector. If you don't need to be accurate in the question, you can avoid the calculation by gaining a rough understanding of the sum of the vectors by using this kind of analogy to understand how they would interact to save time. Using pythagoras' theorem is definitely helpful in some GAMSAT questions. SOHCAHTOA is useful for reasoning (ie. knowing that sine is O/H etc) but they don't expect you to be able to calculate using angles. They usually either give estimates of say sin(20º) or they avoid it entirely
Glad it helped! So in that scenario I'm setting up the condition that the gravitational fields are equal at some point between the two planetary bodies (g1=g2) but at all other distances other than this they would not be equal. It is a real situation in that two objects can have a point between them at which the gravitational field strengths are equal and opposite
just a question on patient on the bed, it makes sense most the weight will be pushed down at point B and less weight pushed down at point A , but I'm confused as to why the weight comes from opposite sides. So why does the weight force go diagonally? It makes sense to go linearly down.
Hey Batu, yep this is a pretty common confusion actually. The best way to think about it is the relative position of the centre of mass from each of the legs. The closer it is, the more weight will push down through that closer leg/support and vice versa. If you check out the pinned comment and my replies I go into a bit more detail and a conceptual way of thinking about it :)
Hi Jesse :) For the bed problem, would it be possible to do: F = m x a Mass for leg A = 90kg/20 x 3 = 13.5kg F = 13.5kg x 10N (gravity) to get 135N Then complete the process with the alternate leg? Thanks so much
Hey Jane, yep absolutely. Effectively this would just be splitting the masses by proportion before multiplying by the gravity but would still be equivalent As you’ve simply factorised out the gravity and just dealt with the mass by proportion. The proof would be: If we let p represent the proportion (3/20) to make it look a bit nicer in text form Force A = mgp = g x (mp) A fancy way to reiterate that when multiplying, order doesn’t matter :)
Hi Jesse, another smashing content 😊 Hey, as you breadth of experience in tutoring year 12's wondering if you have any recommnedation for a level 12 physics book. PS. waiting for the video regarding your reaction about the recent Gamsat. 😊 I sat it in Southbank Melb and all I can say is I miss my bottle of water. I thought it will be allowed in our table, but no 😭 my throat was parched like a desert 🏜
Hey Disha! Yeah I'd say the Heinemann Physics 12 (4th ed) textbook is probably the best one www.pearson.com/store/p/heinemann-physics-12-ebook/GPROG_A103000288118_learnerau-availability/9781488663604 Chapters 1, 2, 5, 7, 8 and a bit of 9 is very relevant to GAMSAT physics This is the textbook I use with my VCE Physics students in tutes too. Yeah the water bottle rule is crazy but I guess it's probably there so people aren't causing distractions constantly fiddling with lids and making noise. I just didn't take my water in with me this time haha
18:42 if a car in motion with a mass of 1000kg hits you and in doing so decelerates at 3m/s^2 it would then apply 3000N over that period of deceleration. I think that's correct. That's how I think of it.
