OCR MEI M1 (Mechanics) 17th June

I did this too. Did you get a value of 4.01 for a?

I think we will probably get 2/4 marks. What do you reckon?

Do people agree with these answers to Q4?

4. a=(0, -8). u= (2, 6).
t=2 speed=10.19ms^-1.
y=3x-x^2 proven.

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.

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)

I did this too. Did you get a value of 4.01 for a?

I think we will probably get 2/4 marks. What do you reckon?

I disagree with your answer to Model C, I think. Would you not have integrated the first equation and found the distance travelled in the first 5 seconds, and then add on the distance travelled at terminal velocity, which was 25*3? I got 233.3m.

That paper was actually the worst thing I've ever seen... Totally couldn't get Q5 and Q7 took me like half the time. Ran out of time before I could get back to Q5. Did anyone get 0.002 for the acceleration when d=6.75 bit? I did but had no time at all to check anything... So annoyed at myself right now - made loads of silly little mistakes all the way through going back and sorting them took half the time...

What do you think the grade boundaries will be ??

exactly same as me, i got 0.0027 for a (2nd last q) and didn't have time to answer the last 2mark explanation. the last q was really hard and took me so long, still didnt finish

I spent about half an hour working with cosbeta and cosalpha values that were wrong - used the 8000N side rather than using pythag! Worked it out but took so long and ran out of time..

Yeah that proof 4 marker took me a good 10 minutes, and i spent way too long trying to explain why ABC are similar and different, so i ran out of time at the end... first ever exam where i havent finished

That's what I did I think:

In first 5 seconds: v = 10t - t^2
Integrates to: s = 5t^2 - 1/3 t^3 (+c)
This gives s = 250/3 = 83.33 when sub in limits t=5 and t=0

Then add on the 25*3 = 75 to get 83.33 + 75 = 158.33m