Question

http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s14_qp_51.pdf Q3ii) Q3iii) Q5ii)

Answer 1

@IrishBoy123 Can you please help me?

Answer 2

yep

3 ii) and iii) can be dealt with by energy equations.

mgh
1/2 k x^2
1/2 m v^2

what did you get for 3 i) ??

Answer 3

Is this right for 3ii)
Energy at the start=Energy at the end
1/2(0.4)v^2+16(0.2)^2/2*0.8=-0.4g(0.4)+16(0.6)^2/2*0.8

Answer 4

in i) i got my answer as 0.2

Answer 5

iii) When the tension becomes slack shouldnt the particle return to its natural length?

Answer 6

Sorry the elastic string.

Answer 7

yes, when tension zero, there is no energy in string

the thing about this that stops this being a straight energy equation is the use of the word "modulus" of elasticity. are you comfortable with fitting that into an energy equation? i'd need to think a bit. Hooke's law is pretty limited, F = kx.

in engineering, for example, you'd want a "proper" equation relation stress to strain, not force to extension; because then it would have universal application.

but if you're happy, i see it as a play on conservation of potential (gravitational and elastic) potential and kinetic energy.

Answer 8

ok, have thought, k = M/L where k is usual Hooke's law, M is the modulus, and L is the 0.8

Answer 9

Yes shouldn't the modulus of elasticity fit into the energy equation in lambdax^2/2l(elastic energy).

Answer 10

Why should we use hook's law here?

Answer 11

how are you calculating the elastic energy stored in the string?

that's all :p

Answer 12

E=λx^2/2l
x=extension λ= modulus of elasticity l=natural length

Answer 13

that makes sense to me

Answer 14

Before the string becomes slack the elastic energy was 16(1.4-0.8)^2/2*0.8 and a potential energy of -0.4g(0.4) is this right

Answer 15

Okay what about Question 5ii) i cant visualize it well in my head

Answer 16

really sorry , dude.

am running around like a loon at the moment. the trick for that last problem is to exploit the symmetry. the CoM, whether frame or actual lamina, lies right in the middle. that will make the math so much easier. if you are using formula for CoM, convert both shapes into point masses and spin them around an axis and take it from there. work all that out and get the 22deg figufe from there.

i can get back tomorrow, but now i must run.

as for the earlier energy stuff, i have not done the arithmetic, but i see how you are approaching it.

Answer 17

its okay,i figured it out. once your free can you please help with question 7 http://freeexampapers.automaticpapers.com/as-a-level/mathematics-9709/9709_s13_qp_51.pdf

Answer 18

well done.

and sure. i will be back in about 2-3 hours from now and can pick up then.

Answer 19

as before i have used a formula for the CoM

let me know if this helps......or if it's wrong :p

i know you know what you are doing so this is not feeding you an answer.

Answer 20

plus some scribbles on the other qu's but you have those licked already.