Post by: nox_fjmoony on May 18, 2010, 04:43:46 pm
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!
Post by: nid404 on May 18, 2010, 05:32:41 pm
ii) it's a smooth plane.
mgsin
gsin
4)iii) The greatest speed of P will be at it's lowest point...
Change in k.e from O to B is given by
1/2 X0.8X82=25.6J
mgh=25.6J
h=3.2 m this is OB
you from AC to O the height is 2.4 sin50
so height of B below AC= 3.2-2.4 sin50
=1.36m
I'll get back with 5 and 6
Post by: cooldude on May 18, 2010, 05:49:29 pm
:>
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
Post by: nid404 on May 18, 2010, 05:50:51 pm
Post by: cooldude on May 18, 2010, 05:52:03 pm
thanks dude...i was just abt to scan my paper...that saved my time..lol thanks
np, sorry cudnt hve posted earlier, was typing all in one go, or i cudve saved u doin the first part also ;)
Post by: nox_fjmoony on May 18, 2010, 07:43:15 pm
Thankyou very very much
:D
Rep+