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# M3 strings watch

1. This is the question

One end of a light elastic string of natural length a and mod of elastic 3mg is fixed to a point A and the other end carries a particle P of mass m. The particle is held at A and then projected vertically downwards with a speed of root(3ga). Find the distance AP when the acceleration of the particle is instantaneously zero.

I've done

KE at A = 0.5m3ga
PE at A = mg(a+x). X is extension of string as far as it will go.

EE at X = 3mg(squared)/(2a)

KE + PE = EE

Leading eventually to 5asqured + 2ax = 3xsqured.

X = 5a/3

The answer in the book is 4a/3

Where am I going wrong please?
2. I have done this quickly on Paint and I believe you are right. Another magical instance when the textbook answers are wrong!
3. (Original post by maggiehodgson)
This is the question

One end of a light elastic string of natural length a and mod of elastic 3mg is fixed to a point A and the other end carries a particle P of mass m. The particle is held at A and then projected vertically downwards with a speed of root(3ga). Find the distance AP when the acceleration of the particle is instantaneously zero.

I've done

KE at A = 0.5m3ga
PE at A = mg(a+x). X is extension of string as far as it will go.

EE at X = 3mg(squared)/(2a)

KE + PE = EE

Leading eventually to 5asqured + 2ax = 3xsqured.

X = 5a/3

The answer in the book is 4a/3

Where am I going wrong please?
You appear to be finding out when the speed is zero. Energy considerations will not help you find out about the acceleration. You need to use Newton's second law (and Hooke's law). The answer in the book is correct.

(Original post by Aspire7736)
I have done this quickly on Paint and I believe you are right. Another magical instance when the textbook answers are wrong!
Perhaps you shouldn't be quite so quick to assume that you are right and the book is wrong (yes I know there are errors in there, which makes people uncertain about who is right).
4. (Original post by tiny hobbit)
You appear to be finding out when the speed is zero. Energy considerations will not help you find out about the acceleration. You need to use Newton's second law (and Hooke's law). The answer in the book is correct.

Perhaps you shouldn't be quite so quick to assume that you are right and the book is wrong (yes I know there are errors in there, which makes people uncertain about who is right).
I didn't know about the energy and acceleration thing so thanks.

So hooked law still applies when something is sent off at speed as well as when it just falls? From that I do get extension a/3 so AP is as required.

The next part of the question asks what is the maximum speed attained during its motion. I have assumed that it is when it drops to taut string but it isn't according to the book.

You've said that I need to use NSL but I don't know how. Could you please explain the motion. I think that it falls under gravity until the string is taut. The it slows down until it reaches extension and then I imagine that it bobs back up but what makes it do that? Oh dear.
Thanks
5. (Original post by maggiehodgson)
I didn't know about the energy and acceleration thing so thanks.

So hooked law still applies when something is sent off at speed as well as when it just falls? From that I do get extension a/3 so AP is as required.

The next part of the question asks what is the maximum speed attained during its motion. I have assumed that it is when it drops to taut string but it isn't according to the book.

You've said that I need to use NSL but I don't know how. Could you please explain the motion. I think that it falls under gravity until the string is taut. The it slows down until it reaches extension and then I imagine that it bobs back up but what makes it do that? Oh dear.
Thanks
NSL effectively came into the first part, as the acceleration was 0, so the Tension = weight.

The maximum speed occurs when the acceleration is 0. So you need to find the speed at the place that was the answer to the first part. To get a connection between speed and position, that's where energy comes in.
6. (Original post by maggiehodgson)
I didn't know about the energy and acceleration thing so thanks.

So hooked law still applies when something is sent off at speed as well as when it just falls? From that I do get extension a/3 so AP is as required.

The next part of the question asks what is the maximum speed attained during its motion. I have assumed that it is when it drops to taut string but it isn't according to the book.

You've said that I need to use NSL but I don't know how. Could you please explain the motion. I think that it falls under gravity until the string is taut. The it slows down until it reaches extension and then I imagine that it bobs back up but what makes it do that? Oh dear.
Thanks
So you're now trying to find the speed when AP = 4a/3, call this place B

Use KE at A + GPE at A = KE at B + EPE at B

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