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the figure below shows the graph of force,F (y axis)acting on a trolley of 0.4kg against time,t (x axis) 20--------- ........._...Asked by cylinder
the figure below shows the graph of force,F (y axis)acting on a trolley of 0.4kg against time,t (x axis)
20---------
........._--
........._...--
........._....---
........._......---
........._.......-----
0........5............9
The velocity of the trolley is constant between t=0ms and t=5ms.
Use the figure above to calculate the impulse acting on the trolley between 0 and 9ms.
THE MARK SCHEME SAYS:
IMPULSE 20 x 5 x 10-3 1/2 x 20 X 4 X 10-3
=0.14 Ns
i DONT UNDERSTAND WHERE THE NUMBERS CAME FROM THE EQUATION USED (AS LEVEL) AND ID LIKE SOMEONE TO STEP BY SEP EXPLAIN THIS TO ME PLEASE
PS: EQUATIONS I KNOW:
p=mv
f X change in t =change in p
if you can use these equations please do it would be helpful
The "10-3" bit especially confuses me as i dont know where those number came from and wat it is for/ wat equation
i figured out that f x change in t =change in p is the equations used and nowi understand that bit so it became:
change in momentum= 20 x 5=100
change in momentum= 1/2(20x4)=40
i just dont get the 10-3 bit
20---------
........._--
........._...--
........._....---
........._......---
........._.......-----
0........5............9
The velocity of the trolley is constant between t=0ms and t=5ms.
Use the figure above to calculate the impulse acting on the trolley between 0 and 9ms.
THE MARK SCHEME SAYS:
IMPULSE 20 x 5 x 10-3 1/2 x 20 X 4 X 10-3
=0.14 Ns
i DONT UNDERSTAND WHERE THE NUMBERS CAME FROM THE EQUATION USED (AS LEVEL) AND ID LIKE SOMEONE TO STEP BY SEP EXPLAIN THIS TO ME PLEASE
PS: EQUATIONS I KNOW:
p=mv
f X change in t =change in p
if you can use these equations please do it would be helpful
The "10-3" bit especially confuses me as i dont know where those number came from and wat it is for/ wat equation
i figured out that f x change in t =change in p is the equations used and nowi understand that bit so it became:
change in momentum= 20 x 5=100
change in momentum= 1/2(20x4)=40
i just dont get the 10-3 bit
Answers
Answered by
bobpursley
do you have any idea what ms means?
that 4 and 5 in your change in momentum is .004, and .005 , and I bet that has something to do with your 10^-3
that 4 and 5 in your change in momentum is .004, and .005 , and I bet that has something to do with your 10^-3
Answered by
cylinder
no its 10-7 as in it equals to 3 AND you minus it, its not a power. Please post hrelp with this in mind
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