Hey Ryan! Sorry about the wait on a reply while I was on my break. So that's almost it, however we need to consider the force and acceleration as vectors. If the acceleration is a deceleration then the acceleration is directed in the opposite direction to it's motion and the force it exerts. This would instead represent the person being hit as exerting 3000N of force on the car, theoretically. This could also be proven mathematically from Newton's second law, F = ma If we take the forward direction of the car as the + direction and the opposite direction as the - direction. m is a scalar quantity and so the direction of the Force vector must be determined by the direction of the acceleration vector If decelerating, a is negative and so F is negative (opposite direction to its travel) In this example, the car is accelerating rather than decelerating so its acceleration and therefore force are directed in the same direction (+ direction) and F = ma = 1000 x 3 = 3000N exerted by the car on the person
@@jesseosbourne Hey Jesse hope you had a good break! Hmmm I'll have to give this more thought and do a few practice problems. I see your point about the person exerting 3000N of force on the car to decelerate it and that is how I would approach these types of questions. I agree that if the car were at rest with the bumper touching the person and then the car accelerated at 3m/s^2 it would apply 3000N of force. I also agree that if the velocity of the person and the velocity of the car (when the bumper first touched the person) were equal and the car was accelerating at 3m/s^2 it would apply 3000N of force. Although in both cases I think the Force would actually be m(car) + m(person) and therefore greater than 3000N but you're probably just considering the mass of the person as negligible (or I could be wrong). Since instead of a person if the car hit a large boulder it would require more force to accelerate it at 3m/s^2. However, if the car were travelling at a non-zero velocity (when the bumper first touched the person) and hit a person (at rest) the acceleration the person would experience would be much greater than 3m/s^2 or the car would decelerate/accelerate more slowly until the velocity of the person matched the velocity of the car. Then the car would continue to apply a 3000N force if it kept accelerating at 3m/s^2. Haha sorry that is a bit of a complicated response. What are your thoughts sir? I may also review momentum. Do you have any good examples of these types of questions which involve momentum and transfer of momentum and how that affects force/acceleration? Oh, something I heard that might interest you. Although I'm not sure how valid it is. I heard that USyd is calculating S1 x 1.25, S2 x 1.25 and S3 x1 with a place offered for total scores >258 so you should be good. ;)
So what this comes down to is separating the collision into two phases. The before the collision phase and the after collision phase. Momentum and energy of the entire system, should be conserved and so velocities of the person and the car will change to adjust for this. This question only considers the force applied at the instant of collision but not after the collision. At the instant of collision, the car is accelerating at 3m/s^2 with a mass of 1000kg and so exerts a 3000N force on the stationary person. In effect there has not yet been a time interval to allow for any deceleration of the car or acceleration of the person. So because we are not comparing the before and after phases, we don't have to look at it as a conservation of energy or momentum question. If however, we do consider what happens after the collision for discussion's sake, the car would technically reduce its velocity as result of some of its kinetic energy being transferred to the person but as you correctly noted, this would be negligible given the difference in mass. The other thing that we would have to assume is that the driver takes their foot off the accelerator at the moment of collision, otherwise they would be continuing to introduce additional energy and therefore momentum into the system (by burning fuel in the engine) which would otherwise allow for the car to continue accelerating at 3m/s^2. This would then mean that the laws of conservation can't be applied. For practice, you could find some free online worksheets by searching 'conservation of energy & momentum worksheet pdf'. Keep in mind that being non-GAMSAT, they may be a little more technical in nature so I'd recommend either using a calculator on things like sine/cosine etc or practising estimating to see how close you can get to the real answer. So long as you can come away with an understanding of how to 'think' about the problems and recognise when it applies, you're good! Here's an example worksheet: www.smcisd.net/cms/lib/TX02215324/Centricity/Domain/1073/unit%20homework%20momentum%20its%20conservation%202016.pdf www.smcisd.net/cms/lib/TX02215324/Centricity/Domain/1073/unit%20homework%20momentum%20its%20conservation%20ans%20key.pdf Thanks, yes I have seen this formula floating around for USyd which brings me a lot of confidence in my previous scores. After researching for the USyd application video though, there's no mention of an algorithm being used and it looks like its a case of section score rankings with equal weightings. The 1.25 weighting appears to have just been correlating with scores for 2022 entry. Hopefully nothing too dramatic changes for this round of applications!
