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    (Original post by faxingstar)
    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
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    (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
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    (Original post by pettolrahc)
    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
    i think mine were t=8a and -t+4g=4a
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    (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?
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    (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)
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    Yeah I got that aswell, what did you work the acceleration out to be then? faxingstar
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    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.
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    (Original post by Alexw1812)
    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 must have got the sign wrong for a then lol. Rest of your answers seem same as mine
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    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
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    For part 7 did the forces = 7500a or was it 7500/G a
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    (Original post by Crozzer24)
    For part 7 did the forces = 7500a or was it 7500/G a
    7500/g as 7500 was in newtons
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    Guys can we make an unofficial mark scheme
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    (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.
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    (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
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    (Original post by pettolrahc)
    7500/g as 7500 was in newtons
    Knew it lol there's another 4 marks lol
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    (Original post by Ohnis)
    That's what I got as well :| I didn't realise it was wrong
    I think 4m is correct?
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    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?
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    Can anyone remember what the angle and force were in question 1?
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    (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?
    Yeah I did the same. Perfectly good method
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    in the last 4 marker, sin(alpha) was 6.75/root(16^2+6.75^2) and sin(beta)was 4/5? then a was (8000sin(beta)+6000sin(alpha)-7500)*9.8 all over 7500?
 
 
 
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