The Student Room Group

M3 Dilemma (Edexcel A2)

A light elastic string AB has natural length l and modulus of elasticity 2mg. A second light elastic string CD has natural length l and modulus of elasticity 4mg. The end B of the first string is connected to the end C of the other, and the end A is attached to a fixed point. A particle of mass m is suspended from end D and is at rest in equilibrium. Find the length AD.

My Working:

l of AB = l of CD = l
1/(combined spring constant)=1/k(AB)+1/k(CD)
k=lambda/l
Let lambda equal the combined modulus of elasticity
1/k=l/lambda=l/4mg+l/2mg
l cancels
1/lambda=1/4mg+1/2mg
1/lambda=1/4mg+2/4mg=3/4mg
lambda=4mg/3

Resolving vertically:
Let T be the combined tensions in the spring combination
T-mg=0; T=mg
mg=(4mg/3)*x/2l
mg=4mgx/6l
6lmg=4mgx
mg cancels
6l=4x
x=1.5l
Therefore the total length is 2l+1.5l=3.5l

Markscheme says 11l/4. How come?
Reply 1
Hi, sorry this isn't related to OP's question, but what does "unit of mass" column mean in this table? (ch.5 statics)
Original post by Nuclear Ghost
A light elastic string AB has natural length l and modulus of elasticity 2mg. A second light elastic string CD has natural length l and modulus of elasticity 4mg. The end B of the first string is connected to the end C of the other, and the end A is attached to a fixed point. A particle of mass m is suspended from end D and is at rest in equilibrium. Find the length AD.

My Working:

l of AB = l of CD = l
1/(combined spring constant)=1/k(AB)+1/k(CD)
k=lambda/l
Let lambda equal the combined modulus of elasticity
1/k=l/lambda=l/4mg+l/2mg
l cancels
1/lambda=1/4mg+1/2mg
1/lambda=1/4mg+2/4mg=3/4mg
lambda=4mg/3

Resolving vertically:
Let T be the combined tensions in the spring combination
T-mg=0; T=mg
mg=(4mg/3)*x/2l
mg=4mgx/6l
6lmg=4mgx
mg cancels
6l=4x
x=1.5l
Therefore the total length is 2l+1.5l=3.5l

Markscheme says 11l/4. How come?


I think you have the wrong idea of where the strings are. Have you drawn a diagram?
Original post by Orand
Hi, sorry this isn't related to OP's question, but what does "unit of mass" column mean in this table? (ch.5 statics)


"Unit of mass" would be better described as "mass ratio". You can (in this example) cancel out rho x pi x 36 x 2 from each of the masses before you put it all into an equation (this is perfectly acceptable with Edexcel, I don't know about the other boards), which makes your working much simpler. You will often be able to cancel out rho x pi, even if you don't notice the numbers.
Original post by Orand
Hi, sorry this isn't related to OP's question, but what does "unit of mass" column mean in this table? (ch.5 statics)


yh we were never told to put the rho in most of the time as it cancelled every time(in m2)


Posted from TSR Mobile
Original post by tiny hobbit
"Unit of mass" would be better described as "mass ratio". You can (in this example) cancel out rho x pi x 36 x 2 from each of the masses before you put it all into an equation (this is perfectly acceptable with Edexcel, I don't know about the other boards), which makes your working much simpler. You will often be able to cancel out rho x pi, even if you don't notice the numbers.



Original post by physicsmaths
yh we were never told to put the rho in most of the time as it cancelled every time(in m2)


Posted from TSR Mobile


The only time that you must put the density in is if different parts of the object have different densities.
Original post by tiny hobbit
The only time that you must put the density in is if different parts of the object have different densities.


What module would that come into


Posted from TSR Mobile
Original post by physicsmaths
What module would that come into


Posted from TSR Mobile


I've seen it in both M2 and M3 questions for Edexcel (not sure if it was in an exam or in their textbooks).

You just need to put, say, rho and 2 rho with the appropriate parts.
Original post by tiny hobbit
I've seen it in both M2 and M3 questions for Edexcel (not sure if it was in an exam or in their textbooks).

You just need to put, say, rho and 2 rho with the appropriate parts.


ph when somethings folded etc. I was getting lost with what you said. Ive done those questions in m2 before


Posted from TSR Mobile
Reply 9
Original post by tiny hobbit
"Unit of mass" would be better described as "mass ratio". You can (in this example) cancel out rho x pi x 36 x 2 from each of the masses before you put it all into an equation (this is perfectly acceptable with Edexcel, I don't know about the other boards), which makes your working much simpler. You will often be able to cancel out rho x pi, even if you don't notice the numbers.


Thanks Hobbit :smile: I got my teacher to explain it today so it makes sense. Are you 100% about leaving out rho in calculations (i know it cancels in the end) since the textbook examples all include it.
Original post by Orand
Thanks Hobbit :smile: I got my teacher to explain it today so it makes sense. Are you 100% about leaving out rho in calculations (i know it cancels in the end) since the textbook examples all include it.


I am for Edexcel, because I have marked both M2 and M3 in my time. If the mass ratio is correct, that's all that matters.

Quick Reply

Latest