31:40 for the downwards lift, why is the 2 taken away from the 10? I'd have thought that you'd add the downwards forces to get a net acceleration of 12? I know your answer is right, I just can't get it to make sense in my head 😅🫠
Yeah this is definitely one of those seemingly counterintuitive scenarios in physics. It comes down to the fact that there is no additional external force that is adding onto the gravitational force. More so, by accelerating downward in the same direction as gravity we are almost 'running away from gravity' and so it's effect on us as the moving body is less. A similar example in a horizontal plane would be a car being towed by a rope. If the tow truck is pulling with 1000N of force then this is the force experienced by the car being towed (assuming no friction etc) but if the towed car begins to move forward on its own a bit, say via 200N of force) then this removes some of the tension in the rope and the net force acting not he towed car is 800N. As a little interesting aside (although not required for GAMSAT) this assumption of force, acceleration and velocity interaction is limited to Newtonian (Classical) Physics - which is what we use in GAMSAT In Einstein's theory of special relativity we can still use this concept for objects but for light we have to make the assumption that light speed is a constant and cannot exceed it's value of 3x10^8 m/s even if classical physics would suggest otherwise in circumstances such as the speed of light from a headlight on a car travelling forwards at 50m/s
@@jesseosbourne thank you so much!!!! That tow truck analogy makes so much sense. You are an absolute lifesaver 🙏🙏🙏 physics is my worst enemy lol 😂 well, maybe after organic chem ... 🤪
can somebody clarify on why the net acceleration is 8 m/s2 when both the gravity and acceleration of the lift are going downwards at 31:18? thank you in advance
Hey William, yeah I'd say they're useful. It's common for them to be provided but not all of them so knowing all five allows you to get to the answer in one step rather than multiple. I have a full list of equations on the 'formula sheet' on my resources page, linked in the video description
Hi Avneet, so here I'm using vector summation. By taking the net horizontal and net vertical forces as vector arrows and placing them 'nose to tail', you can make a simple right angle triangle where the hypotenuse is the direction of the net force. Because the net force has a rightward and upward component, the net force should be pointing in the direction of up and to the right. The actual value of the net force can be calculated using pythagoras' theorem (relatively unlikely in GAMSAT) and the direction could be explained based on the angle that the net force makes with the horizontal solved using trigonometry (not examinable in GAMSAT)
Hi Martine, gravity on Earth is 9.8 N/kg (9.8 m/s^2) but is often simplified to 10 for the sake of estimation. You can always assume g=10 for any calculations involving gravity on the surface of the Earth
Hey Lani, the good news is that you'd never be expected to estimate sin/cos/tan of an angle without some additional given information as there would be too high a degree of variability in people's estimates. Generally the stem glosses over the trigonometry and just says stuff like "let sin(theta) = alpha" and then it becomes an algebra question where you substitute alpha into your expression and continue But, what I would do if I felt I needed to estimate something like this is I would use some outside knowledge that sin(30º) = 1/2 and cos(60º) = 1/2. I could use this as a reference point and then consider a right angle triangle where the angle either increases or decreases from this angle and think about what would happen to the ratio of the sides of the triangle. For example, say we had a right angle triangle and the angle against the horizontal was 40º. If I wanted to know what sin(40) was, I could start by considering sin(30) = 1/2 and then rotating the hypotenuse of the triangle upwards so that the angle increased to to 40º would mean that the opposite side (vertical height) would have to increase and the adjacent side (horizontal) would decrease. Because sin is O/H and the opposite side increased then sin(40) > sin(30) therefore sin(40) > 1/2 so maybe 3/4 as an estimate. The same could be done for cos and tan ratios Another alternative if you were already familiar with it is to think about the basic sin(x), cos(x), tan(x) curves and use reference points from there. Again though, it could be useful to think about this to confirm your reasoning but I wouldn't think it's a required skill for GAMSAT
Hi Jesse, can I ask why the W=F*g=m*g around 42:36? Because F=mg, so if F*g is not possible to qual to m*g, and weight is also not qual to F*g, can you explain it? Thanks in advance
Sorry, that's my handwriting causing trouble there! I'm writing it as g in subscript (F_g) to denote gravitational force but it looks a lot like Fg. So Weight (W) = Gravitational Force (F_g) = mg
Hey Simon, yep so Fg is gravitational force (sometimes referred to as weight force) and this is calculated as mass x gravity. Given that for GAMSAT we always assume gravity to be 10 N/kg rather than the more specific 9.8, all the Fg calculations involve multiplying mass x 10 to calculate the resultant force
Hey Kent! Sorry about the wait on a reply while I was on my break. I use a browser based whiteboard called ziteboard. I started using it at the start of the lockdowns when I had to move all of my highschool tutorials online and it's held up well. ziteboard.com
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Hi Jesse! For the stretcher problem, how come we say 17/20 of the weight pushes towards B and 3/20 of the weight pushes toward A? The 17/20 of the weight is above point A, vice versa. Could you please clarify this?
Hey Acelya! Yep so at first it does seem a little counter intuitive that the side that has the larger length has the smaller force but this is because we consider the entire weight of the object (the patient and the bedding in this case) as a point with its force acting at the point of centre of mass.
So in this case, because the centre of mass sits 1.7m from support A, it is 17/20ths of the way towards Support B meaning that 17/20ths of the weight force is acting through support B.
