5:23 why are the x-components of velocity equal for both before and after it hits the side of the table? You also mention that the x-component of velocity is parallel to the x-axis, but is the y-component also not parallel to the y-axis? So why do we ignore the x-component of velocity?
Yes the y-component is also parallel to the y-axis, but you misunderstood why I specifically said that. So we wrote the equation for y axis forces. That means any force that is parallel to the x-axis will simply be zero, since it's component is lying on the x-axis. In other words, think of a force lying directly or parallel to the x-axis, will it have a y component? No, right? In the same way, if we write an equation for the y axis forces, all we care about are y-components. Does that make sense? :)
i have one doubt in 2nd question, when ball is strike in ground and make another velocity, at that time why we didn't take component, i mean why not only take the vertical velocity of object to calculate maximum height..?
When using the coefficient of restitution at 3:43, does anyone know if the velocities are interchangeable? I'm confused how he knows which velocity is the ground and which is the ball.
The velocity of the ground is zero since it doesn't move, so vg=0. Also, even if you switch the velocities around, you'll still get the same answer since the right side is just a ratio. Make sure if you do switch it around, you switch both top and bottom. 👍
@@QuestionSolutions I see, thanks man. Your videos are super concise and you're bound to find a lot success with the quality you put out. I have no idea how these videos haven't blown up yet.
hello. great video as always. i have a question, could we use the conservation of energy in the question of the ball and the cannon? because i did it, since we have values for the velocity and height to discover the value of the velocity in B (where height=0). but the thing is, if you plug in the values, you get a different answer for velocity in B and therefore a different answer for the final height.. what did i do wrong? thank you.
It's really hard to say without seeing how you got to your answer.😅 Usually, you can solve these problems in more than 1 way, but since this is about the coefficient of restitution, that's the method used.
@@QuestionSolutions i used it when going from the velocity just before it hit and just after. it might be a maths mistake because otherwise i really dont know what is wrong. i dont know how i could show you my procedure so... yhea, thank you nontheless ;)
@@joaobrites5931 The problem lies in the fact that you have used conservation of energy. Coefficient of restitution is symbolic of certain energy loss, and thus, conservation of kinetic energy cannot be applied. However, it should be possible to apply conservation of total energy, so do recheck your answer.
I think you're not understanding how axes work in these problems. VBX1 shouldn't be parallel to the x-axis in the new coordinate system, it would be perpendicular. You need to be able to change your coordinate system without always relying on it to be a certain way. You have to be able to solve problems by piecing in what you found in the previous step and then drawing a new coordinate system at the new location of impact and using the previous information you gained in the new system. You can do it the way you want too, but it'll be tedious to have a central coordinate system.
We assumed the ball is going to move left, and since we picked left to be positive, it is also positive. At the end, if our assumption is wrong, we will get a negative value, which we did, meaning the ball actually traveled to the right. Hope that helps :)
@@QuestionSolutions Same question @2:00, so for the Final velocity both ball A and B, we assume that both of them will move to the right? That is why they are both positive 3VAf AND 2VBf? But what if I assume that ball A hit ball B, then ball A moves to the left and ball B moves to the right, should Ball A be -3VAf then? I know that was a lot of questions but I just want to understand it before the exam LOL
@@MUSICXOFFICIAL01 Yes, so we assumed both will move to the right, and since we arbitrarily picked right to be positive, they are also positive. You can also assume it the way you said, but make sure that assumption holds when you use the coefficient of restitution. I usually find it a lot easier to just pick one side, instead of mixing sides because it makes it easier to keep track of the signs, but its completely up to you. :)
It's based on which side you pick to be positive. You can see that at 1:43, we picked movement to the right to be positive. Ball B is moving to the left, so it's negative. It's the same as forces, if we pick up to be positive, then acceleration due to gravity is negative, and vice versa. It just a establishes a direction.
hello , thanks form video i have a question why we use g negative at 3:19 when it downward ? it supposed to be positive right ? it's upward at vc so we use it negative that's clear
I am a bit confused about your question. When you say "g negative" are you referring to gravity? If so, gravity will be negative since gravity is always downwards and we picked up to be positive. If that isn't what you are asking, could you give me the variable in question so I can help you better? Many thanks!
