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    Did anyone get 15N for the tension on the very last question
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    (Original post by chemari1)
    Was the coefficient of friction 1.57 and was the tension in one of the rods 394?
    I got something like 0.86 or something
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    On question 1, not sure what part, when it said why can't P be moving at half the speed in the opposite direction after the collision, how do you show it? It was 3 marks and I don't think I did enough or even did the right thing at all
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    (Original post by 11234)
    Did anyone get 15N for the tension on the very last question
    Yeah!
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    (Original post by Alex621)
    On question 1, not sure what part, when it said why can't P be moving at half the speed in the opposite direction after the collision, how do you show it? It was 3 marks and I don't think I did enough or even did the right thing at all
    I think you got e=7/3>1 but 0<e<1
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    Im so glad this is done and over with....
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    (Original post by 11234)
    I think you got e=7/3>1 but 0<e<1
    did you work out what the new final velocity of Q would be if P's velocity did half and change direction? at first I just recalculated e using the negative version of P's final velocty, but then I crossed it out and worked out the new velocity of Q. I can't remember what I got for e but it was bigger than 1 :/
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    (Original post by Alex621)
    did you work out what the new final velocity of Q would be if P's velocity did half and change direction? at first I just recalculated e using the negative version of P's final velocty, but then I crossed it out and worked out the new velocity of Q. I can't remember what I got for e but it was bigger than 1 :/
    Yeah I said something like assume its reversed for the sake of contradiction then calculate e based on that which was greater than 1
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    (Original post by 11234)
    Yeah I said something like assume its reversed for the sake of contradiction then calculate e based on that which was greater than 1
    I did that too but i didnt state that e is greater (although my answer of e was greater than 1) instead i said that new e doesnt equal old e (the one we found the previous part) is that acceptable?

    Because e is a constant and it shouldnt change.
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    Answers I got for this paper:
    1: PCLM: speed of Q =7/3 ms^-1.e=1/9. If direction of P was reversed as well as halved, there would be a gianin K.E: 4.375J before and 10.375J after (could also show that e>1). Dropping objectoff back of Q does not change the speed as there is no force in the directionof motion of Q so no impulse. Velocity of p after onject is fired off: using PCLM gives 0.5*2=0.05*(V-10)+0.45V. This gives V=3. The wall cannot be smooth as the velocityparallel to the wall is not the same before and after: 10cos60 is not equal to6cos40.
    2: K.E converts into work doneagainst resistance: 0.5*0.04*2500=0.2F. F=250N. PCLM gives 0.04*50=4*V.V=0.5ms^-1. Change in K.E of bullet= work done against resistance + gain in KEof block: 0.5*0.04*2500-0.5*4*0.25=250d. d=0.198m. Next was the block on theslope.Find the resistance force and mew (sorry for the lack of symbol).91.5*8-0,5*6*(49-1)-9.8*6*8*sin30=WD (against resistance)=352.8J. Frictionforce =WD/distance =352.8/8=44.1N. F=mewR.44.1/(9.8*6*cos30)=0.866025...=0.866. Finally the power of the tension in thestring when moving at 7 ms^-1. p=fv= 91.5*7=640.5W.
    3: centre of mass (0.99,0.42) Rand Q were 180 and 120 (not neccesarily respectively... i am not sure. It was ashow that). To label the diagrams I just added the 180N and 120N and put allthe arrows as tensions. To find X(d) moments about A gives 261N (another showthat). To find the tensions in AB BC and CD, resolve first vertically at b.Gives 468N as tension in AB. Resolve horizontally at B gives 432N (compressionin BC). Resolving horzintally at D (using X(d)=261) gives 435N compression inDC. Resolving vertically at D gives Y(d)=348N. Resolving the exernal forces ofthe framework vertically gives Y(a)=300-348=-48N =48N vertically downwards.
    4: The first two parts were show that’s.The second (for 7 marks involved finding tan(alpha) in terms of h and then solvinga quadratic for h=0.2. The last part of the paper was find T. Moments about theedge about which is will tip (was the edge closest to O and further to theright): T*0.5*cosA-T*0.2*sinA-0.1*42=0. (where tanA=3/4 so a 345 triangle offorces) This gives T=15N. Thats the whole paper!
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    (Original post by Rawsonj)
    x
    Cheers! I think I got all the same as you, except I can't remember what I got for the tensions of the rods for the frameworks question. For Q2, what did the question ask where you have an answer of V=0.5ms^-1, not sure if I missed that part of the question? :eek:
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    thats not correct, to keep is momentum constant, its speed must change to balance out the fact that you lost weight.
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    (Original post by Alex621)
    Cheers! I think I got all the same as you, except I can't remember what I got for the tensions of the rods for the frameworks question. For Q2, what did the question ask where you have an answer of V=0.5ms^-1, not sure if I missed that part of the question? :eek:
    It was a two marker (part ii) telling you to use PCLM to find the velocity of the block (withe the bullet embedded) if the block is on a smooth plane.
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    For the very last one, about finding T, I misread it as 'point of toppling' instead of 'point of turning'. When I worked through, I got T=0 which I knew was not right but didn't have enough time to redo it.
    Do you think I would get any method marks for this?
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    The last question I didn't multiply the tension components by their distance, but I did for the centre of mass.