To better understand this principle, think about the entire weight force acting directly over support A. All the weight force would be acting through support A. But if we began to shift this weight force along the length of the bed, the relative amount of force through A would start to decrease and instead would begin to increase over at support B as the force gets closer. At the halfway point, the force would be split 50-50 through A and B and then as it gets closer to B, the force would continue to increase at B and decrease at A, despite the fact that the length to point A is much larger now.
As a simple summary though, we can just apply the force by proportion of the length on the alternate side to the support. Hopefully this clears things up! :)
@@jesseosbourne Thank you so much! It does make a lot of sense now!
Hi @@jesseosbourne, I am still a little confused about why there is more force on the leg that is closer to the centre of mass. I think my confusion lies with the idea that if the bed is stationary than the forces have to be in equilibrium, so I don't understand how having more force on one leg than the other keeps the bed stationary. Thank you!
Hey Elysia! I can see what you're saying. I'll come back to the equilibrium bit in a second.
First, we can get a different understanding of the weight through each leg pf the stretcher by instead thinking about placing each leg on a set of scales. If we had an empty bed and placed it on the scales, we should expect half the weight to press down on one leg and half on the other. If though, we were to have a patient stand in the exact middle of the bed, then the weight force through each leg would increase due to their weight. Although, the weight shown on each scale would still be the same as it is equally balanced.
Now, what if the patient stepped a little closer to the right side of the bed. Would the scales show different values? We'd expect that the right leg scale would now take on a greater value (greater proportion of the total weight of the bed and patient). If the patient keeps shuffling closer to the right side, that proportion would increase. Does that make sense?
Now, in terms of equilibrium, we can still achieve equilibrium in this state if we also consider the normal force applied by the ground pushing back up against the legs of the stretcher. If we had 1000N of force, but 800N was pressing down through the right leg and only 200 down through the left leg then the normal force applied as a reaction force by the ground (upwards) would be 800N and 200N respectively. So although the weight through each leg is not equal, they are are being cancelled out or counterbalanced by the reaction normal force of the ground pushing upwards, therefore the bed won't move.
If the ground were not there and it were floating above the ground, then it would not be in equilibrium and it would instead be spinning as a result of the unbalanced forces on either end because the ground is not there to provide a normal force.
Hopefully that clears a few things up for you :)
@@jesseosbourne Thank you so much! Your videos are helping me navigate my first GAMSAT and are helping me to feel it is possible to get through it :)
Finding this channel was like stepping on a goldmine, thank you so much for all of these videos Jesse!!!
Fantastic stuff! You're very welcome :)
Hi Jesse. Just want to thank you for creating these tutorials! I'm a biochem grad sitting the GAMSAT for the first time this year in The UK. I can't afford to/don't want to sit it twice and was so extremely intimidated by the thought of tackling physics questions. Really appreciate your time and effort in talking us through these concepts and their applications. Thank you!
"Jesse working hard to educate and inspire future med students" "Doggo in the background also wanting to contribute and demonstrates an example of force and friction by scratching up the carpet". All being said you are a true life saver and miracle worker Jesse! Thank you!
haha I've passed on your comments to let him know that his work is appreciated 😂
Jesse I must begin by saying that you the most wholesome and real person in this gamsat space. And these services you provide are simply invaluable in a time when company prey on vulnerable students with exorbitant prices. I would say that you explain things with such alacrity and your resources are so good. I can see that you truly want to equalise the field and have a great heart. You will be a phenomenal doctor mate amongst many other things
Thanks so much for your comment! I’m
Really glad that people are able to take so much from these videos and avoid falling into the trap of paying for courses
Thank you SO much for providing these!! They seem to take a lot of time and effort but we really do appreciate it them!!
Really glad it's helping! :) Thanks for the support
Mate, you are simply unreal. The value you continue to provide is extraordinary. Greatly looking forward to this series! Keep up the great work.
So good to hear! Thanks for watching :)
Thanks Jesse! Officially the🐐of gamsat prep. sitting first time in september and from nsb so this is truly a goldmine 🤩
Great video Jesse, Thanks. Filling up gaps in my knowledge not through rote learning concepts, instead with Jesse's videos where he makes you understand the concepts.
