M3 Pre-Exam thread (edexcel)

out of curiosity, how did you get a copy?

I may have to kill him if he doesnt

okay just 7

a particle B of mass 0.5kg is attached to one end of a light inextensible string of natural length 0.75m and modulus of elasticity 24.5N the other end of the string is attached to a fixed point A the particle is hanging in equilibrium at the point E vertically below A
show that AE=0.9m (3)

the particle is held at A and released from rest. The particle first comes to instantaneous rest at C
find AC (5)

show that while the string is taught there is SHM (4)

calculate the maximum speed of B (2)

Having it straight after c3 didnt help

Lollol, worried now since I got 5/7 for the k question.


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the angle between AV and the verticle through V can be found by tan-1(1/2) so the angle between the initial verticle through V and the verticle wen it is being hung is 45-tan-1(1/2). the tan of this angle is 1/3. so u now have a right angled triangle with the opposite as the horizontal distance of the centre of mass from O (the bottom of the hole shape) and the adjacent as the distance of the centre of mass from V. by calling the horizontal xbar and the vertical ybar u can work out expressions for these two things, so xbar over ybar =1/3 and then solve for k. hope thats clear enough :P

This is my worked solution for 6b) Can anyone confirm whether or not this matches the answer on the paper?

You had it right on the bottom line, it should stay as 2a/(3root3 - pi)

But it does cancel. 27 + pi^2 = (3rt3 + pi)(3rt3 - pi). And the denominator was 3rt3 - pi.

They had the right answer then rationalised the denominator which I then realised, if you look at what you quoted

Okay, thank you. Although now I'm just frustrated that I didn't get to do it in the exam