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M2 Collisions Confusion!
This one's caught me out and I'm not sure how to even make a start on it.. I tried using some equations of motion and thinking about K.E. but then I get lots of u's and v's and can't seem to make them cancel. Anyway, here goes:
A small smooth ball bearing falls from rest at height H above a smooth horizontal plane. The coefficient of restitution between the ball bearing and the plane is e (e < 1). Show that the total distance travelled by the ball bearing before coming to rest is:
[(1+e^2) / (1-e^2)]H
Rep for any help - TSR has saved me before
A small smooth ball bearing falls from rest at height H above a smooth horizontal plane. The coefficient of restitution between the ball bearing and the plane is e (e < 1). Show that the total distance travelled by the ball bearing before coming to rest is:
[(1+e^2) / (1-e^2)]H
Rep for any help - TSR has saved me before
Right, this is tricky. remember, e = speed of seperation/speed of approach
Ball falls from height h so travels an ititial h. Since it falls from rest its speed as it hits plane is sqrt(2gh) [using]
therefore its speed of seperation is e*sqrt(2gh)
we calculate the height it reaches above the plane to be he^2, so ball falls a total of 2he^2 before hitting the plane again.
Since the ball is now falling from he^2 , its new speed of approach is
e*sqrt(2gh)
therefore its speed of seperation is e^2*sqrt(2gh) , and it reaches a height of he^4 on 2nd bounce, and hence travels 2he^4 before hitting plane again.
This will continue indefinitely to give the infinite sum
Total Distance = h + 2he^2 + 2he^4 + 2he^6 + ...
= h + 2h[ e^2 + e^4 + e^6 + ... ]
= h + 2h[ e^2/(1 - e^2) ] using sum of a Geometric series
= (h - he^2 + 2he^2)/(1 - e^2)
= h(1 + e^2)/(1 - e^2)
as required
Ball falls from height h so travels an ititial h. Since it falls from rest its speed as it hits plane is sqrt(2gh) [using]
therefore its speed of seperation is e*sqrt(2gh)
we calculate the height it reaches above the plane to be he^2, so ball falls a total of 2he^2 before hitting the plane again.
Since the ball is now falling from he^2 , its new speed of approach is
e*sqrt(2gh)
therefore its speed of seperation is e^2*sqrt(2gh) , and it reaches a height of he^4 on 2nd bounce, and hence travels 2he^4 before hitting plane again.
This will continue indefinitely to give the infinite sum
Total Distance = h + 2he^2 + 2he^4 + 2he^6 + ...
= h + 2h[ e^2 + e^4 + e^6 + ... ]
= h + 2h[ e^2/(1 - e^2) ] using sum of a Geometric series
= (h - he^2 + 2he^2)/(1 - e^2)
= h(1 + e^2)/(1 - e^2)
as required
use v^2 - u^2 = 2as again. with u = e*sqrt(2gh) and a = -g
then s = -2ghe^2/-2g
then s = -2ghe^2/-2g
Don't worry about these sort of questions in an exam! They usually come in many parts with hints along the way, for example a) what is the coefficient of restitution in terms of speeds of seperation/approach? b) how high does the ball bounce after first hitting the plane? c) what distance has it travelled after n bounces etc.
Original post by Faaip De Oiad
Right, this is tricky. remember, e = speed of seperation/speed of approach
Ball falls from height h so travels an ititial h. Since it falls from rest its speed as it hits plane is sqrt(2gh) [using]
therefore its speed of seperation is e*sqrt(2gh)
we calculate the height it reaches above the plane to be he^2, so ball falls a total of 2he^2 before hitting the plane again.
Since the ball is now falling from he^2 , its new speed of approach is
e*sqrt(2gh)
therefore its speed of seperation is e^2*sqrt(2gh) , and it reaches a height of he^4 on 2nd bounce, and hence travels 2he^4 before hitting plane again.
This will continue indefinitely to give the infinite sum
Total Distance = h + 2he^2 + 2he^4 + 2he^6 + ...
= h + 2h[ e^2 + e^4 + e^6 + ... ]
= h + 2h[ e^2/(1 - e^2) ] using sum of a Geometric series
= (h - he^2 + 2he^2)/(1 - e^2)
= h(1 + e^2)/(1 - e^2)
as required
Ball falls from height h so travels an ititial h. Since it falls from rest its speed as it hits plane is sqrt(2gh) [using]
therefore its speed of seperation is e*sqrt(2gh)
we calculate the height it reaches above the plane to be he^2, so ball falls a total of 2he^2 before hitting the plane again.
Since the ball is now falling from he^2 , its new speed of approach is
e*sqrt(2gh)
therefore its speed of seperation is e^2*sqrt(2gh) , and it reaches a height of he^4 on 2nd bounce, and hence travels 2he^4 before hitting plane again.
This will continue indefinitely to give the infinite sum
Total Distance = h + 2he^2 + 2he^4 + 2he^6 + ...
= h + 2h[ e^2 + e^4 + e^6 + ... ]
= h + 2h[ e^2/(1 - e^2) ] using sum of a Geometric series
= (h - he^2 + 2he^2)/(1 - e^2)
= h(1 + e^2)/(1 - e^2)
as required
Could u please explain the last 2 steps
Original post by Lahiru7
Could u please explain the last 2 steps
Please don't bump a really old thread. Start a new thread with your questions.
Original post by Vunce
How did u calculate the height after the 1st bounce
Please start another thread with your question, bumping a old thread won’t help.
Original post by Vunce
How did u calculate the height after the 1st bounce
Original post by Black Water
Please start another thread with your question, bumping a old thread won’t help.
Original post by DFranklin
If the velocity before the bounce is v, the velocity after the bounce is scaled by a factor e. Height reached is proportional to velocity^2, so the new height will be scaled by a factor e^2
I'm prepared to defer to a mod on this, but although the general rule of not bumping old threads is a good one, in a case like this where someone wants to discuss the original problem (rather than adding a comment directed to someone from 7 years ago), I'm not sure starting a new thread is helpful.
I'm prepared to defer to a mod on this, but although the general rule of not bumping old threads is a good one, in a case like this where someone wants to discuss the original problem (rather than adding a comment directed to someone from 7 years ago), I'm not sure starting a new thread is helpful.
Ok thank you. I’ll keep this in mind for next time.
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