@@ahmedhassanin2155 No, gravity is always negative because its pointing downwards. I can't think of a situation where gravity can be positive, unless you pick downwards to be positive. The equation is written assuming up is positive, so gravity must be negative. Does that make sense?
@@ahmedhassanin2155 I assume your professor said "downwards is positive" otherwise, gravity would be negative. I am unsure of another case, please ask your professor again, I don't want to give the wrong answer here. :)
Hello. I dont understand why the velocities are the same after hitting the table in q3. What does the velocity being parallel to the axis have to do with their values. Thank you
@@ibrahimhashmi6716 So I guess a way to think about it to realize that when the ball hits at an angle, only one component of the velocity changes. In other words, whenever an object makes contact with another object and the direction changes, the parallel side to the axis of reference won't change. It's the perpendicular side that changes. This changing of one component effects the overall magnitude of velocity along with the direction. I am having a hard time explaining this with words, because you have to sort of visualize this. If you have a marble at home, you can roll it against something so that it hits at an angle and then try to visualize how the velocity of the ball changes with respect to each of the components.
Can you elaborate on your question? When you say plane of impact, do you mean to say that if an object impacts another object with velocity vectors in the x-y plane only, then it's constrained to just the x-y plane? In real life, I don't know how easy it would be to actually have an object constrained like that, but in questions, yes.
@@QuestionSolutions means In restitution problems I never considered signs for moving objects when colliding as per x and y coordinates but in this video in every problem you considered positive negative
@@saiprasadsatya3677 You should always keep positives and negatives in mind, it's a very good habit to develop. Sometimes, without considering positives and negatives, you might get the correct answer, but that's not the case with most problems. Instead, always establish a proper coordinate system when you draw your free body diagrams, and follow through your positives and negatives until the end. It leads to less confusion and at the end, if you get a negative answer, you know your assumption was wrong. 👍
The x-component is parallel to the x-axis, which means only the y-component will change. Just like how when the ball hits point B, it's the y-component that is parallel to the y-axis, which means that won't change.
you made a small mistake by switching the components of v in the last problem and the second collision, but greater teacher I always understand everything so not complaining about it just so other people don't get confused.
@@BrownMInc Thank you for taking the time to give me a timestamp. Really hard to figure things out without one with all the comments on different videos. Keep up the great work and best wishes with your studies :)
For the last question, the answer is just Vi (2.5) time e (0.6) = 1.5. I'm very interested in why this is the case, also maybe you did not need to do any of the complicated trigonometry since the answer was right there. Somehow you have actually made me interested in a math equation! :)
@@QuestionSolutions So at the very end, at 6:42 your answer is 1.5 for Vb. You didn't have to do all that complicated mathematics the answer is just 2.5 * 0.6. Since the two collisions, the X one and the Y one, both add to make a normal collision like with a normal object, the problem is easily solved.
@@markmalyshev2031 The process is to familiarize students using the topics learned. Not to get to an answer quickest way possible. Do you think if I wrote "2.5*0.6" as the answer, that would help students learn what the topic is about? 😅 Probably not. The process is incredibly important as a proof to explain why this happens. You can't simply say "things cancel out so its just 2.5*0.6", you must show the process to prove it. 👍
@@QuestionSolutions Good point, The way you did it is important for leanring. Also, my way would only work in rare occasions and you taught the principle entirely. Thanks for your work, I'll see you next exam!
@@markmalyshev2031 Yes, your way works when the angles are equal and cancel the components out :) I wish it's easy to write one line and say "here you go, this is the answer" but I really want students to learn the steps to get to it, so they don't look for patterns but are able to solve any question they face. 👍 Best of luck with your future endeavors!