Sounds like you're on top of it! Great stuff :)
Keep up the great work mate, we need an online community to help us all decipher the GAMSAT and prep efficiently. People like you are paving the way.
Exactly what I'm hoping to achieve! Hopefully more change is coming :) Best of luck with the study!
You explain everything so well, thanks Jesse
Jesse you're an absolute legend
This is my first time sitting the Gammy in March and this was so helpful ! Thank you :)))
Awesome to hear! :)
Thank you so much for this video! It was the first time I FINALLY understood these concepts! Coming from a non-science background has been super overwhelming but this video has made me feel as though I can overcome the GAMSAT - so THANK YOU!!!!
This is fantastic to hear, Jessie! This is the best feedback I could receive so I'm really glad it's been helpful. Best of luck!
Hi Jesse! Thanks for your guidance! I really struggle with question types alike to the final one on this video. How do you practice developing this type of skill set? I understand as I watch you, but feel this type of thinking is what is needed to unlock the gamsat. How can I work on this type of logic/rationale?
Hi Jesse!! Randomly read a comment on reddit that your crash courses were amazing, but turns out that it's MORE than amazing!! Thank you sooo much for all the valuable resources you are providing!! Really appreciate it. I am planning to finish all three crash courses and move on to more of your other video playlists as well. Wish I could know you earlier but well, glad to know you now :) Good luck on the March exam coming soon!
Hey Jenn! Super happy to hear that the videos are already making an impact. Yep, I'd say that working through the crash courses as a starting point is a good plan if you are in the early stages of your study. Will you be sitting in September?
Yes! and the March one this week too haha😂 Not prepared much tbh, but think it’s better to take the exam than not to. Gaining one more real-exam experience!
The best practice you can get is an actual sitting! Good luck for March! :)
Thank you so much for these extraordinary videos, appreciate your time and effort Jesse!!!
You're very welcome! Glad they're helping :)
I have been watching your videos on section 3 crash courses and solving bunch of questions related to each topic. You are doing us a huge favour. Thank you so much.
Great to hear!
Great video to watch for a recap today, thank you!
This is wonderful! Thanks so so much 🙏🏽🙏🏽🙏🏽🙏🏽
Glad it was helpful!
hey Jesse! thank you for the videos and the extra explanations in the comments section, makes so much sense! Just wondering if you have ever done math olympia or things like that? you are soooo smart!!!
Nope, never done any kind of competition stuff. It sounds very cliche but I've never liked the idea of education/knowledge becoming a competition, I see it as a communal resource. I did always love a good 'times tables race' in primary school though haha
Hi again Jesse, physics is somewhat alien to me, particularly with vectors. In your first example we wanted to find an unknown side, so we could use pythagorus theorem, for the second example I'm not quite sure what we were trying to find. You mentioned at the end about two tennis balls colliding, so perhaps it was the impact of two vectors colliding together? and for the 3rd example we wanted to know the net direction and movement of the box. Is there a set of skills that would be helpful to know in all the vector kind of questions? e.g. pythagorus theorum, SOHCAHTOA etc. Physics and math concepts seem to overlap quite a bit, and I'm a bit rusty on math concepts too haha
Yep that's right! So I'm effectively explaining how to add two vectors together and using the tennis ball example as an analogy to give a more conceptual understanding of what is happening.
Adding vectors in physics involves breaking them down into horizontal and vertical components first and then summing these together to get the net vector. If you don't need to be accurate in the question, you can avoid the calculation by gaining a rough understanding of the sum of the vectors by using this kind of analogy to understand how they would interact to save time.
Using pythagoras' theorem is definitely helpful in some GAMSAT questions. SOHCAHTOA is useful for reasoning (ie. knowing that sine is O/H etc) but they don't expect you to be able to calculate using angles. They usually either give estimates of say sin(20º) or they avoid it entirely
This was so helpful! Thank you🙏🏿 I just wanted to know if at 48:58 you made up a scenario where g1=g2. Or is that an actual physics principle?
