Author Topic: TSSM Exam 1 2009 (Read 3688 times) Tweet Share

Hooligan · « **Reply #15 on:** October 28, 2009, 02:56:50 pm »

Quote from: TrueTears on October 27, 2009, 11:36:33 pm

With regards to your other question, yes normally when something stops with v = 0 the question should specify something like this: "...comes to rest" or else it is really ambiguous.

Quote from: Hooligan
# First, a=g, t=0.4s, s=8, u=?, v=0.... using $s = ut + \frac{1}{2}at^{2}$ I get $u = 20 - \frac{g}{5} m/s^{2}$ or $\frac{100-g}{5} m/s^{2}$

Your way is fine

$8 = u(0.4) + (0.5)(g)(0.4)^2$

$u = \frac{100-g}{5}$

$a = g$ $u = 0$ $v = \frac{100-g}{5}$

$v^2 = u^2 + 2ax$

$\left(\frac{100-g}{5}\right)^2 = 2gx$

$x = \frac{(g-100)^2}{50g}$

then $\frac{(g-100)^2}{50g} + 8 = \frac{(g+100)^2}{50g}$

Uhhmm.... I get everything until the last line...
You don't have to add 8 at the end, because my calculation of 'x' already takes into account the 8 metres (i.e my calculation calculates the whole height... doesn't it?)
I don't get how $\frac{(g-100)^2}{50g} + 8 = \frac{(g+100)^2}{50g}$ ? how does adding an 8 to make the 100 in the bracket suddenly positive?

TrueTears · « **Reply #16 on:** October 28, 2009, 03:01:49 pm »

No it does not take into consideration the whole motion. After you worked out the 'u' it is the speed which the object is travelling when it is exactly 8 m above ground.

So after you worked that out, from the top of the tower you have a = g, u = 0 and now $v = \frac{100-g}{5}$ . Now you can work out the distance travelled from the top of the tower until the object is 8 m above the ground.

Also $\frac{(g-100)^2}{50g} + 8 = \frac{g^2 - 200g + 100^2}{50g} + \frac{400g}{50g} = \frac{g^2 + 200g + 100^2}{50g} = \frac{(g+100)^2}{50g}$

Hooligan · « **Reply #17 on:** October 28, 2009, 03:10:49 pm »

Quote from: TrueTears on October 28, 2009, 03:01:49 pm

No it does not take into consideration the whole motion. After you worked out the 'u' it is the speed which the object is travelling when it is exactly 8 m above ground.

So after you worked that out, from the top of the tower you have a = g, u = 0 and now $v = \frac{100-g}{5}$ . Now you can work out the distance travelled from the top of the tower until the object is 8 m above the ground.

Also $\frac{(g-100)^2}{50g} + 8 = \frac{g^2 - 200g + 100^2}{50g} + \frac{400g}{50g} = \frac{g^2 + 200g + 100^2}{50g} = \frac{(g+100)^2}{50g}$

Beautiful! Thanks TT!! Get it now!

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