At 4:27, there is a typo. The acceleration should be negative in that step, since we picked up to be positive. However, the next step is correct and this does not affect the final answer.
You are by far the best dynamics online instructor I've found! your videos are simple to understand and tackle some hard problems. PLEASE keep up the great work you're amazing!!
For question 2, if anyone finds it challenging to identify the integrals, I found that using the work-energy method is more intuitive. It avoids dealing with integrals and is simplified as: 1/2(Ma+Mb)(v)^2 - (N)(k)(s) = 0 and solve for s
love to watch your videos thankyou for creating such a nice content for students that struggling in their studies hope to see more videos like these thankyou buddy again
I never really comment but Damn! This is perfect, I was searching for this kinda channel , thankyou for clearing all our doubts! Loads of love ❤️❤️❤️❤️❤️
I paused the video and solved it that way and got exactly 3.996m and then played it and saw he did something else and it scared me XD then he got the same answer as I did! i was looking for somebody who also did this and it looks like i found you. good thought!
@@mebawubeshet6729 For question 2, if anyone finds it challenging to identify the integrals, I found that using the work-energy method is more intuitive. It avoids dealing with integrals and is simplified as: v = 3.9604 as solved for in part 1 1/2(Ma+Mb)(v)^2 - (N)(k)(s) = 0 and solve for s
hey guys In 2:56 question, if you use work energy principle then it will be far more easier. T1+U=T2 (;T2=0) we have all value except the distance traveled, easy! Also thanks to you I can pass my dynamic exam.
A lot of these questions can be solved in more than one way. I only show case a single method, usually one pertaining to a chapter. Best wishes with your dynamics exam! :)
Hello, I don't get why in the first example, we applied conservation of energy on the system of particles (car A and B) and not only on A. Doesn't the spring become an internal force? In other examples like this, we only applied conservation of energy on the car with the spring attached on it. How do we know when to do that? Also, when does max compression occur in this example? does it stay compressed to the max even after the collision?
I can't answer your question since I don't know what other examples you are referring to. It might be different, or you used a different method at getting the answer. There are usually more than one way of getting to same answer so it's hard to say, and if you get the same answer, then your method is also just as valid as what's shown here. I only showcase a single method. For max compression, the train will couple when it hits another car, and the spring will stay constantly compressed due to the coupling, which is why they both have the same velocity after they have the impact. I show this at 0:52 :)
Why is S_B/A in the last example 0.5 m, isn't that S_A because it originated from 0.5 m away from point O? I'm having trouble visualizing what the 0.5 m means in the diagram, it looks like it means the width of the triangle B. Thanks
So an easier way to think about it is to imagine the origin at block A. In that case, block B "moves" 0.5 m. It's really just the difference in the starting points of A and B, hence B/A.
can you solve that question A5kg mass moving with velocity 10m/s collides with 10kg mass moving with velocity of 7m/s along the same line . if the two masses join together on impact find their common velocity if they were moving a) in opposite direction b)in the same direction
Sorry but I don't solve questions like this because then I have to solve everyone's questions and it's not doable :( Please ask your professors/TAs during their office hours. You can also ask on forums for help.
On the last problem at 7:25, I think it should be V a/b because we know from B's perspective A moved .5 meters. We don't know yet what b/a is. If we get it wrong in a case would it make a difference if we keep our signs correct? I can't figure that out. Thanks for your hard work.
Did you see this video first? ua-cam.com/video/opVKNCedkRo/v-deo.html The whole answer will change if you use a/b vs b/a. It's very important you distinguish which you use because the equation changes based on it. If you have the time, please watch that video.
You can solve these questions in more than one way, sometimes, it might even be easier/faster to use another method, I just show this method because it pertains to the topic we are talking about. 😅
In the last example, why is the small box’s velocity pointed tot the left? And not pointed in the direction of the 30 degree plain? I find it difficult to see that.
Take note of the notations used. It shows V_Ax, which represents the x-component of the velocity, which will be parallel to the x-axis, in this case, the grey base.
For problem 2, using linear impulse and momentum, would it be correct if you instead did s=integral (t to 0) 3.196-1.962t dt? And then set s=0 and solve for time? Or would you have to find the bounds of integration for time?