Glad it helped! So in that scenario I'm setting up the condition that the gravitational fields are equal at some point between the two planetary bodies (g1=g2) but at all other distances other than this they would not be equal. It is a real situation in that two objects can have a point between them at which the gravitational field strengths are equal and opposite
hehe scratchy scratchy at 14:30
haha he is relentless
just a question on patient on the bed, it makes sense most the weight will be pushed down at point B and less weight pushed down at point A , but I'm confused as to why the weight comes from opposite sides. So why does the weight force go diagonally? It makes sense to go linearly down.
Hey Batu, yep this is a pretty common confusion actually. The best way to think about it is the relative position of the centre of mass from each of the legs. The closer it is, the more weight will push down through that closer leg/support and vice versa. If you check out the pinned comment and my replies I go into a bit more detail and a conceptual way of thinking about it :)
Thanks for that, the pinned comments really helped.
Hi Jesse :) For the bed problem, would it be possible to do:
F = m x a
Mass for leg A = 90kg/20 x 3 = 13.5kg
F = 13.5kg x 10N (gravity) to get 135N
Then complete the process with the alternate leg?
Thanks so much
Hey Jane, yep absolutely. Effectively this would just be splitting the masses by proportion before multiplying by the gravity but would still be equivalent As you’ve simply factorised out the gravity and just dealt with the mass by proportion.
The proof would be:
If we let p represent the proportion (3/20) to make it look a bit nicer in text form
Force A = mgp = g x (mp)
A fancy way to reiterate that when multiplying, order doesn’t matter :)
Hi Jesse, another smashing content 😊 Hey, as you breadth of experience in tutoring year 12's wondering if you have any recommnedation for a level 12 physics book.
PS. waiting for the video regarding your reaction about the recent Gamsat. 😊 I sat it in Southbank Melb and all I can say is I miss my bottle of water. I thought it will be allowed in our table, but no 😭 my throat was parched like a desert 🏜
Hey Disha! Yeah I'd say the Heinemann Physics 12 (4th ed) textbook is probably the best one
www.pearson.com/store/p/heinemann-physics-12-ebook/GPROG_A103000288118_learnerau-availability/9781488663604
Chapters 1, 2, 5, 7, 8 and a bit of 9 is very relevant to GAMSAT physics
This is the textbook I use with my VCE Physics students in tutes too.
Yeah the water bottle rule is crazy but I guess it's probably there so people aren't causing distractions constantly fiddling with lids and making noise. I just didn't take my water in with me this time haha
18:42 if a car in motion with a mass of 1000kg hits you and in doing so decelerates at 3m/s^2 it would then apply 3000N over that period of deceleration. I think that's correct. That's how I think of it.
Hey Ryan! Sorry about the wait on a reply while I was on my break.
So that's almost it, however we need to consider the force and acceleration as vectors. If the acceleration is a deceleration then the acceleration is directed in the opposite direction to it's motion and the force it exerts. This would instead represent the person being hit as exerting 3000N of force on the car, theoretically.
This could also be proven mathematically from Newton's second law, F = ma
If we take the forward direction of the car as the + direction and the opposite direction as the - direction.
m is a scalar quantity and so the direction of the Force vector must be determined by the direction of the acceleration vector
If decelerating, a is negative and so F is negative (opposite direction to its travel)
In this example, the car is accelerating rather than decelerating so its acceleration and therefore force are directed in the same direction (+ direction) and F = ma = 1000 x 3 = 3000N exerted by the car on the person
@@jesseosbourne Hey Jesse hope you had a good break!
Hmmm I'll have to give this more thought and do a few practice problems. I see your point about the person exerting 3000N of force on the car to decelerate it and that is how I would approach these types of questions.
I agree that if the car were at rest with the bumper touching the person and then the car accelerated at 3m/s^2 it would apply 3000N of force. I also agree that if the velocity of the person and the velocity of the car (when the bumper first touched the person) were equal and the car was accelerating at 3m/s^2 it would apply 3000N of force. Although in both cases I think the Force would actually be m(car) + m(person) and therefore greater than 3000N but you're probably just considering the mass of the person as negligible (or I could be wrong). Since instead of a person if the car hit a large boulder it would require more force to accelerate it at 3m/s^2.
