Wouldn't this be above the horizontal but still a push
nope draw the horizontal line from the start of the force, not the end
if your force was acting such that it had a vertical component acting in an opposite direction to the reaction force, it's wrong - I don't think ocr will allow two different answers on account of ambiguity because it very clearly states 'above the horizontal'.
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.
Uh. For the last part of Q6, I added 68 to 54 and got 112 and then divided by 2 to get 61 :/ I didn't think the negative sign on the velocity mattered.
I can go through it if someone has the question, the whole paper was posted before but I think it's been deleted, or at least I can't find it.
I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2
Then the questions were Find the velocity of P when t = 3 Find the displacement of OP when t = 3
I think it was to do with particle P passing through a point O and the velocity of the particle when it passes O is 2ms^-1 and the acceleration at the time it passes O is (4+12t)ms^-2
Then the questions were Find the velocity of P when t = 3 Find the displacement of OP when t = 3
That sounds familiar (i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m. (ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.
What kind of grade do you reckon 55 marks would be? Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
That sounds familiar (i) To get from acceleration to velocity, you integrate (as acceleration is change in velocity over time, a=dv/dt so to get back to v to integrate). So integrating 4+12t gives us 4t+6t2+c. When t=0, v=2 as it states in the question. Therefore 2=4(0)+6(0)2+c => c=2. You end up with v=4t+6t2+2. At t=3, sub in 3 to get v=4(3)+6(3)2+2 = 68m. (ii) To get from velocity to displacement, you integrate again. Integrating 4t+6t2+2 gives you s=2t2+2t3+2t+c. When t=0, s=0 so c=0. Sub in t=3 to get s=2(3)2+2(3)3+2(3) = 78m.
Ah thank you, it all makes sense now, I forgot the plus C coefficient :///
What kind of grade do you reckon 55 marks would be? Also, can someone please go through question 7iv)b. because I got 2.32 and I'm not sure where/why I went wrong
Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii). (a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N. (b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N. Not sure about the grade sorry!
Starting from (a) because it's easier. Using deceleration = 0.9 from part (iii). (a) F=ma for P, gives T-0.3gsin30=0.3*-0.9 => T=0.3*-0.9+0.3gsin30 => T=1.2N. (b) Tension must be constant throughout the string, so using F=ma for Q => 0.4gsin30-T-Fr=0.4*-0.9 => Fr=0.4gsin30-1.2-0.4*-0.9 (using T=1.2N from above) => Fr=1.12N. Not sure about the grade sorry!
Ooooh thank you! What a ridiculous mistake to make damn
I would just like to thank all of you who have helped me over the past few days. I have got from a U to a somewhat good grade. Wasn't the biggest fan of M1 as our teacher wasn't very good.
Also, can someone please do Q4 because I got 13.3 and 25.1 degrees I think. How did people get like 10 or something?
4) i. I used the cosine rule: R^2 = 10^2 + 6^2 -2 x 10 x 6 x cos(70) which gives R = 9.74 (3 s.f) then I used the sine rule to find the angle: sin(x)/6 = sin(70)/9.74 which gives x = 35.4 (3 s.f) 4) ii. I did: 20 - 9.74 = 10.3 (3 s.f) because the particle is at rest so I figured the resultant force must be acting directly upward (90 degrees above the horizontal) 4) iii. I then did: 90 - 35.4 = 54.6