OCR MEI M1 (discussion)

How do you work out the last question, it really confused me! x

Also, for Q3 (Work out the values of y and z, and the acceleration and magnitude of acceleration? I didnt have a clue! x

This is what I think is right.

Because they were parallel, they were proportionate. This means the ratio between -2 and 3 is the same as that between -6 and y, or alternatively the ratio between -2 and -6 is the same as that between 3 and y. I.e. -6/-2 = y/3 or 3/-2 = y/-6. You can then rearrange and form similar equations to find y.

Use F = ma to find acceleration: add vectors up and divide by 5.

Magnitude is the square root of the squares of the vector's components.

This is what I did.

Essentially, you needed to treat the trolleys as a whole to find acceleration, using F = ma. Going gown the slope you had the two weights of the trolleys acting downwards, so drawing in the angles that you find 10gsin5 and 8gsin5. Going upwards you have 15N + 5N of resistance. The mass is 10kg + 8kg.

Then, with acceleration you can work out tension using two separate equations, which was acting up the slope for trolley D and down the slope for C.

Here's what I completely forgot to do because I am an idiot!

Once you have acceleration you can substitute into a suvat equation to find velocity after two seconds. s = ?, u = 3 ms-1, v = v ms-1, a = a ms-2 (worked out), t = 2s.

I believe I got 1.41ms-2 for the acceleration, and T = 3.89 N, but I can't swear to that. Had I not been so stupid and forgotten to found the velocity as it asked for in the question, I suppose I would have gotten 5.82ms-1.

Edit: actually that acceleration was for a different question, never mind!

What I did:
10(9.8)sin5+8(9.8)sin5-20=18a
That gives you a = -0.26ms^-2 as far as I remember.
Therefore using v=u+at => v=3-0.26(2) => v=2.48ms^-2

For all of you out there wondering how you got on, the answers to the paper are as follows:-

1) 18.6ms-1
2) u=-1.75ms-1, a=0.125ms-2]
3) y=9, z=-12
a=0.2i - 0.4j -k; magnitude = 1.1ms-1
4i) showing P, R (Normal Reaction) and W
4ii) 53.5N
5)Time of flight = 3.51s, Time of sound wave = 0.25s
Total time = 3.76s
6i) v = 4i - 2jt s = 4i t - 2jt^2 r = 3j + 4t i - t^2 j (or r = 4t i + (3 - t^2) j)
6ii) 141.3 degrees

7)
i) -10ms- 12(accel, not velocity)
ii) 20m and 13.8m
iii) -12ms-2
iv) -0.5ms-1
v) 12m

8)
i) 25N (sign not required as question asks for magnitude)
ii) 45.3N (sign not required as question asks for magnitude)
iii) 57.27N
iv) 1.41ms-2
v) v = 0.92ms-1
T = 3.86N (Thrust)
(a = -0.257ms-2 (up the slope)) (treat the system as a whole and use F=ma)
v = 2.49ms-1 (down the slope)
T = -3.89N or 3.89N in compression


Hope you all did well

i ****ed up q4 ii) + q8 v) =/ i ended up with a tension of 53N and v = 22.6 can anyone explain question 8v) to me?

Quite a few mistakes above, have corrected where necessary.

Quite a few mistakes above, have corrected where necessary.

Firstly apply F=ma parallel to the slope for the whole system:

F=ma
18g sin5 - 20 = 18a
=> a = -.257ms^-2

Then apply this acceleration to suvat:
a = -0.257, u=3, t=2, v=? (all in standard units)

v=u+at
v = 3 - 0.257 x 2
= 2.49ms^-1


To find tension you then then to consider F=ma for one of the objects. I used the 8kg object:

F=ma
T + 8g sin5 - 5 = 8 x _0.257
=> T = -3.89N

In my resultant force I assumed that that the bar was in tension (thus acting down the slope for the 8kg object), therefore my negative value means the force is acting up the slope, thus the bar is in compression.

Hope this is of help.

how did you get z=8? I thought it was -12?

EDIT: And shouldn't y=9?

http://www.onlinemathlearning.com/parallel-vectors.html

The examples given in this website are similar to this question

how did you get z=8? I thought it was -12?

EDIT: And shouldn't y=9?

http://www.onlinemathlearning.com/parallel-vectors.html

The examples given in this website are similar to this question

how did you get z=8? I thought it was -12?

EDIT: And shouldn't y=9?

http://www.onlinemathlearning.com/parallel-vectors.html

The examples given in this website are similar to this question

I definitely don't agree with your answer to 7iv) and also in question 2, if i did the right method but read the q wrong and thus got different answers how many marks do you reckon i would get?

checked and updated 7iv, oops (shouldn't have completed this late at night!)

Q2 you may get some method marks if you accidently misread a figure (e.g. 22.5 instead of 2.5) but applied SUVAT correctly with your incorrect/misread value. But if you confused u and v and have used incorrect signs it will be hard to gain marks in my opinion.

many thanks, made a honest mistake, have updated.

Also, for Q3 (Work out the values of y and z, and the acceleration and magnitude of acceleration? I didnt have a clue! x