However, if the car were travelling at a non-zero velocity (when the bumper first touched the person) and hit a person (at rest) the acceleration the person would experience would be much greater than 3m/s^2 or the car would decelerate/accelerate more slowly until the velocity of the person matched the velocity of the car. Then the car would continue to apply a 3000N force if it kept accelerating at 3m/s^2.
Haha sorry that is a bit of a complicated response. What are your thoughts sir?
I may also review momentum. Do you have any good examples of these types of questions which involve momentum and transfer of momentum and how that affects force/acceleration?
Oh, something I heard that might interest you. Although I'm not sure how valid it is. I heard that USyd is calculating S1 x 1.25, S2 x 1.25 and S3 x1 with a place offered for total scores >258 so you should be good. ;)
So what this comes down to is separating the collision into two phases. The before the collision phase and the after collision phase. Momentum and energy of the entire system, should be conserved and so velocities of the person and the car will change to adjust for this.
This question only considers the force applied at the instant of collision but not after the collision.
At the instant of collision, the car is accelerating at 3m/s^2 with a mass of 1000kg and so exerts a 3000N force on the stationary person. In effect there has not yet been a time interval to allow for any deceleration of the car or acceleration of the person. So because we are not comparing the before and after phases, we don't have to look at it as a conservation of energy or momentum question.
If however, we do consider what happens after the collision for discussion's sake, the car would technically reduce its velocity as result of some of its kinetic energy being transferred to the person but as you correctly noted, this would be negligible given the difference in mass. The other thing that we would have to assume is that the driver takes their foot off the accelerator at the moment of collision, otherwise they would be continuing to introduce additional energy and therefore momentum into the system (by burning fuel in the engine) which would otherwise allow for the car to continue accelerating at 3m/s^2. This would then mean that the laws of conservation can't be applied.
For practice, you could find some free online worksheets by searching 'conservation of energy & momentum worksheet pdf'. Keep in mind that being non-GAMSAT, they may be a little more technical in nature so I'd recommend either using a calculator on things like sine/cosine etc or practising estimating to see how close you can get to the real answer. So long as you can come away with an understanding of how to 'think' about the problems and recognise when it applies, you're good!
Here's an example worksheet:
www.smcisd.net/cms/lib/TX02215324/Centricity/Domain/1073/unit%20homework%20momentum%20its%20conservation%202016.pdf
www.smcisd.net/cms/lib/TX02215324/Centricity/Domain/1073/unit%20homework%20momentum%20its%20conservation%20ans%20key.pdf
Thanks, yes I have seen this formula floating around for USyd which brings me a lot of confidence in my previous scores. After researching for the USyd application video though, there's no mention of an algorithm being used and it looks like its a case of section score rankings with equal weightings. The 1.25 weighting appears to have just been correlating with scores for 2022 entry. Hopefully nothing too dramatic changes for this round of applications!
Thanks Jesse.
You’re gonna get in for sure. :)
I’ll have some S2 questions for you in the next few days.
31:40 for the downwards lift, why is the 2 taken away from the 10? I'd have thought that you'd add the downwards forces to get a net acceleration of 12? I know your answer is right, I just can't get it to make sense in my head 😅🫠
Yeah this is definitely one of those seemingly counterintuitive scenarios in physics. It comes down to the fact that there is no additional external force that is adding onto the gravitational force. More so, by accelerating downward in the same direction as gravity we are almost 'running away from gravity' and so it's effect on us as the moving body is less.
A similar example in a horizontal plane would be a car being towed by a rope. If the tow truck is pulling with 1000N of force then this is the force experienced by the car being towed (assuming no friction etc) but if the towed car begins to move forward on its own a bit, say via 200N of force) then this removes some of the tension in the rope and the net force acting not he towed car is 800N.
As a little interesting aside (although not required for GAMSAT) this assumption of force, acceleration and velocity interaction is limited to Newtonian (Classical) Physics - which is what we use in GAMSAT
In Einstein's theory of special relativity we can still use this concept for objects but for light we have to make the assumption that light speed is a constant and cannot exceed it's value of 3x10^8 m/s even if classical physics would suggest otherwise in circumstances such as the speed of light from a headlight on a car travelling forwards at 50m/s
@@jesseosbourne thank you so much!!!! That tow truck analogy makes so much sense. You are an absolute lifesaver 🙏🙏🙏 physics is my worst enemy lol 😂 well, maybe after organic chem ... 🤪
can somebody clarify on why the net acceleration is 8 m/s2 when both the gravity and acceleration of the lift are going downwards at 31:18? thank you in advance
Hi Jesse,
Would you recommend learning newtons equations of motion and all their variations? Or would these sort of formulae be given?
