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

[saadgujjar]

plz telll me the answers of q4 in which S/2 and T/2 was given

 

[tanixx]

what was acceleration in question 1?

 

[tanixx]

plz telll me the answers of q4 in which S/2 and T/2 was given

s was 13m

 

[saadgujjar]

what was acceleration in question 1?

i dont remember but i thinh it was 2.46

 

[raiyanfarrukh]

here are the steps how to do it!! if any doubts ask me!

 

[saadgujjar]

what was acceleration in question 1?

3.46

 

[0Louis0]

here are the steps how to do it!! if any doubts ask me! View attachment 26044View attachment 26045View attachment 26044View attachment 26045

I approve
P.S: Instead of integrating Acceleration you could've gotten the equation of the line 20 < t < 26 using y=mx+c
don't know which is faster tho

 

[Your-Blood]

in question 7 answer was 0.47 m
in q 4 last part speed was 2.65 and in first paprt distanc was 13 .tension in pb was 13..P was 12KW and steady speed was 13.3
w work done against resistance was 750 and the question we have to show kinetic energy
plz tell me other questions,i forgot them

Power was 20KW

 

[0Louis0]

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)

 

[talha196]

hey tell me about question no 4 wat was the frst part..... was it like wats the speed wen the distance is s/2 ????

 

[talha196]

man i am talking about q 4 that 1/2 S ones..... wat were the answers

 

[rose1700]

I think in question 4 initial speed was zero and at the end of the slope speed was 8

 

[Intisar Hasnain]

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????

 

[rose1700]

first part was to calculate s and time taken
second part was to calculate s when t is t/2

 

[Ariel Robert]

I approve
P.S: Instead of integrating Acceleration you could've gotten the equation of the line 20 < t < 26 using y=mx+c
don't know which is faster tho

either way works you just have to prove...

 

[Ariel Robert]

Power was 20KW

yes it was 20KW not 20000KW, one of my friend rote 20000KW and went like this was the answer...

 

[talha196]

hey wat were the masses of particle a and b in last question?????

 

[talha196]

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

 

[raiyanfarrukh]

I approve
P.S: Instead of integrating Acceleration you could've gotten the equation of the line 20 < t < 26 using y=mx+c
don't know which is faster tho

but i can assure u that this required less thinking than y=mx+c

 

[rose1700]

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