You are confusing different topics together. When you need the coefficient of restitution, then you will need to use that equation. Here, it's only the conservation of energy, so in this topic, we are excluding things like noise, heat, friction...
hi really helpful big thanks...but i have i question when we use (linear impulse and momentum or conservation of energy or others) cuz i really confused.
If you look at the 2 equations given, you can see when to use it. Impulse has to do with time, so if time is given, you can use it with an equation of linear impulse and momentum. If it's just velocity and mass, you can use conservation of momentum. 👍 The more questions you do for each topic, the faster you will realize which equations to use in what situations :)
@@QuestionSolutions really thx...im sorry ,but I'm going ask another question (in the example number 2 there is no time?) it just masses ,velocities and foresees...and you used linear impulse and momentum thank you
@@slimlady9923 Don't be sorry about asking questions, the more you ask, the more you learn. I was able to use impulse and momentum because time would get cancelled out. So notice how I started off with a conservation of energy equation because we had mass and velocity. After that, we still didn't get the answer we need, so we then switch to a linear impulse and momentum equation. You can also notice that we are looking for distance, and distance is related to speed, which is related to time. So you sort of have to work through these steps in your mind and see what can be done. Even if you take the wrong path, you will see that you're not getting the value you need, so you can retrace your steps and come back to an equation that works. In more advanced problems, you will definitely have to use both to get to an answer. This is just a tip, and in no way a solid factor, but in most final exams, the questions your professor will create will require you to use at least 2 of the main equations, if not more to arrive at an answer.
@@slimlady9923 Really, you don't have to be sorry for asking questions. Feel free to ask, and I'll do my best to explain. I wish you the best of luck with your exam! 👍
For finding the distance on the second part of the problem can't you just use the principle of work and energy which is easier than first finding t. I did that and got 4.03m which is the same answer.
You can use many different methods to arrive at the answer. I just use the methods pertaining to the topic. There are many ways to get to the same answer with these problems, some are much easier than others, some take the long route, but I try to stick to whatever is relating to a chapter, otherwise, it's sort of pointless to do a video on that topic 😅
Thanks for your great videos. Note that after I calculated v = 3.96 m/s (as you did), I then applied the following simple suvat equation to determine the stopping distance: final velocity squared = initial velocity squared + 2*acceleration (negative, due to the friction force)*s. Note final velocity squared = 0. My answer is 4.036 m, which is very close to your answer of 3.996 m. It looks like my method is simpler and quicker in this case. Was I correct in applying this formula, or was it a coincidence that I got the correct answer (or one very close to the correct answer)?
There is usually multiple ways to get to the same answer, I only showcase what's pertaining to a certain chapter. But in general, there are tons of ways to get to it. 👍
Hello. once again a excellent video. I have a question: in the last problem, could we say that the normal force exerted by B in A as a contrary force exerted by A in B, because of newtons third law? and thus we would apply the similarity of triangles and discover the X component of this force. then we would find out that that component is in fact the total force aplied in B ( the forces in Y axis cancel), therefore we would need to calculate the distance A did ( by applying trignometry ) and with the kinematics equations for A discover the time it took A to reach the end. With that value of time and the value for acceleration of B ( applying newtons second law, we have the total force and the mass) we plug in the values in the kinematics equations for B and discover the distance he covered in the X axis in that time period. The problem is, if you do so, you get a absurd value of 2.02m. what did i do wrong?? Thank you and keep up with the great job.
That seems like a lot of steps to get to this answer, doesn't it? I mean was your method faster? I have no idea 😅 Again, I don't know what you did wrong without seeing your steps :) With this method, I think you can solve the problem easier and faster than going through a lot of trigonometry and all the other parts you did. Though I do like how you always try to solve the questions in different ways, since there is always more than 1 way to solve them. The only problem is that sometimes, there aren't enough values given, or some just don't help out.
@@QuestionSolutions they were indeed a lot of steps... i suppose your method is much better but i did not see it crystal clear at first.. nevertheless i continue to be confused on what i did wrong, because something must be wrong with my way of thinking.. i dont know how i could show you just like i said in the other comment. but thank you a lot!
