I have this Further Mechanics question which my teacher didn't go into detail about so I don't entirely understand the method (and therefore how to apply it to other questions). I know it has something to do with the sum of finite / infinite geometric series.

A ball is dropped from a height hm. The coefficient of restitution between the ball and the ground is "e". What is the total distance travelled by the ball before it comes to rest permanently?

Any help would be greatly appreciated!

A ball is dropped from a height hm. The coefficient of restitution between the ball and the ground is "e". What is the total distance travelled by the ball before it comes to rest permanently?

Any help would be greatly appreciated!

Original post by TwisterBlade596

I have this Further Mechanics question which my teacher didn't go into detail about so I don't entirely understand the method (and therefore how to apply it to other questions). I know it has something to do with the sum of finite / infinite geometric series.

A ball is dropped from a height hm. The coefficient of restitution between the ball and the ground is "e". What is the total distance travelled by the ball before it comes to rest permanently?

Any help would be greatly appreciated!

A ball is dropped from a height hm. The coefficient of restitution between the ball and the ground is "e". What is the total distance travelled by the ball before it comes to rest permanently?

Any help would be greatly appreciated!

Just do restitution for each bounce and note that the speed leaving the ground after a bounce will be the same as the speed it hits the ground again on the next bounce, so its (speed and height) a geometric sequence which can be summed (infinite series). Try writing it down and upload if you get stuck.

Original post by mqb2766

Just do restitution for each bounce and note that the speed leaving the ground after a bounce will be the same as the speed it hits the ground again on the next bounce, so its (speed and height) a geometric sequence which can be summed (infinite series). Try writing it down and upload if you get stuck.

Would it be sum to infinity = h / 1 - e^2??

Original post by twisterblade596

Would it be sum to infinity = h / 1 - e^2??

Looks mostly right, though it would help to see your working. Im guessing youre modelling heights rather than distance travelled and you should be careful about the first drop as thats only 1/2 a normal bounce.

For stuff like this, a sketch (of a bouncing ball) really helps to make the arguments clear.

(edited 3 months ago)

This is a diagram I drew.

To consider the first height and the distance travelled by the ball, would the expression have to be changed to: h + 2 x he^2 / 1 - e^2?

To consider the first height and the distance travelled by the ball, would the expression have to be changed to: h + 2 x he^2 / 1 - e^2?

Original post by twisterblade596

This is a diagram I drew.

To consider the first height and the distance travelled by the ball, would the expression have to be changed to: h + 2 x he^2 / 1 - e^2?

To consider the first height and the distance travelled by the ball, would the expression have to be changed to: h + 2 x he^2 / 1 - e^2?

Looks right. Key things to note are that restitution gives

v^+ = e*v^-

so the speed after the bounce is e multplied by the speed before the bounce so it forms a geometric sequence with ratio e. Then as max height is h = v^2/2g, it forms a geometric sequence with ratio e^2. Probably slightly simpler to write the total distance travelled as

h(2/(1-e^2) - 1)

or

h(1+e^2)/(1-e^2)

but there isnt much in it.

(edited 3 months ago)

Original post by mqb2766

Looks right. Key things to note are that restitution gives

v^+ = e*v^-

so the speed after the bounce is e multplied by the speed before the bounce so it forms a geometric sequence with ratio e. Then as max height is h = v^2/2g, it forms a geometric sequence with ratio e^2. Probably slightly simpler to write the total distance travelled as

h(2/(1-e^2) - 1)

or

h(1+e^2)/(1-e^2)

but there isnt much in it.

v^+ = e*v^-

so the speed after the bounce is e multplied by the speed before the bounce so it forms a geometric sequence with ratio e. Then as max height is h = v^2/2g, it forms a geometric sequence with ratio e^2. Probably slightly simpler to write the total distance travelled as

h(2/(1-e^2) - 1)

or

h(1+e^2)/(1-e^2)

but there isnt much in it.

1.

Why did you multiply by h rather than adding h to the expression?

2.

Also, would the same principle apply when finding the total time take before an object comes to rest

3.

For these types of scenarios, when would you have to use the sum of finite geometric series rather than the sum of infinite geometric series? Are there any examples?

Original post by TwisterBlade596

OK thank you. I have a few questions though:

1.

Why did you multiply by h rather than adding h to the expression?

2.

Also, would the same principle apply when finding the total time take before an object comes to rest

3.

For these types of scenarios, when would you have to use the sum of finite geometric series rather than the sum of infinite geometric series? Are there any examples?

What do you mean by adding h to the expression instead of multiplying it?

Original post by TwisterBlade596

OK thank you. I have a few questions though:

1.

Why did you multiply by h rather than adding h to the expression?

2.

Also, would the same principle apply when finding the total time take before an object comes to rest

3.

For these types of scenarios, when would you have to use the sum of finite geometric series rather than the sum of infinite geometric series? Are there any examples?

1) The expressions are equivalent to the one you did. I simply used 2h as the iniital distance travelled, then subtracted h at the end (first one) and then simply combined into a single term (second one).

2) total time is obviously similar (zenos paradox). For each bounce you can use 2v = gt so it basically similar to the speed geometric sequence

3) You could use the finite series if you had to find the number of bounces in a certain time or how many bounces would occur if you travelled xxx or ....

Original post by π/2=Σk!/(2k+1)!!

What do you mean by adding h to the expression instead of multiplying it?

I meant adding h to the whole expression (as the initial drop didn’t represent a full bounce so it had to be considered differently) but I understand now.

Original post by mqb2766

1) The expressions are equivalent to the one you did. I simply used 2h as the iniital distance travelled, then subtracted h at the end (first one) and then simply combined into a single term (second one).

2) total time is obviously similar (zenos paradox). For each bounce you can use 2v = gt so it basically similar to the speed geometric sequence

3) You could use the finite series if you had to find the number of bounces in a certain time or how many bounces would occur if you travelled xxx or ....

2) total time is obviously similar (zenos paradox). For each bounce you can use 2v = gt so it basically similar to the speed geometric sequence

3) You could use the finite series if you had to find the number of bounces in a certain time or how many bounces would occur if you travelled xxx or ....

I get it now. Thank you so much!!

Original post by twisterblade596

I get it now. Thank you so much!!

NP. Something that twigged after the previous post was that if you had the velocity series so the sum of v^+ say (upwards velocity after each bounce), then you could get the total time by simply doubling that and dividing by g, rather than having to form a new sequence/series for time (with the small initial correction appropriately handled). Should be obvious from the algebra, but easy to overlook when setting up the problem and similar to the famous problem

https://www.flyingcoloursmaths.co.uk/two-trains-and-a-fly/

Purely for interest.

(edited 3 months ago)

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