You're solving 2 equations with 2 unknowns, so you can use the method of elimination, substitution, etc. So you can isolate for VA in the 2nd equation, plug it into the first equation and solve. 👍
I dont know how to use elimination in this, can you tell me how did you get the VA by using the 2nd equation? Like more precise answer please I really need this for my class😅
@@linkbchygg Sure, here it is solved: bit.ly/3ryVc5t I am not trying to be mean or anything, but you will most likely have a hard time in your course if you are having trouble solving linear equations. I really recommend brushing up on your linear algebra, especially simultaneous equation solving since that is going to be required for most of your courses :)
@@dexamiequijoy9201 Add the left side of the equation in your calculator. 8 x 3 = 24, then you have 2 x -4 = -8. So 24 - 8 = 16. We picked right to be positive, and ball B is moving to the left, so it's negative.
So you're just solving 2 equations with 2 unknowns. Isolate for a variable in one equation, plug that into the second equation and you'll get your answers :)
Interesting, I found out that the next bounce height is the maximum height of the particle (hmax) * e^2 by just testing if it gave the same answer, I am too lazy to derive it though.
why do not we just apply the e formula for both x and y i do not get it why the x component does not change. please give me explanation and do not simply said it is parllel to x axis
@@mebawubeshet6729 Please speak to your professor or TA during their office hours. I am unable to answer all your questions (on multiple videos), you are missing a lot of fundamental knowledge required for this course :(
I think your notations can be better and I think it would easier to imagine if you flip the axis to always be perpendicular to the boundary of collision
It's solving 2 equations with 2 unknowns. If you are having a hard time with that, please consider dedicating some time to understanding how to solve them because in the future, you will have to solve 3 equations with 3 unknowns and even equations with trig functions in them. It's very important that you get the fundamentals solid or it'll become tough. 😅Here it is solved step by step: bit.ly/3ryVc5t
It would be good if you gave constructive criticism so that I can improve videos for others in the future. Saying "bad video" does nothing for me or others.
This is only channel that I don't watch at 2X speed
😅 I try my best to make them as concise as possible.
I do xD my exam is in 36 hours gotta go fast. Love these videos
@@Emma-dd1fc I wish you the best on your exams!
It's pretty satisfying in my opinion. Is like a voice bringing calm to my mind's voices, terrified by the exams coming
Thanks bro I've been learning a lot with you
Really glad to hear :)
5:23 why are the x-components of velocity equal for both before and after it hits the side of the table? You also mention that the x-component of velocity is parallel to the x-axis, but is the y-component also not parallel to the y-axis? So why do we ignore the x-component of velocity?
Yes the y-component is also parallel to the y-axis, but you misunderstood why I specifically said that. So we wrote the equation for y axis forces. That means any force that is parallel to the x-axis will simply be zero, since it's component is lying on the x-axis. In other words, think of a force lying directly or parallel to the x-axis, will it have a y component? No, right? In the same way, if we write an equation for the y axis forces, all we care about are y-components. Does that make sense? :)
@@QuestionSolutions yes, thank you 🙏🏼
@@QuestionSolutions Does that mean that we can solve the problem using equation for x-axis instead of equation for y-axis?
Clarified it all!! Thank you!
You're very welcome!
i have one doubt in 2nd question, when ball is strike in ground and make another velocity, at that time why we didn't take component, i mean why not only take the vertical velocity of object to calculate maximum height..?
Can you provide me with a timestamp please?
When using the coefficient of restitution at 3:43, does anyone know if the velocities are interchangeable? I'm confused how he knows which velocity is the ground and which is the ball.
The velocity of the ground is zero since it doesn't move, so vg=0. Also, even if you switch the velocities around, you'll still get the same answer since the right side is just a ratio. Make sure if you do switch it around, you switch both top and bottom. 👍
@@QuestionSolutions I see, thanks man. Your videos are super concise and you're bound to find a lot success with the quality you put out. I have no idea how these videos haven't blown up yet.
@@ryanbragg2274 Thank you so much for your kind comment :) Best of luck with your studies!
hello. great video as always. i have a question, could we use the conservation of energy in the question of the ball and the cannon? because i did it, since we have values for the velocity and height to discover the value of the velocity in B (where height=0). but the thing is, if you plug in the values, you get a different answer for velocity in B and therefore a different answer for the final height.. what did i do wrong? thank you.
It's really hard to say without seeing how you got to your answer.😅 Usually, you can solve these problems in more than 1 way, but since this is about the coefficient of restitution, that's the method used.
