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Okay I was solving may/june-2008 and uh I've got problems with the following questions:
1)ii), 4)iii), 5)i and ii and 6)i,ii, and iii.
I'm not especially understanding how they arrived to the answer in 5. It says that the block B is released, and so the system starts to move. So does the system moves to the left or the right? Cause, if it does move to the left, then I think I got it. But if it doesn't...
Oh well.
Any help would be appreciated!
1ii) a=2.5, let alpha=x
mgsinx=ma
10sinx=2.5
x=alpha=14.5
4iii) speed of P at A-->
PE=0.8*10*2.4sin50, KE=0.5*0.8*(v^2)
v at A=6.06 m/s
therefore gain in kinetic energy from A to B-->
0.5*0.8*(8^2-6.06^2)=10.9
therefore 10.9=mgh
10.9=0.8*10*h
h=1.36 m
5 i) it moves to the right because the tension in the string than the friction.
for B-->
let coefficient of friction=c
therefore T-F=ma, F=cR, R=6
T-3=0.6a
for A-->
mg-T=ma
4-T=0.4a
solve simultaneously-->
a=1, T=3.6
ii) s=ut+0.5at^2
3=0.5*1*t^2
t=2.45 sec
6)
i) v^2=u^2+2as
0=5.2^2+2*-10.4*s
s=1.3m
therefore dist above ground=1.3+6.2=7.5m
ii) v^2=u^2+2as
v^2=2*9.6*7.5
v=12 m/s
iii) on the upward journey-->
ke=0.5*m*v^2=0.5*0.6*5.2^2=8.112
pe=1.3*10*0.6=7.8
therefore work done against air resistance=8.112-7.8=0.312
on the downward journey-->
ke=0.5*0.6*12^2=43.2
pe=7.5*10*0.6=45
therfore work done against resistance=45-43.2=1.8
therfore total work done=0.312+1.8=2.112
