# Thread:

1. ## 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

2. 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
now try it.

3. 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

$t = \dfrac{2 \cdot v_0}{g} = \dfrac{6}{7} \, sec$

4. 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

5. 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?

6. 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

7. 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!
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9595471620559692, "perplexity_flag": "middle"}