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OCR MEI M1 (Mechanics) 17th June

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i can only remember that i got the angle to be 30?

I got the angle of 30 too but i struggles with the first parts of the question, i think my equations were: 4*9.8-T=4a and 4*9.8+T=8a and then got a as 12. something

Overall I didn't like this exam especially the beginning of section A and the end of section B

Reply 41

Crozzer24

Original post by Akashbabuta

I thought it was 6m as v=0 when it changes direction no?

To be brutally honest I think I've done it wrong, I used s = 12 + AB, u = 8, t = 4 and a = 2

Reply 42

faxingstar

Original post by pettolrahc

i think mine were t=8a and -t+4g=4a

Reply 43

pettolrahc

Original post by faxingstar

i think mine were t=8a and -t+4g=4a

they seem right- i think i must have gone wrong, maybe i'll get ecf marks for the acceleration?

Reply 44

Alexw1812

Original post by Crozzer24

I think I got distance as 20m, took t as 4, because it took 2 seconds to get to A and 6 seconds to get back to A.

I'm pretty sure u was 8, a was -2 and AB was 4m. He did hit the window (0.975m from ground) frictions was 12.5N, acceleration was 49/15 for a couple of them. Resultant acceleration when d=6.75 was 0ms-2. (Just some answers I could remember, ask me if you know the question you got stuck on and I'll probs remember the answer)

Reply 45

h.xlx

Yeah I got that aswell, what did you work the acceleration out to be then? @faxingstar

(edited 7 years ago)

Reply 46

otherdan

What I can remember:

1. Frictional force = 12.5N

2. u = 8 m/s, a = -2 m/s/s
AB = 4m

3. a = 3.27 m/s
theta = 30 degrees

4. a = (0, -8) u = (2,6)
t=2, v = (2, -10) = 10.19 m/s
Rearrange to sub x into y equation for show that.

5. Did not reach pigeon since height reached (3.27m) less than 4.0m
Did hit window since 0.925m at x = 22.5m is in 0.8 to 2.0m range

6. Model A estimate = 320m (not in range of local data)
Derived from s = ut + 1/2at^2 with u = 0 and a = 10
Model B estimate = 275m (not in range of local data)
Model A assumed constant acceleration up to t=8; model B assumed terminal velocity reached at t=5
Model C estimate = 158.33m (in range of local data)
Models B and C both assumed terminal velocity reached; B assumed initial constant acceleration, C did not.

7. cos alpha and beta = 4/5 and 3/5
6000 * 0.8 = 8000 * 0.6 = 4800 therefore horizontal equilibrium so no horizontal acceleration
Use horizontal equilibrium; Tcosalpha = 8000cosbeta; rearrange
Use Pythagoras to use d^2 in formula instead of cos alpha/beta for show that.
Vertical acceleration = 3.27 m/s/s
When d = 6.75, vertical a = 0 (since total of vertical components of tensions were 7500 - equal to weight)
Cannot be in equilibrium at P since no vertical component of tension but weight of bomb.

(edited 7 years ago)

Reply 47

Crozzer24

Original post by Alexw1812

I must have got the sign wrong for a then lol. Rest of your answers seem same as mine

Reply 48

pettolrahc

what was all the compare this to the local records bit about on the mine question? thought i'd signed up for maths not writing 3 essays on comparisons

Reply 49

Crozzer24

For part 7 did the forces = 7500a or was it 7500/G a

Reply 50

pettolrahc

Original post by Crozzer24

For part 7 did the forces = 7500a or was it 7500/G a

7500/g as 7500 was in newtons

Reply 51

Synmatic

Guys can we make an unofficial mark scheme

Reply 52

username1357089

Original post by otherdan

What I can remember:

1. Frictional force = 12.5N

2. u = 8 m/s, a = -2 m/s/s
AB = 4m

3. a = 3.27 m/s
theta = 30 degrees

4. Don't fully remember the question.

5. Did not reach pigeon since height reached (3.27m?) less than 4.0m
Did hit window since 0.925m at x = 22.5m is in 0.8 to 2.0m range

6. Model A estimate = 320m (not in range of local data)
Derived from s = ut + 1/2at^2 with u = 0 and a = 10
Model B estimate = 275m?
Model A assumed constant acceleration up to t=8; model B assumed terminal velocity reached at t=5
Model C estimate = 158.33m
Models B and C both assumed terminal velocity reached; B assumed initial constant acceleration, C did not.

7. cos alpha and beta = 4/5 and 3/5
6000 * 0.8 = 8000 * 0.6 = 4800 therefore horizontal equilibrium so no horizontal acceleration
Use horizontal equilibrium; Tcosalpha = 8000cosbeta; rearrange
Use Pythagoras to use d^2 in formula instead of cos alpha/beta for show that.
Vertical acceleration = 3.27 m/s/s
When d = 6.75, vertical a = 0 (since total of vertical components of tensions were 7500 - equal to weight)
Cannot be in equilibrium at P since no vertical component of tension but weight of bomb.

I agree with these. I found the second part of the 'show that' question in 7 difficult, although someone has tried to explain it since, I thought the paper was OK if I'm honest. Question 7 was the most challenging part for me.

(edited 7 years ago)

Reply 53

Ohnis

Original post by Qwertykeyboard15

I got distance of AB to be 4m... I completely messed the first question up 😣

That's what I got as well :| I didn't realise it was wrong

Reply 54

Crozzer24

Original post by pettolrahc

7500/g as 7500 was in newtons

Knew it lol there's another 4 marks lol

Reply 55

username1357089

Original post by Ohnis

That's what I got as well :| I didn't realise it was wrong

I think 4m is correct?

Reply 56

prepdream

For the pigeon question, I set like 4=ut+1/2at^2, tried to solve it and said that there is no real root for the equation so it cant reach the pigeon? How do you guys solve this one?

Reply 57

Connorbwfc

Can anyone remember what the angle and force were in question 1?

Reply 58

Crozzer24

Original post by prepdream

For the pigeon question, I set like 4=ut+1/2at^2, tried to solve it and said that there is no real root for the equation so it cant reach the pigeon? How do you guys solve this one?