# M3 questionWatch

This discussion is closed.
#1
30. A light elastic string of natural length a and modulus 7mg has a particle P of mass m attached to one end. The other end of the string is fixed to the bsae of a vertical wall. The particle P lies on a rough hoirzontal surface, and is released from rest at a distance (4/3)a from the wall. The coefficient of friction between P and the surface is (1/4).
Use the work-energy principle
(a) to show that P will hit the wall
(b) to find, in terms of a and g, the speed of P when it hits the wall.

For (a) I showed that when the extension in the string is zero, the velocity of P > 0. Then I tried to solve that to get the value of v for (b), but I can't get the right answer (v = 1/3 rt[ga])....
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14 years ago
#2
(Original post by shift3)
30. A light elastic string of natural length a and modulus 7mg has a particle P of mass m attached to one end. The other end of the string is fixed to the bsae of a vertical wall. The particle P lies on a rough hoirzontal surface, and is released from rest at a distance (4/3)a from the wall. The coefficient of friction between P and the surface is (1/4).
Use the work-energy principle
(a) to show that P will hit the wall
(b) to find, in terms of a and g, the speed of P when it hits the wall.

For (a) I showed that when the extension in the string is zero, the velocity of P > 0. Then I tried to solve that to get the value of v for (b), but I can't get the right answer (v = 1/3 rt[ga])....
I'll try to do this without a sketch. Note it's always a good idea to sketch though.

Work against friction = Force x Distance = (1/4)(mg)(4a/3) = (1/3)mga
KE gain = 0.5mv^2
EPE loss = (7mg)(a/3)^2 / 2a = (7mg/2a)(a^2 / 9) = 7mga/18.

0.5mv^2 + (1/3)mga = 7mga/18
0.5mv^2 = 7mga/18 - mga/3
0.5v^2 = 7ga/18 - ga/3
v^2 = 7ga/9 - 2ga/3
v^2 = (1/9)(ga)
v=(1/3)rt[ga]

Are you sitting the exam tomorrow too? Good luck if so =)
What are your favourite and least favourite parts of M3?
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#3
Thanks! I forgot to calculute work against friction and just put friction down... Is my reasoning correct for part a?

I am sitting it tomorrow. You too? Good luck to the both of us.
I like dynamics, kinematics and elasticity, but I really hate statics and SHM.
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14 years ago
#4
(Original post by shift3)
Thanks! I forgot to calculute work against friction and just put friction down... Is my reasoning correct for part a?
It would be if you factored in the work done but using the same method. As v^2 > 0 upon hitting the wall then there is a velocity and v is not imaginary or 0.

I am sitting it tomorrow. You too? Good luck to the both of us.
Same. I hope it goes well. I'd like a confidence booster before the rest of my exams. I've done lots of questions on unit so i'll be surprised if something comes up that i haven't seen.

I like dynamics, kinematics and elasticity, but I really hate statics and SHM.
I used to dislike all of statics but it's got better as i practiced. Now i mainly just dislike the more complex equilibrium questions such as a 3d object attached to a string, but i should be able to manage it.

I find it to mainly be about luck now! It all boils down to not making any silly mistakes in the exam. I find it useful where some of the final parts say 'show' or 'prove' a statement, which gives confidence or enables you to correct mistakes.

I'm hoping to gain some extra UMS to ease the pressure of June, where i take P2-P6 and M4

Best of luck!
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14 years ago
#5
for me its the circular motion
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#6
(Original post by Gaz031)
I find it to mainly be about luck now! It all boils down to not making any silly mistakes in the exam. I find it useful where some of the final parts say 'show' or 'prove' a statement, which gives confidence or enables you to correct mistakes.
Exactly! I really like it when the last part is a proof question, because I make LOTS and LOTS of stupid mistakes (like adding up wrong stuff, copying wrong numbers from my calculator, adding random negative signs... the works!). It's really cool to see that you got the entire question correct and gathered up 10+ marks. Unless of course you can't find your mistakes or don't know how to prove the statement.

I'm hoping it'll go smoothly as last June's was a bit tough.

And good luck with the rest of your modules! I'm almost doing the same ones.. P5-P6 & M4-M5 this June, and P4 a couple of weeks from now.
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14 years ago
#7
(Original post by amo1)
for me its the circular motion
Best of luck to you too
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14 years ago
#8
Exactly! I really like it when the last part is a proof question, because I make LOTS and LOTS of stupid mistakes (like adding up wrong stuff, copying wrong numbers from my calculator, adding random negative signs... the works!). It's really cool to see that you got the entire question correct and gathered up 10+ marks. Unless of course you can't find your mistakes or don't know how to prove the statement.
Precisely, the possibility of silly errors are the only things that annoy me about exams as they don't really indicate your true ability.

I'm hoping it'll go smoothly as last June's was a bit tough.
Was that the one with the skier? I think i did the past paper.

And good luck with the rest of your modules! I'm almost doing the same ones.. P5-P6 & M4-M5 this June, and P4 a couple of weeks from now
Thanks, have you done any M4 yet? How do you find it?
The other person at my school doing FM is taking P4 in Jan (though i couldn't for some reason). The problem is though that he hasn't properly covered P2 or done P3 yet =/ Hope he does ok though.
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#9
(Original post by Gaz031)
Was that the one with the skier? I think i did the past paper.
Yeah.

Thanks, have you done any M4 yet? How do you find it?
I've finished it. It's easier than M3, that's for sure. Although I'm not really big on the SHM chapter.
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#10
I've got another question:
A particle of mass m moves along the positive x-axis. It is acted upon by a force directed towards the origin and of magnitude mn²x. Write down an equation governing the motion. Given at time t=0 the particle is at the origin and that the maximum distance of the particle from the origin is a, find x in terms of a, n and t.
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14 years ago
#11
(Original post by shift3)
I've got another question:
A particle of mass m moves along the positive x-axis. It is acted upon by a force directed towards the origin and of magnitude mn²x. Write down an equation governing the motion. Given at time t=0 the particle is at the origin and that the maximum distance of the particle from the origin is a, find x in terms of a, n and t.
mass m, f=-m(n^2)x as the force is directed towards the origin.
f=ma
-m(n^2)x = ma
a = -(n^2)x. Hence the motion is simple harmonic.
t=0, x=0. xmax=a.
x=asinnt as t=0, x=0, amp=a.
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