OCR M2 19th May 2014

Original post by Niamh45

Wow that means we must have got those 12 marks

or it means we all dropped those marks :wink:

Reply 62

Bloxorus

Original post by Niamh45

For V I got 15.9 and then for the direction of motion I got that it was going down by like 14 degrees

I got this too :smile:

Reply 63

Original post by Niamh45

Really?! Im so happy, I thought I would be the only one who got 14, thanks

I checked both of the answers a couple of ways, so I'm pretty much completely sure they're right! :smile:

Reply 64

can someone post the unofficial markscheme for this paper? :biggrin:

Reply 65

Kimberley123

My answers:

20.2 degrees
26.5
2.30 cm
0<= u < 0.384
7.43 cm
6.52 degrees
16.5N
13.2
u= 0.841
Prove
3.60 seconds
A 1.04 m/s and B 1.36 m/s
18.1m J
M = 0.25kg
2.49N
1.94m/s
Prove
16.5 degrees and 2.? N
14.2 degrees below the horizontal
V= 15.9

I can't remember all the units and they might not be in the completely right order :smile:

Can anyone else confirm?

(edited 9 years ago)

Reply 66

Original post by Kimberley123

My answers:

20.2 degrees
26.5
2.30 cm
0<= u < 0.384
7.43
6.52
16.5
13.2
0.841
Prove
3.60 seconds
A 1.04 m/s and B 1.36 m/s
18.1m J
0.25kg
2.49N
1.94m/s
Prove
16.5
14.2 degrees below the horizontal
V= 15.9

I can't remember all the units and they might not be in the completely right order :smile:

Can anyone else confirm?

I just put u<tan(21) so I might drop a mark or 2 there but they look pretty similar to what I got. With the change in energy I didn't round my answer but Idk if they'll penalise that.

Reply 67

Original post by Kimberley123

Can anyone else confirm?

I got all the same (I think) other than 3 (if it's the one I'm thinking of - centre of mass -) I got 6.3cm from the line AE.

Reply 68

Original post by alow

I got all the same (I think) other than 3 (if it's the one I'm thinking of - centre of mass -) I got 6.3cm from the line AE.

i got 7 point something oh dear

Reply 69

ofhaofodso

the value of M was also 4, for the collision question

Reply 70

Original post by AMDD

i got 7 point something oh dear

I might have not got that. How did you calculate the CoM of the triangle lamina? I just used

\frac13

the distance from BD to C (then obviously put all that into the overall equation.

(edited 9 years ago)

Reply 71

Original post by ofhaofodso

the value of M was also 4, for the collision question

I got 0.25kg (so did Kimberley 123 by the looks of it).

Reply 72

Original post by alow

I got all the same (I think) other than 3 (if it's the one I'm thinking of - centre of mass -) I got 6.3cm from the line AE.

I got 7.43

Reply 73

Original post by alow

I might have not got that. How did you calculate the CoM of the triangle lamina? I just used

\frac13

the distance from BD to C (then obviously put all that into the overall equation.

yeah i did a third of the height which was 12 i think from BD and added that to 8 to get the distance of the cofm of the triangle from AE

Reply 74

Original post by alow

I got 0.25kg (so did Kimberley 123 by the looks of it).

i got 0.25 kg

Reply 75

Original post by furthermathsman

I got 7.43

i think i may have got that too

Reply 76