Thanks so much for posting these! A question about the second question. I used the equations of motion in y to find the Frictional force, then I used them again in x to find acceleration. Plugged it into the kinematic equation a ds = v dv, integrated and solved. I got the correct answer but I was wondering if it's another valid method or just a fluke. Thanks again!
More often than not, there is more than one way to solve a problem. You used kinematics to solve the problem, which works as well. Try to solve it in both ways so you learn the new concepts as well :)
You can solve these questions anyway you like, but this video is about linear momentum, so that's what's shown here. Usually, if time is given, it's going to involve impulse.
Did you try it with that method? So the goal in this video is to use conservation of linear momentum, which is what I showcase. But you can use any method you like to get to an answer, since there are multiple ways to solving the same problem.
@@vigneshwaranrm6277 Then I am thinking maybe there is a numerical error or a part is missing on your equation? I am not sure, it's hard to say since I didn't solve it with conservation of energy. 😅 See your professor during office hours and show him your method so he can check over your work.
Wouldn't it be easier to solve the second question using the work energy equations after finding N , since it directly provides the horizontal displacment.
Yes, it's possible, along with the equation already shown on the video, you'll need one more, an equation of work and energy, or conservation of energy. Because right now, we only have 1 equation with 2 unknowns. That should allow you to find the speed of B when the box reaches the ground.
@@QuestionSolutions like friction has a force in the principle equation but the force applied by the bullet isn’t in the equation. I’m assuming it’s because this moment is already after the impact
So we are only interested in the relative displacement, so in other words, where "A" is with respect to "B". Another way to think about it is to realize that in the horizontal plane, "A" moves 0.5m when it slides down to the bottom.
At 4:27, there is a typo. The acceleration should be negative in that step, since we picked up to be positive. However, the next step is correct and this does not affect the final answer.
You are by far the best dynamics online instructor I've found! your videos are simple to understand and tackle some hard problems. PLEASE keep up the great work you're amazing!!
Thank you very much for your kind comment. I really appreciate it. I wish you the best with your studies and to you also, keep up the great work!
This deserves a million views. Thanks a lot, Sir!
You're very welcome!
I finished Dynamics! This channel did help me!
I am really glad to hear that. Well done with your Dynamics class!
This is my favorite channel to prepare for exams! Please do lots more example videos on work, conservation of energy, and impulse and momentum
I am really glad to hear that. I hope your exams go well! :)
I am almost done my first year of engineering and would like to thank you for getting me through dynamics :D
That's awesome to hear! I wish you the best on your upcoming years, may success follow you everywhere. :D
same here
@@abdallababikir9154 isn't it the greatest felling in the world first year is very rough
@@jadkhalil9263 hey, so are you done now??😭😭
The work that is done on these video is unreal, keep it up :)
Many thanks :)
For question 2, if anyone finds it challenging to identify the integrals, I found that using the work-energy method is more intuitive. It avoids dealing with integrals and is simplified as:
1/2(Ma+Mb)(v)^2 - (N)(k)(s) = 0
and solve for s
👍
this is so clear and helpful, thank you!
You're very welcome!
@4:27 when you calculate the normal force N, you said that upwards was positive so I think the equation would have been N(t) - 9.81(2.02)(t) = 0.
Yes! Good catch, it's a typo. 😅
@@QuestionSolutions you should mention that in the title just because.
@@justinbeck560 I pinned a comment saying it in case anyone was wondering.
love to watch your videos thankyou for creating such a nice content for students that struggling in their studies hope to see more videos like these thankyou buddy again
You're very welcome, and thank you for taking the time to write a nice comment like this. I appreciate it and I wish you the best with your studies.
thanks a lot for this video, GOD Bless
You're very welcome!
Very nice pedagogy and all concepts in a nutshell
Many thanks!
I never really comment but Damn! This is perfect, I was searching for this kinda channel , thankyou for clearing all our doubts! Loads of love ❤️❤️❤️❤️❤️
I am glad to hear this was helpful. I hope all of the videos help you out and wish you the best with your studies! ❤
For the bullet block problem. wouldn't it be much easier to use work energy principle?
