# Problem with pulley

• May 8th 2011, 10:29 AM
chartsy
Problem with pulley
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 (Doh)
• May 8th 2011, 10:53 AM
abhishekkgp
Quote:

Originally Posted by chartsy
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)
this is wrong. the above equation is true for a particle in free fall. P is not in a free fall
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 (Doh)

now try it.
• May 8th 2011, 10:58 AM
skeeter
http://clip2net.com/clip/m0/13048755....png?nocache=1

when P hits the ground, Q is still rising at speed $v = at = 4.2 \, m/s$

time for it to rise and fall to its original position is

$t = \dfrac{2 \cdot v_0}{g} = \dfrac{6}{7} \, sec$
• May 8th 2011, 11:36 AM
chartsy
so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time
• May 8th 2011, 12:30 PM
chartsy
what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?
• May 8th 2011, 12:54 PM
topsquark
Quote:

Originally Posted by chartsy
so...you kind of combine the upmove and the downmove into the one equation v=u+at? i didn't know you could do that , i thought you could only use equations of motion one direction at a time

You aren't combining them. You are using them one at a time. You need to find the speed that Q is moving upward initially (the speed it gets to after P hits the ground). That gives you the v0 for the problem after P is on the ground.

-Dan
• May 8th 2011, 01:22 PM
chartsy
Quote:

Originally Posted by chartsy
what i also don't get is, why should the speed stay at 4.2 m/s? i mean, if Q falls back to where it was when P hit the ground, surely ,given that -9.8/ms^2 is greater than -4.2ms^2, the speed V would be bigger than 4.2?

I get it! it's symmetrical because it decelerates as fast as it accelrates!