I think the rotational moments questions were easier, but I would think that the other questions were a bit harder than usual, grade boundaries will probably be roughly the same.

Reply 149

Nikulp68

1

Original post by talentedlobster

It is. I got Mr²/2 + Mh²/3 though.

That's the one ¡

Reply 150

username1162972

18

Original post by Euclidean

A cylindrical shell is a hollow cylinder with no caps on right?

I got 1/3 •M•h^2 but I got so confused

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I just did it with ends aswell tbh. **** knows. Did both ways.
I definately dropped marks tbh. The simple pendulum i am pretty sure i got wrong ,i found stable equilibrium pint so where the COM was then did some next geoemtry moments

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Reply 151

username1162972

18

Original post by Bolts

Did people find that easier than last years?

Na it was harder i predict 2 marks lower then last yrs boundaries.

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Reply 152

Bolts

2

Original post by physicsmaths

I just did it with ends aswell tbh. **** knows. Did both ways.
I definately dropped marks tbh. The simple pendulum i am pretty sure i got wrong ,i found stable equilibrium pint so where the COM was then did some next geoemtry moments

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For the COM did you have it a/2 from the bend to the right and a/2 up?

Reply 153

username1162972

18

Original post by Bolts

For the COM did you have it a/2 from the bend to the right and a/2 up?

Yep.
I made a mistake in MI question did
MI about diameter of disc is 1/2mr^2 😂.
Lol cares, got my 85.

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Reply 154

Euclidean

11

Original post by physicsmaths

I just did it with ends aswell tbh. **** knows. Did both ways.
I definately dropped marks tbh. The simple pendulum i am pretty sure i got wrong ,i found stable equilibrium pint so where the COM was then did some next geoemtry moments

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I think I f****d up. :facepalm:

Anyone have any numerical answers?

Q1: -6 i + 25/2 j

Q2: cos (t) i - 2sin (2t) j

Q3: (-3, 0, 1) but this was different to the resultant x the vector they give so I probably messed up

Q4: Really messed up

Q5: Assuming this is the variable mass

v = u•ln (5) - 19.6

Q6: a) omega = 3J/2mga I think
b) T = 20pi/3 • root {g/5a}

Q7: b) Y = 7/8 mg
But I think I was supposed to take 3mg/8 away from mg/2 not add it... FFS

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Reply 155

Euclidean

11

Original post by Mathemagicien

I recognise these numbers

right so I probably got 4/7 right then 😂

Praying for more method marks than accuracy marks :redface:

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Reply 156

Nikulp68

1

Original post by Mathemagicien

I did, although of course maybe I just made loads of mistakes :biggrin:

Hmm, I got Mr^2/4 + Mh^2/3

You did remember that M.I. of circle about its diameter is mr^2/4, not mr^2/2? I remember I initially made that mistake

It was hollow though so you need the MI of a circular arc about its diameter which is mr^2/2

Reply 157

username1162972

18

Original post by Nikulp68

It was hollow though so you need the MI of a circular arc about its diameter which is mr^2/2

It was completely hollow? I don't think it was.
It was either both ends covered or none since hey said about 1 end implying that it is symmetric.

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Reply 158

Nikulp68

1

Original post by physicsmaths

It was completely hollow? I don't think it was.
It was either both ends covered or none since hey said about 1 end implying that it is symmetric.

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Shoot. My bad. How many marks would that be? Marks lost, that is.