Connected particles : Greatest height reached by a particle when the other hits the ground
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- čas přidán 22. 10. 2013
- Tutorial on how to find the maximum height reached by a particle when the other hits the ground when they are connected particles.
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This was a beast of a question
@Leonel Axl Oh wow
It's so simple when you explain it; breaking it down into bit size chunks really helps my old brain. Sincere thanks
Very clear and understandable. This is how it’s supposed to be. Thanks
THANKS!! Missed this lesson in college!
Hi,
I love this video. I was just curious how do you know particle Q would hit the pulley but not simply stay in middle air ?
Many thanks.
thank you so much for this video!! you are a legend!!! plz keep posting exam solutions , and I'd be quiet thankful if you answered me , when do we take the component of weight in negative ??? that was in January 2007 q#7 , part f , but I just don get it
thank you, i was struggling to get this concept in my head!!!
You are amazing.. such smooth answer .. thanks a lot
Hi Sir, I appreciate your work so so much and its helping me a lot in my As level mechanics. I am aiming for an A and with your help I am very sure that i can achieve my goal thank you so so so much
MsTehnoobworld Good luck and I hope you get there. Just stick at it and believe in yourself and anything is possible. Let us know how you get on.
@@ExamSolutions_Maths I love you
wow ik its been 6 years but how did it go
How’d ur As exam go
OMG.This was so helpful. thank you sir.
No problem - pleased to hear you found it useful. Best wishes.
Helped me a lottt.thanksss
thanks so much
Why is the acceleration for Q ‘-g’ and not -3/7g?
The string had now gone slack, no tension so it was if you like disconnected from Q and now moving freely under gravity, so acceleration = -g. I hope that makes sense.
wonder how did you make the animation. many thanks
Hi, why does the particle move beyond the first blue line ? Is this is maximum height reached by the particle under free gravity after the other particle has hit the ground? So, when the particle moves back down to the blue line, is that when the string becomes taut again? Thank you.
thank you so much
Great teacher
when you change the direction of positive for Q does why does that not affect your calculations. What instances can you change it?
What will be the acceleration of the particle at inst rest on a smooth inclined plane ??
Hi sir, Thanks a lot for all your videos! I have literally watched every video in your mei playlist on c1234, s12, m12, and fp12. And theyve literally saved my life.LOL! Anyway i was wondering why does the speed that p hits the ground at equals the speed that q rises??? I would really appreciate if you answer.
+Mohammed D It is because they are connected by a non elastic string. They must always travel at the same speed whilst connected to a taut string.
Makes sense. i guess that is why they keep the case where the string is elastic till mechanics 3 since its harder
Surely the string won’t go slack and Q won’t rise further if it’s supposed to be ‘inextensible’?
The Clear Summer inextensible initially when the particles are at rest
Do you know what inextensible means lol
Why are the speeds Vp and Uq the same? I though the speeds on both sides of pulleys are always different?
The speeds are the same but the velocities are different. Velocity is a vector quantity with speed as it's magnitude.
at around 12:45, why does one consider the acceleration to be '-g' and ignore the previous value of acceleration of the particles which was 3/7(g) or 4.2 m/s^2?
Is it just because it's the acceleration due to gravity as it's downwards?
Sorry to bother but why do you find v?
Sir I would to know if the question is asking in this way..how high can a particle rest inside a hollow sphere of radius a and coefficient of friction is one upon a square root of three?
This is best asked here facebook.com/groups/mathsrevision.examsolutions/
thank you sir
No problem. Pleased it helped.
I am an older guy with a physics background doing the Alevels for college. Would it qualify as a full mark solution in the exam answering this problem using Energy Conservation Theorem ? @ExamSolutions
I would have thought so unless the exam board specifically asked for a non energy solution.
hi,
i was wondering why the 7m was not added to the three?, dosnt it ask for the greatest height after it hits the ground? so would it not be 10m?
please help thanks
Ishak Ussen It says "Find the height that Q rises after P hits the ground" This is how much further does Q rise at the point when P hits the ground. Not how high is Q above the ground.
ExamSolutions oh right, i see. they are quite confusing when it comes to that part of the question. i was wondering if there were any hints to spot the diffrence? thanks
When considering Q, why is the initial velocity the same as the final velocity of P? (root6g)
Thanks
Ben Irwin Because the acceleration of the two particles are the same due to an inextensible string being used. If you do the SUVAT for particle Q, you'll get the same answer.
My guess before watching the video:
The reason the particle reaches a "maximum height" rather than stopping is because it had been accelerating upwards until the moment the larger mass hits the ground. The instant the large mass hits the ground, the smaller particle maintains its velocity upwards (Newtons 1st Law) and beings decelerate because a force equal to -gm is acting on it. Once the vertical velocity of the particle decreases 0 m/s, it has reached its maximum height.
If we can find the upwards accelerating of the particle and we know the height of the system above the ground, we can use suvat to find the velocity of the particle the moment the larger particle hits the ground. Then we can look only at the smaller particle and treat it as a regular suvat problem with g as the downward acceleration and the previously found maximum velocity as the initial velocity. We can set the final velocity to equal 0 and solve for the displacement using suvat.
There might be an easier way but this is how I would do it having never seen a problem like this before.
"It's not difficult but it's complicated" - someone, probably
I hate maths.
The Adjutor fuck you
Irbid GT fucka you
who else got 14 m
thank you so much for this video!! you are a legend!!! plz keep posting exam solutions , and I'd be quiet thankful if you answered me , when do we take the component of weight in negative ??? that was in January 2007 q#7 , part f , but I just don get it