Pulley problem

May 2011
1
0
Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .

I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as

v^2=0+2x(9.8x3.15)
v=7.857 m/2

so i used V=U+at
giving t=0.8017

What is wrong with my method? Would appreciate help, been bugging me all day
 

topsquark

Forum Staff
Apr 2008
2,978
631
On the dance floor, baby!
Hey, i have the mark scheme for this question but i simply do not understand it.the question is here : maths problem , part E . the answer is here:answer .

I don't understand, what i did was use equations of motion , from the point when Q reverses and falls 3.15 m, therefore giving me the value of V from v^=u^2+2as

v^2=0+2x(9.8x3.15)
v=7.857 m/2

so i used V=U+at
giving t=0.8017

What is wrong with my method? Would appreciate help, been bugging me all day
Look at your answer file. The v in the first calculation is the upward speed that Q is moving at when P hits the ground. So the acceleration here is 2.8 m/s^2http://www.mathhelpforum.com/math-help/members/abhishekkgp/. Then we use this number, 4.2 m/s, as the initial speed for the part of the problem where Q is now in free-fall. Also see here.

-Dan
 
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