@@QuestionSolutions i used it when going from the velocity just before it hit and just after. it might be a maths mistake because otherwise i really dont know what is wrong. i dont know how i could show you my procedure so... yhea, thank you nontheless ;)
@@joaobrites5931 The problem lies in the fact that you have used conservation of energy. Coefficient of restitution is symbolic of certain energy loss, and thus, conservation of kinetic energy cannot be applied. However, it should be possible to apply conservation of total energy, so do recheck your answer.
have u done vid on solving the two equations with 2 unknowns im kinda lost at that point
nvm im slow i got it
Yes, please see: ruclips.net/user/shorts86uENomd53U?feature=share
ruclips.net/user/shortsHe7lrJEB04U?feature=share
ruclips.net/user/shorts4euH1289_Kg?feature=share
ruclips.net/user/shortsrAlhrq5hWFc?feature=share
Hello can I ask how did you get the VA= -0.16 and Vb= 8.24??
Please kindly provide a timestamp so I know where to look, many thanks!
2:31
At 6:28 why is your VBX2 and VBX1 on different axis? Like your VBX1 isn't parallel to your newly drawn x-axis
I think you're not understanding how axes work in these problems. VBX1 shouldn't be parallel to the x-axis in the new coordinate system, it would be perpendicular. You need to be able to change your coordinate system without always relying on it to be a certain way. You have to be able to solve problems by piecing in what you found in the previous step and then drawing a new coordinate system at the new location of impact and using the previous information you gained in the new system. You can do it the way you want too, but it'll be tedious to have a central coordinate system.
hi I have a question, @2:00 why isn't 3VA be negative?
We assumed the ball is going to move left, and since we picked left to be positive, it is also positive. At the end, if our assumption is wrong, we will get a negative value, which we did, meaning the ball actually traveled to the right. Hope that helps :)
@@QuestionSolutions thank you.
You're very welcome!
@@QuestionSolutions Same question @2:00, so for the Final velocity both ball A and B, we assume that both of them will move to the right? That is why they are both positive 3VAf AND 2VBf? But what if I assume that ball A hit ball B, then ball A moves to the left and ball B moves to the right, should Ball A be -3VAf then? I know that was a lot of questions but I just want to understand it before the exam LOL
@@MUSICXOFFICIAL01 Yes, so we assumed both will move to the right, and since we arbitrarily picked right to be positive, they are also positive. You can also assume it the way you said, but make sure that assumption holds when you use the coefficient of restitution. I usually find it a lot easier to just pick one side, instead of mixing sides because it makes it easier to keep track of the signs, but its completely up to you. :)
In the first example, why do you take B to have a negative velocity when in the question it says the velocity is +4ms-¹
It's based on which side you pick to be positive. You can see that at 1:43, we picked movement to the right to be positive. Ball B is moving to the left, so it's negative. It's the same as forces, if we pick up to be positive, then acceleration due to gravity is negative, and vice versa. It just a establishes a direction.
hello , thanks form video
i have a question why we use g negative at 3:19
when it downward ? it supposed to be positive right ?
it's upward at vc so we use it negative that's clear
I am a bit confused about your question. When you say "g negative" are you referring to gravity? If so, gravity will be negative since gravity is always downwards and we picked up to be positive. If that isn't what you are asking, could you give me the variable in question so I can help you better? Many thanks!
@@QuestionSolutions yes gravity
when upward we use it negative
but you use it negative at 3:21
when it downward too
@@ahmedhassanin2155 No, gravity is always negative because its pointing downwards. I can't think of a situation where gravity can be positive, unless you pick downwards to be positive. The equation is written assuming up is positive, so gravity must be negative. Does that make sense?
@@QuestionSolutions thanks
but we studied at physics when the object is falling the acceleration (gravity ) is positive and vise versa
@@ahmedhassanin2155 I assume your professor said "downwards is positive" otherwise, gravity would be negative. I am unsure of another case, please ask your professor again, I don't want to give the wrong answer here. :)
The magnitude of Va is 2.6 m/s in last example? You got Vb as 1.499 m/s, but I did not see Va?
I think I forgot to do VA, but it's the same way as how VB was found. 😅 Good catch!
