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How was Mechanics Paper 42!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

How was the paper

  • Easy

    Votes: 27 19.0%
  • Hard

    Votes: 40 28.2%
  • Not Hard and Not easy

    Votes: 75 52.8%

  • Total voters
    142
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d
dont you think the initial velocity was zero and final velocity was 8 m/s
No, after B touched the floor A started another journey with the V of B then stopped moving (v=0), you had to calculate that "journey" distance and add the distance given in the question (which was B's heigh above the floor)
 
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hey tell me about question no 4 wat was the frst part..... was it like wats the speed wen the distance is s/2 ????
 
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No, after B touched the floor A started another journey with the V of B then stopped moving (v=0), you had to calculate that "journey" distance and add the distance given in the question (which was B's heigh above the floor)

i got.. 1.58 m and was pretty sure that's correct.. as i finfished 15 minutes earlier i sat and recalculated the whole thing... cant remember the full, but the last part i found that:
due to B's motion, A was pulled from rest to the same speed of B, v^2=0^2+2as, so distance = 's'..
Then A travelled a bit more due to subsequent motion( or continuing motion but against gravity until it stopped): 0^2=u^2-2gs (here u of A = final speed of B just before hitting ground,) hence a futher 's' travelled.
total dist= s+s, which came 0.92m by A towards P for me.
A travelled 0.92 m towards P.
and dist from P (2.5-0.92) = 1.58 m... plz share wat others did..
*****And wat was the MARKS for this part????
 
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first part was to calculate s and time taken
second part was to calculate s when t is t/2
 
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Q4 a part was like wats the speed of particle when when it has covered a distance of 1/2 S
and second part was like the speed of particle at 1/2 T
 
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i got.. 1.58 m and was pretty sure that's correct.. as i finfished 15 minutes earlier i sat and recalculated the whole thing... cant remember the full, but the last part i found that:
due to B's motion, A was pulled from rest to the same speed of B, v^2=0^2+2as, so distance = 's'..
Then A travelled a bit more due to subsequent motion( or continuing motion but against gravity until it stopped): 0^2=u^2-2gs (here u of A = final speed of B just before hitting ground,) hence a futher 's' travelled.
total dist= s+s, which came 0.92m by A towards P for me.
A travelled 0.92 m towards P.
and dist from P (2.5-0.92) = 1.58 m... plz share wat others did..
*****And wat was the MARKS for this part????

7 marks. you did totally wrong
 
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