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    This exam just became my "great enemy" 😐
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    I t the wrong values of cos(a) and cos(b) and worked out horizontal acceleration using F-r=ma which I manipulated to equal 0 .... hopefully that will at least gain method marks...
    In 7 where did the 4500 come from ??
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    (Original post by Connorbwfc)
    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 did something similar, i worked out the new value for T and then used Pythagoras with 8000 then -7500 and got a as 4.04 i think

    Hopefully we get some error carried forward marks or something
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    (Original post by Connorbwfc)
    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.
    I did X= the top row
    Y= the bottom row

    I rearranged X's eqn to make T the subject then sub T in in the Y eqn and it rearranged to the eqn in the question x
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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. 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 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.
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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)
    Yeah I got 0ms-2 for acceleration and thought I was wrong because 0 is an odd answer but my confidence is boosted slightly if you got the same!
    Got the same answer for the window question (stone hit window) .

    I also got u=8 and a=-2 in the other question. Can't remember AB though!
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    (Original post by Connorbwfc)
    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 think I got the acceleration as this... What did we do wrong 😣
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    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...
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    (Original post by thelion786)
    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'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
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    What do you think the grade boundaries will be ??
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    UNOFFICIAL MARK SCHEME DOC HERE:

    https://docs.google.com/document/d/1...it?usp=sharing

    Please contribute!!
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    (Original post by tizzaclaire98)
    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...
    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
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    (Original post by alicereading99)
    What do you think the grade boundaries will be ??
    Like 55 for an A
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    (Original post by MilindS99)
    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..
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    https://docs.google.com/document/d/1...it?usp=sharing


    ^ editable mark scheme ^
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    (Original post by tizzaclaire98)
    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
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    (Original post by MilindS99)
    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
    Same...
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    (Original post by otherdan)
    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
    That looks about right
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    https://docs.google.com/document/d/1...QgSNcDOQ/edit#
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    I swear they didn't ask for u when t=0, just the expression in terms of t??
 
 
 
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