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Symmetry in vectors? (M2 June 2012 7c)

Hey there folks,

So I understand exactly how to solve this question (after briefly going over it with my teacher - he had to go elsewhere and it was the last day of sixth form so couldn't catch him later).

What I don't understand is why the initial vertical component of speed going up on leaving the dashed line, 14 ms-1 is not the same as that when going down at the point it meets the dashed line again. Instead it's 10.5 ms-1 (Again, I know how this is calculated).

I thought this was as a result of symmetry, as the Exam Solutions guy taught it.... what am I missing?

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Thank you!!!!
(edited 10 years ago)
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piguy
8
???? They are symmetrical!

In calculating the initial projection speed:
[br]v=0,u=?,s=10,a=−10[br][br]v2=u2+2as[br][br]u=200=14[br] [br]v = 0 , u = ?, s = 10, a = -10[br][br]v^2 = u^2 + 2as[br][br]u = \sqrt{200} = 14[br]

Calculating the speed when it's at the dotted line the second time round uses pretty much the same equations, just that v and u are swapped, and acceleration is positive: \

[br]v=?,u=0,s=10,a=10[br][br]v2=u2+2as[br][br]v=200=14[br] [br]v = ?, u = 0, s = 10, a = 10[br][br]v^2 = u^2 + 2as[br][br]v = \sqrt{200} = 14[br]

Equations are symmetrical! and the speed at the dotted line is 14, not 10.5
(edited 10 years ago)
Original post by Piguy
???? They are symmetrical!

In calculating the initial projection speed:
[br]v=0,u=?,s=10,a=−10[br][br]v2=u2+2as[br][br]u=200=14[br] [br]v = 0 , u = ?, s = 10, a = -10[br][br]v^2 = u^2 + 2as[br][br]u = \sqrt{200} = 14[br]

Calculating the speed when it's at the dotted line the second time round uses pretty much the same equations, just that v and u are swapped, and acceleration is positive: \

[br]v=?,u=0,s=10,a=10[br][br]v2=u2+2as[br][br]v=200=14[br] [br]v = ?, u = 0, s = 10, a = 10[br][br]v^2 = u^2 + 2as[br][br]v = \sqrt{200} = 14[br]

Equations are symmetrical! and the speed at the dotted line is 14, not 10.5


Why do you need to find this speed anyway?
It is totally superfluous.

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