Hello. I dont understand why the velocities are the same after hitting the table in q3. What does the velocity being parallel to the axis have to do with their values. Thank you
Can you give me a timestamp as to the location you're referring to? Many thanks!
@@QuestionSolutions at 5:20
@@ibrahimhashmi6716 So I guess a way to think about it to realize that when the ball hits at an angle, only one component of the velocity changes. In other words, whenever an object makes contact with another object and the direction changes, the parallel side to the axis of reference won't change. It's the perpendicular side that changes. This changing of one component effects the overall magnitude of velocity along with the direction. I am having a hard time explaining this with words, because you have to sort of visualize this. If you have a marble at home, you can roll it against something so that it hits at an angle and then try to visualize how the velocity of the ball changes with respect to each of the components.
is momentum conserved on both line of impact and plane of impact?
Can you elaborate on your question? When you say plane of impact, do you mean to say that if an object impacts another object with velocity vectors in the x-y plane only, then it's constrained to just the x-y plane? In real life, I don't know how easy it would be to actually have an object constrained like that, but in questions, yes.
Is it mandatory to take positive and negative values for different directions
Could you give me an example of what you're asking? I don't really understand your question :( Sorry!
@@QuestionSolutions means In restitution problems I never considered signs for moving objects when colliding as per x and y coordinates but in this video in every problem you considered positive negative
@@saiprasadsatya3677 You should always keep positives and negatives in mind, it's a very good habit to develop. Sometimes, without considering positives and negatives, you might get the correct answer, but that's not the case with most problems. Instead, always establish a proper coordinate system when you draw your free body diagrams, and follow through your positives and negatives until the end. It leads to less confusion and at the end, if you get a negative answer, you know your assumption was wrong. 👍
@@QuestionSolutions thank you for your valuable suggestion
at 5:56 why is the initial velocity on the x axis(vax) the same as the after collison velocity on the x axis
The x-component is parallel to the x-axis, which means only the y-component will change. Just like how when the ball hits point B, it's the y-component that is parallel to the y-axis, which means that won't change.
@@QuestionSolutions why thhis the case please do more explanation not give nut shell
teacher, we should get Vax2 because it impacts x component but you got Vay2 why? at 5.30
It's a typo 😅
you made a small mistake by switching the components of v in the last problem and the second collision, but greater teacher I always understand everything so not complaining about it just so other people don't get confused.
Could you please provide me with a timestamp to the location I made a mistake? I can verify it and then pin a comment for others to see. Thanks!
@@QuestionSolutionshello 👋, I believe the error is roughly at 6:15 on the diagram. Namely, the component before collision
@@BrownMInc Yes, I see it, there is a typo with Vbx1 and Vby1, should be flipped, however the numerical solution is correct.
@@QuestionSolutions oh yes I know, I just believe that was the error they were pointing out. And like they said, lovely content as always!!
@@BrownMInc Thank you for taking the time to give me a timestamp. Really hard to figure things out without one with all the comments on different videos. Keep up the great work and best wishes with your studies :)
For the last question, the answer is just Vi (2.5) time e (0.6) = 1.5. I'm very interested in why this is the case, also maybe you did not need to do any of the complicated trigonometry since the answer was right there. Somehow you have actually made me interested in a math equation! :)
Sorry, could you give me a timestamp as to where you're referring to? :)
@@QuestionSolutions So at the very end, at 6:42 your answer is 1.5 for Vb. You didn't have to do all that complicated mathematics the answer is just 2.5 * 0.6. Since the two collisions, the X one and the Y one, both add to make a normal collision like with a normal object, the problem is easily solved.
@@markmalyshev2031 The process is to familiarize students using the topics learned. Not to get to an answer quickest way possible. Do you think if I wrote "2.5*0.6" as the answer, that would help students learn what the topic is about? 😅 Probably not. The process is incredibly important as a proof to explain why this happens. You can't simply say "things cancel out so its just 2.5*0.6", you must show the process to prove it. 👍
@@QuestionSolutions Good point, The way you did it is important for leanring. Also, my way would only work in rare occasions and you taught the principle entirely. Thanks for your work, I'll see you next exam!
