OCR M3 (non MEI) Wednesday 8th June

2/7 seconds from suvat. t = root(0.8/9.8)

3/4 of a period so 3/4 * 2pi/root98 = 0.476 s

the first bit from 0.075 above to 0 = sin-1(0.075/0.225)/root98 = 0.034s

most likely something wrong, how did you do it?

One of the was the linear motion with variable forces question, the other was either impact question or SHM question


Posted from TSR Mobile

One of the was the linear motion with variable forces question, the other was either impact question or SHM question


Posted from TSR Mobile

and do you remember the variable forces result? I remember getting a minimum and maximum velocity of like -3.5 and +11.5 or something

I used a suvat to find the time for Q to reach P, which was t=2/7Then I used x=0.225cos(wt), where w=7root(2)with x=-0.075 to find the time to reach the bottom, which gave t=0.193and x=-0.1 to get to the point where it is no longer in SHM, which gave t=0.205v^2=98(0.225^2-0.1^2) to find the velocity leaving SHM, which gave v=2.00Then another suvat to find the time until it is instantaneously at rest, which gave t=0.204The sum of the above is t=0.888

Soz I meant the elastic thingy question, yeah I got those numbers as well


Posted from TSR Mobile

do you remember what question 2 and 3 were on?

Think Q2 might've been m=0.2, F=3cos(2t).

Think the final answer was 8.77m/s, think Vmax was 11.5 and Vmin was -3.5.

Q3 was P between two strings, max KE was 1.04 I think, x=1/6 where x is extension of left string. Couldn't easily be wrong

that's exactly what I got... I messsed up the one with the tension in a circle going slack though, on the part where you gotta find the range of U

Yeah it was tricky I messed that up slightly, didn't write a lower bound for U so that it reaches the horizon. And in part i I got cos(theta) is 2/3 but then manged to get the height wrong somehow! Tricky business

I'd say I'm on around 57/72.. enough for 80 UMS do you think?

I used a suvat to find the time for Q to reach P, which was t=2/7

Then I used x=0.225cos(wt), where w=7root(2)

with x=-0.075 to find the time to reach the bottom, which gave t=0.193

and x=-0.1 to get to the point where it is no longer in SHM, which gave t=0.205

v^2=98(0.225^2-0.1^2) to find the velocity leaving SHM, which gave v=2.00

Then another suvat to find the time until it is instantaneously at rest, which gave t=0.204

The sum of the above is t=0.888

I reckon thats about 80-85 UMS yh, maybe more, A* has been as low as 57

I know this probably sounds weird but how did everyone's classmates find it? I was the only person doing it at my school so didn't really get the perspective. I don't really think people on the StudentRoom is a fair sample of how difficult people found the paper generally haha!

I know this probably sounds weird but how did everyone's classmates find it? I was the only person doing it at my school so didn't really get the perspective. I don't really think people on the StudentRoom is a fair sample of how difficult people found the paper generally haha!

I know this probably sounds weird but how did everyone's classmates find it? I was the only person doing it at my school so didn't really get the perspective. I don't really think people on the StudentRoom is a fair sample of how difficult people found the paper generally haha!

woudnt the last question involve a phase shift, so a extra constant in the sin bracket since it didnt start at the extreme extension?

i got the amp and other things wrong so il only get about 6 marks total for last question but im pretty sure you needed a phase shift,and im pretty sure it was the first time its ever came up on ocr