You can use any method you like. You can get the answer to these problems in so many different ways. I just showcase a single method per video. :)
I paused the video and solved it that way and got exactly 3.996m and then played it and saw he did something else and it scared me XD then he got the same answer as I did! i was looking for somebody who also did this and it looks like i found you. good thought!
@@aidenbrown7200 how do u solve it. what do u use for the intial kinetic , the friction forcse
@@mebawubeshet6729 For question 2, if anyone finds it challenging to identify the integrals, I found that using the work-energy method is more intuitive. It avoids dealing with integrals and is simplified as:
v = 3.9604 as solved for in part 1
1/2(Ma+Mb)(v)^2 - (N)(k)(s) = 0
and solve for s
hey guys In 2:56 question, if you use work energy principle then it will be far more easier. T1+U=T2 (;T2=0) we have all value except the distance traveled, easy!
Also thanks to you I can pass my dynamic exam.
A lot of these questions can be solved in more than one way. I only show case a single method, usually one pertaining to a chapter. Best wishes with your dynamics exam! :)
Hello, I don't get why in the first example, we applied conservation of energy on the system of particles (car A and B) and not only on A. Doesn't the spring become an internal force? In other examples like this, we only applied conservation of energy on the car with the spring attached on it. How do we know when to do that? Also, when does max compression occur in this example? does it stay compressed to the max even after the collision?
I can't answer your question since I don't know what other examples you are referring to. It might be different, or you used a different method at getting the answer. There are usually more than one way of getting to same answer so it's hard to say, and if you get the same answer, then your method is also just as valid as what's shown here. I only showcase a single method. For max compression, the train will couple when it hits another car, and the spring will stay constantly compressed due to the coupling, which is why they both have the same velocity after they have the impact. I show this at 0:52 :)
Why is S_B/A in the last example 0.5 m, isn't that S_A because it originated from 0.5 m away from point O? I'm having trouble visualizing what the 0.5 m means in the diagram, it looks like it means the width of the triangle B. Thanks
So an easier way to think about it is to imagine the origin at block A. In that case, block B "moves" 0.5 m. It's really just the difference in the starting points of A and B, hence B/A.
can you solve that question A5kg mass moving with velocity 10m/s collides with 10kg mass moving with velocity of 7m/s along the same line . if the two masses join together on impact find their common velocity if they were moving
a) in opposite direction
b)in the same direction
Sorry but I don't solve questions like this because then I have to solve everyone's questions and it's not doable :( Please ask your professors/TAs during their office hours. You can also ask on forums for help.
On the last problem at 7:25, I think it should be V a/b because we know from B's perspective A moved .5 meters. We don't know yet what b/a is. If we get it wrong in a case would it make a difference if we keep our signs correct? I can't figure that out. Thanks for your hard work.
Did you see this video first? ua-cam.com/video/opVKNCedkRo/v-deo.html
The whole answer will change if you use a/b vs b/a. It's very important you distinguish which you use because the equation changes based on it. If you have the time, please watch that video.
3:43 We can also determine the distance using energy method right? Rather than determining the time (t).
You can solve these questions in more than one way, sometimes, it might even be easier/faster to use another method, I just show this method because it pertains to the topic we are talking about. 😅
Thanks you for huge inf video you made sir.
You're very welcome!
Bro , really you are so amazing thank you ❤
You're very welcome! ❤
In the last example, why is the small box’s velocity pointed tot the left? And not pointed in the direction of the 30 degree plain? I find it difficult to see that.
Take note of the notations used. It shows V_Ax, which represents the x-component of the velocity, which will be parallel to the x-axis, in this case, the grey base.
thank U soooo much
You are very welcomeeee.
For problem 2, using linear impulse and momentum, would it be correct if you instead did s=integral (t to 0) 3.196-1.962t dt? And then set s=0 and solve for time? Or would you have to find the bounds of integration for time?
You need the bounds of the integral to solve the problem. Otherwise, you will have final distance as a variable and final time as a variable.
2:08 if this is an inelastic collision, why do we use the conservation of energy if energy isn't conserved in an inelastic collision?
