ALL CIE PHYSICS DOUBTS HERE !!!

[$!$RatJumper$!$]

Thankx for all ur help +rep

umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?

[$!$RatJumper$!$]

Q4 (a)(ii)- why doesnt it work when I find the area under of the triangle for the first 0.3 sec then times it by 2

4aii) It should work the way you did it. I get about 2*1/2*0.3*0.10 =0.03 by APPROXIMATING THE slope by a line

why did u use 0.1

[TJ-56]

Thankx for all ur help +rep

umm and could you also explain why 3(c) direction is 35 below the horizontal. isnt it just 90 degrees?

I think you just need to state that the direction is horizontal, but its not 90 degrees, its 90 degrees when its in the center (i.e. the weight acts through the pivot), but no need since they dont want the angle

[$!$RatJumper$!$]

ok.. and can u explain 4(a)(ii) .. why should we use 0.1

[TJ-56]

ok.. and can u explain 4(a)(ii) .. why should we use 0.1

Ok i didnt understand the calculation that much, but heres what i did, ur way, calc till 0.3 seconds then multiplied the value by 2:
Divided the area into 2 triangles 2 squares and 1 trapezium (thats alot but its the only way it matched the ms value): so respectively:
triangles : 0.1*4*0.5 + 0.1*2.9*0.5
squares: .2*4 + 0.1*2
trapezium: 0.1 * (0.9+2 / 2)
Multiply the sum by 2, then by 10^-2 (in metres) gives us 0.0298 which is 0.030 rounded to 2 sf

[$!$RatJumper$!$]

thank you very much. +rep

[TJ-56]

thank you very much. +rep

Anytime
thanx for the rep

[$!$RatJumper$!$]

S06 Q6 please explain

[astarmathsandphysics]

S06 Q6 please explain
which paper

[Deadly_king]

S06 Q6 please explain
which paper

I think it should be paper 2 since he is sitting for AS.

[$!$RatJumper$!$]

yes p2 please

[joel]

How do you do Question No. 3, part (a), from May/June 2001 Paper 2 (CIE)

paper found on freeexampapers.com

[ashish]

How do you do Question No. 3, part (a), from May/June 2001 Paper 2 (CIE)

paper found on freeexampapers.com

a) assuming no energy is loss
loss in PE = gain in KE
mgh=1/2 mv²
m cancel out
(10*0.13*2)^1/2=v
v=1.6 ms^-1

b) since momentum is conserved
m*1.6 = m*v + m*0.9
m cancel out
1.6=v+0.9
v=0.7

c) inelastic collision since KE has not been conserved

[$!$RatJumper$!$]

S06 Q6 P2 and W07 Q 5(a) P2

Please explain.

[TJ-56]

S06 Q6 P2 and W07 Q 5(a) P2

Please explain.

W07 Q 5(a) P2
phi/360=AB/lambda
(60/360) * lambda = AB
AB = 1/6 * 6
     = 1
So it leads or lags behind (question didnt state which) by 1
see attachment

S06 Q6 P2
What part didnt u understand exactly?