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M2 Work done question: where did I go wrong?

creativebuzz

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Reply 1

nerak99

The gravitational force and the resistance act in the same direction and hence the RHS is added with +1.24 rather than -1.24. I think that is where you have gone wrong. What is the correct answer? (between 3 and 4?)

Reply 2

Zacken

Original post by creativebuzz

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Surely the word done against resistance is simply

8 \times 1.24

and not

-8 \times 1.24

?

Remember that gravitational potential energy is just "work done against gravity" and yet you still took that as positive (which you should) as you should take "work done against resistance" as positive.

Reply 3

Zacken

Original post by nerak99

The correct answer would be

v = 12 \, \text{ms}^{-1}

if I'm not wrong.

Reply 4

nerak99

Original post by Zacken

The correct answer would be

v = 12 \, \text{ms}^{-1}

if I'm not wrong.

That seems a bit high? We have increased the force by more than half and with no extra force v<16

Do you mean root 12?

(edited 7 years ago)

Reply 5

creativebuzz

Original post by nerak99

That seems a bit high? We have increased the force by more than half and with no extra force v<16

Do you mean root 12?

No, Zacken is spot on!

Reply 6

nerak99

Original post by creativebuzz

No, Zacken is spot on!

Yep I forgot the 0.1 and got root 14.4. Still, surprised that in 8m at mg+1.24 we only get down to 12 from 20.

(edited 7 years ago)

Reply 7

creativebuzz

Original post by Zacken

Surely the word done against resistance is simply

8 \times 1.24

and not

-8 \times 1.24

Ah of course! I got the answer correct now, thanks :smile:

I feel really bad but would you mind giving me a hand on these questions! I've been doing really well on M2 so far (got 100% last time) but I just did the June 2013 Reserved paper and it killed me!

For part b, why is incorrect to use s=dt by finding the distance travelled in 2 secs and then doubling it (as when t=2, the ball turns around)

For part B, why did they put the normal reaction coming out of the cyclinder in the same direction as the reaction from the floor? Surely the normal reaction should be coming out diagonally as it's supposed to be perpendicular to the cylinder?

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6a (Hardest question in the whole thing) but for this question the mark scheme found moments when surely it's supposed to be treated like a lamina so find I split the trapezium into two mini triangles and a rectangle and found the CoM and area of them etc

ucles sending.jpg

Part 7d, I just don't see where I went wrong

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Reply 8

Zacken

Original post by creativebuzz

Ah of course! I got the answer correct now, thanks :smile:

For this - you need only sketch the graph:

You know that the area between the velocity time graph and the time axis is the distance travelled. The area between 2 and 4 is clearly much smaller than the area between 0 and 2. Hence you can't just double it. You can only double it if the graph was symmetrical about about t=2.

Had they asked you to find the distance between "2 and 4" then you could have done distance from 2 to 3 and then doubled it because the graph is symmetrical about t=3. But this is not so for "0 to 4".

Reply 9

Zacken

Original post by creativebuzz

Look at the diagram properly. The reaction force does come out diagonally/perpendicular to the cylinder, as you've rightly said; but in their diagram, they've drawn two arrows resolving that perpendicular/diagonal force horizontally and vertically. I've zoomed it up for you:

As you can see, they've split the normal perpendicular force into the horizontal and the vertical component.

Reply 10

Zacken

Original post by creativebuzz

I would do it that way as well (edit: to clarify - I would do it the way you have), I'm not quite sure what went wrong with your method?

(Also, the attachments for the rest of your questions don't seem to be showing up)

(edited 7 years ago)

Reply 11

ghostwalker

Original post by creativebuzz

It's far easier, IMO, to consider the masses of the given triangles and the trapezium.

mass x centre of mass of large triangle = mass x cofm small triangle + mass x cofm trapezium.

Taking measurements from the top vertex.

Reply 12

creativebuzz

Original post by ghostwalker

I agree and I've literally tried this question 10 times and still can't get the right answer..

Share2016-04-26-b7d70b7fe4bea1262dbdfc0c645423ad912b1d17f43d7deff5e1010e2a2707e9-Picture.jpg

Share2016-04-26-b7d70b7fe4bea1262dbdfc0c645423ad912b1d17f43d7deff5e1010e2a2707e9-Picture.jpg

Reply 13

ghostwalker

Original post by creativebuzz

I agree

Why not do it that way then?

and I've literally tried this question 10 times and still can't get the right answer..

Share2016-04-26-b7d70b7fe4bea1262dbdfc0c645423ad912b1d17f43d7deff5e1010e2a2707e9-Picture.jpg

You have the MR of the triangles incorrect - should be half what you've got.

Not at home, so can't post again for a bit.

Reply 14

creativebuzz

Original post by ghostwalker

Why not do it that way then?

You have the MR of the triangles incorrect - should be half what you've got.

Not at home, so can't post again for a bit.

OMG YES I FINALLY GOT IT RIGHT! I redid that question soo many times, I can't believe my mistake was so small! Thank you

Reply 15

Zacken

Original post by creativebuzz

OMG YES I FINALLY GOT IT RIGHT! I redid that question soo many times, I can't believe my mistake was so small! Thank you

Well done.

Reply 16

creativebuzz

Original post by Zacken

Well done.

Thanks!

Did you spot where I went wrong in q7d?

Reply 17

Zacken

Original post by creativebuzz

Thanks!

Did you spot where I went wrong in q7d?

Like I said, your attachments aren't showing.

Reply 18

Hjyu1

What papers are these questions from ?

Reply 19

Zacken

I feel really bad but would you mind giving me a hand on these questions! I've been doing really well on M2 so far (got 100% last time) but I just did the June 2013 Reserved paper and it killed me!

...

Original post by Hjyu1

What papers are these questions from ?