You are confusing different topics together. When you need the coefficient of restitution, then you will need to use that equation. Here, it's only the conservation of energy, so in this topic, we are excluding things like noise, heat, friction...
hi really helpful big thanks...but i have i question when we use (linear impulse and momentum or conservation of energy or others) cuz i really confused.
If you look at the 2 equations given, you can see when to use it. Impulse has to do with time, so if time is given, you can use it with an equation of linear impulse and momentum. If it's just velocity and mass, you can use conservation of momentum. 👍 The more questions you do for each topic, the faster you will realize which equations to use in what situations :)
@@QuestionSolutions really thx...im sorry ,but I'm going ask another question (in the example number 2 there is no time?) it just masses ,velocities and foresees...and you used linear impulse and momentum
thank you
@@slimlady9923 Don't be sorry about asking questions, the more you ask, the more you learn. I was able to use impulse and momentum because time would get cancelled out. So notice how I started off with a conservation of energy equation because we had mass and velocity. After that, we still didn't get the answer we need, so we then switch to a linear impulse and momentum equation. You can also notice that we are looking for distance, and distance is related to speed, which is related to time. So you sort of have to work through these steps in your mind and see what can be done. Even if you take the wrong path, you will see that you're not getting the value you need, so you can retrace your steps and come back to an equation that works. In more advanced problems, you will definitely have to use both to get to an answer. This is just a tip, and in no way a solid factor, but in most final exams, the questions your professor will create will require you to use at least 2 of the main equations, if not more to arrive at an answer.
@@QuestionSolutions thanks and sorry again , wish me luck i have an exam
@@slimlady9923 Really, you don't have to be sorry for asking questions. Feel free to ask, and I'll do my best to explain. I wish you the best of luck with your exam! 👍
For finding the distance on the second part of the problem can't you just use the principle of work and energy which is easier than first finding t. I did that and got 4.03m which is the same answer.
You can use many different methods to arrive at the answer. I just use the methods pertaining to the topic. There are many ways to get to the same answer with these problems, some are much easier than others, some take the long route, but I try to stick to whatever is relating to a chapter, otherwise, it's sort of pointless to do a video on that topic 😅
@@QuestionSolutions true. Thanks for these videos. They are great. Im using them to study for the FE.
@@Robtmmartine77 You're welcome, I wish you the best!
Thanks for your great videos. Note that after I calculated v = 3.96 m/s (as you did), I then applied the following simple suvat equation to determine the stopping distance: final velocity squared = initial velocity squared + 2*acceleration (negative, due to the friction force)*s. Note final velocity squared = 0. My answer is 4.036 m, which is very close to your answer of 3.996 m. It looks like my method is simpler and quicker in this case. Was I correct in applying this formula, or was it a coincidence that I got the correct answer (or one very close to the correct answer)?
There is usually multiple ways to get to the same answer, I only showcase what's pertaining to a certain chapter. But in general, there are tons of ways to get to it. 👍
I did the same method as you did and got 3.996m. I think you rounded off the acceleration too much.. I used a = -1.962
Hello. once again a excellent video. I have a question: in the last problem, could we say that the normal force exerted by B in A as a contrary force exerted by A in B, because of newtons third law? and thus we would apply the similarity of triangles and discover the X component of this force. then we would find out that that component is in fact the total force aplied in B ( the forces in Y axis cancel), therefore we would need to calculate the distance A did ( by applying trignometry ) and with the kinematics equations for A discover the time it took A to reach the end. With that value of time and the value for acceleration of B ( applying newtons second law, we have the total force and the mass) we plug in the values in the kinematics equations for B and discover the distance he covered in the X axis in that time period. The problem is, if you do so, you get a absurd value of 2.02m. what did i do wrong?? Thank you and keep up with the great job.
That seems like a lot of steps to get to this answer, doesn't it? I mean was your method faster? I have no idea 😅 Again, I don't know what you did wrong without seeing your steps :) With this method, I think you can solve the problem easier and faster than going through a lot of trigonometry and all the other parts you did. Though I do like how you always try to solve the questions in different ways, since there is always more than 1 way to solve them. The only problem is that sometimes, there aren't enough values given, or some just don't help out.
