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# Mechanics 3 - edexcel Watch

1. This book http://www.amazon.co.uk/Edexcel-Leve.../dp/0435519182

on m3 I think has a few values wrong in one of the examples I would like clarification with.

Chapter 2 - page 32, Example 13, the Question reads:

A light elastic spring, of natural length 1m, and modulus of elasticity 10N, has one end attached to a fixed point A. A particle of mass 2kg is attached to the other end of the spring and is held at a point B which is 0.8m vertically below A. The particle is projected vertically downwards from B with speed 2 ms -1. Find the distance it falls before first coming to rest.

This is the books solution for the example:

KE loss + PE loss = E.PE gain

0.5 x 2 x 2^2 + 2g(0.2 + x) = 20/2 *x^2 - [ (20 x 0.2^2) /2 ]

What I don't get, is that the question said mod elasticity is 10N, so why use 20??!?!

also now the working out:

4 + 3.92 + 19.6x = 10x^2 - 0.4

0 = 10x^2 - 19.6x - 7.52 ( I have severe problems with the rearranging of this equation as I thought -0.4 - 4 - 3.92 = -8.32?????????? )

it goes on to solve the quadratic to get x = 2.288 or -0.328

but the problem is that more than 1 value has been mistaken for. And the KE+PE losses = EPE Gain equation has a change in sign when rearranging the equation into a quadratic on the RHS.

CAN SOMEONE PLEASE POST THE CORRECT VALUES TO THIS QUESTION AND THE CORRECT METHOD I WOULD GREATLY APPRECIATE IT AS I AM SELF TEACHING THIS MODULE AND WOULD LIKE TO HAVE SOME CORRECT FIGURES TO WORK WITH!!

THANKS SO MUCH!!!
2. (Original post by Coral Reafs)
This book http://www.amazon.co.uk/Edexcel-Leve.../dp/0435519182

on m3 I think has a few values wrong in one of the examples I would like clarification with.

Chapter 2 - page 32, Example 13, the Question reads:

A light elastic spring, of natural length 1m, and modulus of elasticity 10N, has one end attached to a fixed point A. A particle of mass 2kg is attached to the other end of the spring and is held at a point B which is 0.8m vertically below A. The particle is projected vertically downwards from B with speed 2 ms -1. Find the distance it falls before first coming to rest.

This is the books solution for the example:

KE loss + PE loss = E.PE gain

0.5 x 2 x 2^2 + 2g(0.2 + x) = 20/2 *x^2 - [ (20 x 0.2^2) /2 ]

What I don't get, is that the question said mod elasticity is 10N, so why use 20??!?!

also now the working out:

4 + 3.92 + 19.6x = 10x^2 - 0.4

0 = 10x^2 - 19.6x - 7.52 ( I have severe problems with the rearranging of this equation as I thought -0.4 - 4 - 3.92 = -8.32?????????? )

it goes on to solve the quadratic to get x = 2.288 or -0.328

but the problem is that more than 1 value has been mistaken for. And the KE+PE losses = EPE Gain equation has a change in sign when rearranging the equation into a quadratic on the RHS.

CAN SOMEONE PLEASE POST THE CORRECT VALUES TO THIS QUESTION AND THE CORRECT METHOD I WOULD GREATLY APPRECIATE IT AS I AM SELF TEACHING THIS MODULE AND WOULD LIKE TO HAVE SOME CORRECT FIGURES TO WORK WITH!!

THANKS SO MUCH!!!
In example 13, there's a misprint in the question - it should say modulus of elasticity of 20N. (I also self-taught this module and corrected my copy of the textbook.)

The rearrangement is wrong, as you said.

So replace 7.52 with 8.32 wherever it appears. Their method is otherwise correct.

Then x=2.33 (or -0.359)

So distance fallen=2.52 (this still equals 2.5m to 2sf)

Posted from TSR Mobile
3. (Original post by Claree)
In example 13, there's a misprint in the question - it should say modulus of elasticity of 20N. (I also self-taught this module and corrected my copy of the textbook.)

The rearrangement is wrong, as you said.

So replace 7.52 with 8.32 wherever it appears. Their method is otherwise correct.

Then x=2.33 (or -0.359)

So distance fallen=2.52 (this still equals 2.5m to 2sf)

Posted from TSR Mobile

In terms of the answers, I got the very same

Thanks again!!
4. Mechanics 3 Edexcel

Help! It's less than 30 days til the exam and i just can't for the life of me get my head around the answer to chapter 5 exercise D question 4 of the Edexcel textbook! The question is: http://tinypic.com/r/2rpa4cg/8

The answer they give says take moments about point A when the cone is on verge of topple as the friction and weight will act from/through A. This I get! But then! They take moments about A and somehow get both the horizontal and vertical components of P but with different distances as well! I am so confused and would be so grateful to anyone who could help! The working the disc gives is:

http://tinypic.com/r/2h3at5s/8

Thanks so much for helping! I think I'm going out of my mind on this question!
5. The moments bit might be confusing you depending on how you've put the P force into it's components.
This is not a full solution, but this picture might help get you unstuck.
6. (Original post by Mr T Pities You)
The moments bit might be confusing you depending on how you've put the P force into it's components.
This is not a full solution, but this picture might help get you unstuck.
Could you look at Exercise 3F Question 10, and for part c, explain to me how the amplitude is 0.3? I will attach the CD solution in a word document, it's near the bottom of the page and I can't get my head around how they worked out the amplitude.
Attached Files
7. M3 Further Dynamics.docx (133.9 KB, 47 views)
8. (Original post by Coral Reafs)
Could you look at Exercise 3F Question 10, and for part c, explain to me how the amplitude is 0.3? I will attach the CD solution in a word document, it's near the bottom of the page and I can't get my head around how they worked out the amplitude.
The string has a natural length of 1.2m, but it's been stretched between A and B, which are 2m apart. If the partical was just left alone, how far along AB would it be? It's not actually been left alone, so where was it released from?
Did you draw a diagram in your solution? This is the single most important thing in mechanics modules IMO.
Hope that helps!

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