@@markmalyshev2031 Yes, your way works when the angles are equal and cancel the components out :) I wish it's easy to write one line and say "here you go, this is the answer" but I really want students to learn the steps to get to it, so they don't look for patterns but are able to solve any question they face. 👍 Best of luck with your future endeavors!
um hello I got kinda lost. How did you get the 8.4 = (Vb)-(Va) in 2:24
8-(-4) = 12 ==> 12 x 0.7 = 8.4. ==> 8.4 = v_B - v_A (top is untouched). I hope that helps.
@@QuestionSolutions thankyou very much
@@emswoooo You're very welcome!
Thanks a bunch!
You're very welcome!
For the second example why cant you use the conservation of energy equation?
You can use many different methods to find the answer to these problems. I just show case the methods required for the specific section.
@@QuestionSolutions Oh ok thanks for the quick response you are truly a gift from god
@@jadkhalil9263 😅You're very welcome!
How did you get the final velocity a and b? At 2:30
You're solving 2 equations with 2 unknowns, so you can use the method of elimination, substitution, etc. So you can isolate for VA in the 2nd equation, plug it into the first equation and solve. 👍
I dont know how to use elimination in this, can you tell me how did you get the VA by using the 2nd equation? Like more precise answer please I really need this for my class😅
@@linkbchygg Sure, here it is solved: bit.ly/3ryVc5t
I am not trying to be mean or anything, but you will most likely have a hard time in your course if you are having trouble solving linear equations. I really recommend brushing up on your linear algebra, especially simultaneous equation solving since that is going to be required for most of your courses :)
Thank you for your patience with me and thank you for the great feedback, it really helps me to find the thing I should practice😅
@@linkbchygg You're very welcome! Best of luck with your studies.
If you are asked to find the energy loss, how would you calculate that? @Question Solutions. Also, thanks for the video!
What type of energy loss?
Please how did you get the 3 va + 2 vb = 16
Sorry, I don't understand your question. Are you asking where the 16 came from, or where the 3 and the 2 came from?
The 16
And why the vb of Ball B is -4 not positive 4?
@@dexamiequijoy9201 Add the left side of the equation in your calculator. 8 x 3 = 24, then you have 2 x -4 = -8. So 24 - 8 = 16. We picked right to be positive, and ball B is moving to the left, so it's negative.
@@QuestionSolutions Thank you
how did u get va as -0.16 and vb as 8.24m/s
So you're just solving 2 equations with 2 unknowns. Isolate for a variable in one equation, plug that into the second equation and you'll get your answers :)
Interesting, I found out that the next bounce height is the maximum height of the particle (hmax) * e^2 by just testing if it gave the same answer, I am too lazy to derive it though.
Awesome! 👍
Outstanding 😇
Thank you! 😇
why do not we just apply the e formula for both x and y i do not get it why the x component does not change. please give me explanation and do not simply said it is parllel to x axis
6:15
and does it mean the e will affect one of the components
and why don't we just apply the e formula with out decomposing to its component
@@mebawubeshet6729 Please speak to your professor or TA during their office hours. I am unable to answer all your questions (on multiple videos), you are missing a lot of fundamental knowledge required for this course :(
I think your notations can be better and I think it would easier to imagine if you flip the axis to always be perpendicular to the boundary of collision
Thanks for the feedback 👍
how did you solve for two unknowns
time 2:30
@@AdamWatts-or9vg Please see: www.cymath.com/answer?q=3a%2B2b%3D16%2C%208.4%3Db-a
I love this 🤝🤝
Thank you 🤝🤝
i dont get how u solve the -.16 and 8.24
It's solving 2 equations with 2 unknowns. If you are having a hard time with that, please consider dedicating some time to understanding how to solve them because in the future, you will have to solve 3 equations with 3 unknowns and even equations with trig functions in them. It's very important that you get the fundamentals solid or it'll become tough. 😅Here it is solved step by step: bit.ly/3ryVc5t
🎉❤❤❤
❤❤
Little apple
🤔
Bad video
It would be good if you gave constructive criticism so that I can improve videos for others in the future. Saying "bad video" does nothing for me or others.
Cuz u r not here to learn.