    NOOOOOOOOOOOOOOOOOOOOooooooooooo ooooooooooOOOOOOOOOOOOOOOOO.
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    (Original post by Rawsonj)
    Answers I got for this paper:
    1: PCLM: speed of Q =7/3 ms^-1.e=1/9. If direction of P was reversed as well as halved, there would be a gianin K.E: 4.375J before and 10.375J after (could also show that e>1). Dropping objectoff back of Q does not change the speed as there is no force in the directionof motion of Q so no impulse. Velocity of p after onject is fired off: using PCLM gives 0.5*2=0.05*(V-10)+0.45V. This gives V=3. The wall cannot be smooth as the velocityparallel to the wall is not the same before and after: 10cos60 is not equal to6cos40.
    2: K.E converts into work doneagainst resistance: 0.5*0.04*2500=0.2F. F=250N. PCLM gives 0.04*50=4*V.V=0.5ms^-1. Change in K.E of bullet= work done against resistance + gain in KEof block: 0.5*0.04*2500-0.5*4*0.25=250d. d=0.198m. Next was the block on theslope.Find the resistance force and mew (sorry for the lack of symbol).91.5*8-0,5*6*(49-1)-9.8*6*8*sin30=WD (against resistance)=352.8J. Frictionforce =WD/distance =352.8/8=44.1N. F=mewR.44.1/(9.8*6*cos30)=0.866025...=0.866. Finally the power of the tension in thestring when moving at 7 ms^-1. p=fv= 91.5*7=640.5W.
    3: centre of mass (0.99,0.42) Rand Q were 180 and 120 (not neccesarily respectively... i am not sure. It was ashow that). To label the diagrams I just added the 180N and 120N and put allthe arrows as tensions. To find X(d) moments about A gives 261N (another showthat). To find the tensions in AB BC and CD, resolve first vertically at b.Gives 468N as tension in AB. Resolve horizontally at B gives 432N (compressionin BC). Resolving horzintally at D (using X(d)=261) gives 435N compression inDC. Resolving vertically at D gives Y(d)=348N. Resolving the exernal forces ofthe framework vertically gives Y(a)=300-348=-48N =48N vertically downwards.
    4: The first two parts were show that’s.The second (for 7 marks involved finding tan(alpha) in terms of h and then solvinga quadratic for h=0.2. The last part of the paper was find T. Moments about theedge about which is will tip (was the edge closest to O and further to theright): T*0.5*cosA-T*0.2*sinA-0.1*42=0. (where tanA=3/4 so a 345 triangle offorces) This gives T=15N. Thats the whole paper!
    I think I got the same answers!!! What do you think boundary wise it will be this year
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    If I put two resasons with the main body being the one I think is right and the secondary one not, will I lose marks
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    I'm so glad that that's over! I can't believe I actually found that paper ok! It seems like I have similar answers to everyone however on the one abot how far the bullet goes...what was the original distance before they included friction


    The only question I really stuggled with was q4) i! I'm suprised I got ii and iii!
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    (Original post by 11234)
    I think I got the same answers!!! What do you think boundary wise it will be this year
    Guessing around 58. But wouldn't be surprised if it was higher for an A.
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    65/66 what UMS do u think? ( made really stupid mistakes that I could kick myself for!)
 
 
 
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