Official M1 OCR 2016 Thread

Didn't you have to consider the fact that it was inclined? Meaning that g was acting at an angle?

Yeah. So when released,
Force on Q = 0.2g - Tension

Force on P = Tension - Weight component - Friction
= T - 0.2gsin30 - 0.4 x 0.2gcos30 = T - 1.659

P and Q have the same mass and the same acceleration, so the forces acting on them are equal, hence:

T - 1.659 = 0.2g - T
2T = 0.2g + 1.659

T = 1.81 N

Reply 41

duncant

2

what did everyone get for the coalesce particles having maximum velocity? 6.75m/s? part iii

Reply 42

alborghetti

Anyone got the unofficial mark scheme

Reply 43

drandy76

13

Original post by ixguard

Yeah. So when released,
Force on Q = 0.2g - Tension

Force on P = Tension - Weight component - Friction
= T - 0.2gsin30 - 0.4 x 0.2gcos30 = T - 1.659

P and Q have the same mass and the same acceleration, so the forces acting on them are equal, hence:

T - 1.659 = 0.2g - T
2T = 0.2g + 1.659

T = 1.81 N

Same as me fam

Posted from TSR Mobile

Reply 44

AlfieH

2

Original post by duncant

what did everyone get for the coalesce particles having maximum velocity? 6.75m/s? part iii

I got that then changed it to 4.75.

Reply 45

swagmister

11

Original post by duncant

what did everyone get for the coalesce particles having maximum velocity? 6.75m/s? part iii

I got that too but I don't know if it's right

Reply 46

ixguard

1

Original post by AlfieH

I got that then changed it to 4.75.

me also

Reply 47

Karmen99

5

Original post by ixguard

Yeah. So when released,
Force on Q = 0.2g - Tension

Force on P = Tension - Weight component - Friction
= T - 0.2gsin30 - 0.4 x 0.2gcos30 = T - 1.659

P and Q have the same mass and the same acceleration, so the forces acting on them are equal, hence:

T - 1.659 = 0.2g - T
2T = 0.2g + 1.659

T = 1.81 N

Got the same for the force on P but I considered P for before release as well, which i probably wrong... Thank you anyway! 😊