Thanks
Hey William, yeah I'd say they're useful. It's common for them to be provided but not all of them so knowing all five allows you to get to the answer in one step rather than multiple.
I have a full list of equations on the 'formula sheet' on my resources page, linked in the video description
Hi Jesse, Can you please explain @1438, how did u come out with the net direction of the box. Thanks
Hi Avneet, so here I'm using vector summation. By taking the net horizontal and net vertical forces as vector arrows and placing them 'nose to tail', you can make a simple right angle triangle where the hypotenuse is the direction of the net force. Because the net force has a rightward and upward component, the net force should be pointing in the direction of up and to the right.
The actual value of the net force can be calculated using pythagoras' theorem (relatively unlikely in GAMSAT) and the direction could be explained based on the angle that the net force makes with the horizontal solved using trigonometry (not examinable in GAMSAT)
I might have missed this but how did you know that g=10 in the patient on the bed question?
Hi Martine, gravity on Earth is 9.8 N/kg (9.8 m/s^2) but is often simplified to 10 for the sake of estimation. You can always assume g=10 for any calculations involving gravity on the surface of the Earth
@@jesseosbourne thankyou
Hi Jesse, just wondering with 13:20 and hypothetically solving it, how would we in the exam calculate sin/cos/tan without a calculator quickly?
Hey Lani, the good news is that you'd never be expected to estimate sin/cos/tan of an angle without some additional given information as there would be too high a degree of variability in people's estimates. Generally the stem glosses over the trigonometry and just says stuff like "let sin(theta) = alpha" and then it becomes an algebra question where you substitute alpha into your expression and continue
But, what I would do if I felt I needed to estimate something like this is I would use some outside knowledge that sin(30º) = 1/2 and cos(60º) = 1/2. I could use this as a reference point and then consider a right angle triangle where the angle either increases or decreases from this angle and think about what would happen to the ratio of the sides of the triangle.
For example, say we had a right angle triangle and the angle against the horizontal was 40º. If I wanted to know what sin(40) was, I could start by considering sin(30) = 1/2 and then rotating the hypotenuse of the triangle upwards so that the angle increased to to 40º would mean that the opposite side (vertical height) would have to increase and the adjacent side (horizontal) would decrease. Because sin is O/H and the opposite side increased then sin(40) > sin(30) therefore sin(40) > 1/2 so maybe 3/4 as an estimate. The same could be done for cos and tan ratios
Another alternative if you were already familiar with it is to think about the basic sin(x), cos(x), tan(x) curves and use reference points from there.
Again though, it could be useful to think about this to confirm your reasoning but I wouldn't think it's a required skill for GAMSAT
@@jesseosbourne Thank you for your detailed reply!! I had to reread about 30 times until I understood it but I got there haha
Hi Jesse, can I ask why the W=F*g=m*g around 42:36? Because F=mg, so if F*g is not possible to qual to m*g, and weight is also not qual to F*g, can you explain it? Thanks in advance
Sorry, that's my handwriting causing trouble there! I'm writing it as g in subscript (F_g) to denote gravitational force but it looks a lot like Fg. So Weight (W) = Gravitational Force (F_g) = mg
Hey Jesse, how come the fg calculations is always multiplied by 10?
Hey Simon, yep so Fg is gravitational force (sometimes referred to as weight force) and this is calculated as mass x gravity.
Given that for GAMSAT we always assume gravity to be 10 N/kg rather than the more specific 9.8, all the Fg calculations involve multiplying mass x 10 to calculate the resultant force
Hi Jesse! Bit of a random one but what software program are you using in the video?
Hey Kent! Sorry about the wait on a reply while I was on my break.
I use a browser based whiteboard called ziteboard. I started using it at the start of the lockdowns when I had to move all of my highschool tutorials online and it's held up well.
ziteboard.com
@@jesseosbourne Awesome thanks mate! Keep up the great work. It’s really helping me a lot 😊