@@QuestionSolutions they were indeed a lot of steps... i suppose your method is much better but i did not see it crystal clear at first.. nevertheless i continue to be confused on what i did wrong, because something must be wrong with my way of thinking.. i dont know how i could show you just like i said in the other comment. but thank you a lot!
Thanks so much for posting these!
A question about the second question. I used the equations of motion in y to find the Frictional force, then I used them again in x to find acceleration. Plugged it into the kinematic equation a ds = v dv, integrated and solved. I got the correct answer but I was wondering if it's another valid method or just a fluke. Thanks again!
More often than not, there is more than one way to solve a problem. You used kinematics to solve the problem, which works as well. Try to solve it in both ways so you learn the new concepts as well :)
yup got the same. You have done correctly
On first thought, I thought about using conservation of energy and that also worked well👍🏻
Thanks
You're very welcome!
What would change if there was friction in between A and B?
Please give me a timestamp so I know where you're referring to. Thanks!
@@QuestionSolutions No need anymore, I just finished the class!
nice explanation (:
Many thanks!
why did we use linear ımpulse and momentum equation at 4.25 question. why didnt we use t1+v1 = t2+v2 ? how will we distinguish?
You can solve these questions anyway you like, but this video is about linear momentum, so that's what's shown here. Usually, if time is given, it's going to involve impulse.
@7:14 why can't we proceed by conservation of energy change in potential energy of block A = change in K.E of block B + change in K.E of block A ??
Did you try it with that method? So the goal in this video is to use conservation of linear momentum, which is what I showcase. But you can use any method you like to get to an answer, since there are multiple ways to solving the same problem.
@@QuestionSolutions yes but there is variation in ans...i am getting 0.095m according to your solution it is 0.07m.
@@vigneshwaranrm6277 Then I am thinking maybe there is a numerical error or a part is missing on your equation? I am not sure, it's hard to say since I didn't solve it with conservation of energy. 😅 See your professor during office hours and show him your method so he can check over your work.
Please make videos on complex streeses
Normally, complex stresses aren't covered in 1st year courses, so I didn't make a video on it. But I will keep it in mind for the future.
@@QuestionSolutions thank you,and your explanation is great
Wouldn't it be easier to solve the second question using the work energy equations after finding N , since it directly provides the horizontal displacment.
You can solve these problems in so many ways and get to the same answer 👍
Is it be possible in the last question to find the speed of B when the box reaches the ground? If yes. how is it possible? Thank you
Yes, it's possible, along with the equation already shown on the video, you'll need one more, an equation of work and energy, or conservation of energy. Because right now, we only have 1 equation with 2 unknowns. That should allow you to find the speed of B when the box reaches the ground.
Great
Thanks!
4:16 in the y-direction why weight is not negative?
It should be negative, good catch, it's a typo :)
solved the second question using conservation of momentum & then work energy principle got the same distance
You can use many different methods to arrive at the same answer. 👍
@@QuestionSolutions yea true. thanks for the reply, great videos & content tbh
@@ElCrankoPunko You're very welcome!
4:44 why is the force of the bullet not in this equation?
What do you mean by force here? Are you referring to energy instead? Work?
@@QuestionSolutions like friction has a force in the principle equation but the force applied by the bullet isn’t in the equation. I’m assuming it’s because this moment is already after the impact
why we did not put zero for velocity directly at 4:59
We needed to find the velocity time equation to figure out the distance travelled in the next step.
Thanks a lot sir
@@deyardeyar8163 You're very welcome!
7:40 why is S B/A= 0.5? Isn’t that the displacement of B with respect to A? Why isn’t it 0.5+SB?
So we are only interested in the relative displacement, so in other words, where "A" is with respect to "B". Another way to think about it is to realize that in the horizontal plane, "A" moves 0.5m when it slides down to the bottom.
Ohhhhh that makes sense. Thank you for your help!
@@迪安-z6i You're very welcome!
good
👍👍👍👍
Hi, mmm what is the text?
Any book used is always listed in the description.
👍👍
👍👍