TSSM 09 Question

[Shark 774]

I totally disagree with the answer that TSSM gave to quesion 4, anybody else with me? (I attached their answer too)

[forumguy]

I've always thought it was =mgtan(theta)

SO i got that wrong, i did it this afternoon

[onur369]

The solution is incorrect.  you should use v=squareroot(radius x gravity x tantheta)

[Shark 774]

The solution is incorrect.  you should use v=squareroot(radius x gravity x tantheta)

Bingo, that's exactly what I did. Cheers!

[LinusX]

Awesome
Was getting worried by this.
Does anyone have the actual answer?

[Vincezor]

Awesome
Was getting worried by this.
Does anyone have the actual answer?

 Should be approx 37.42m/s

[LinusX]

Another question for this exam (Question 11).

I used v^2 = u^2 + 2ax to work out the final velocity as it started from rest and is only acted upon by
g. When I work it out however I get a different answer to the one they give us in the solutions. How is this possible? Is my method wrong?

In the solutions they use change in gravitational potential = kinetic energy

EDIT: Could someone run through that series of questions? I cant seem to to get question 12 either

[yawho]

the velocity you obtained using v^2 = u^2 + 2ax is the vertical component

[LinusX]

Yes I know its the vertical component but there is no horizontal component for this question...I think I'm missing something here..

[xZero]

Would be nice if you post the original question up

[LinusX]

question attached

[xZero]

v^2=u^2+2ax, let a=10 for easy calculation, x=1, u=0
v^2=2*10*1
v^2=20
v=sqrt(20)ms^-1

using energy conservation
1/2mv^2=mg delta h, delta h = 1
1/2mv^2=2*10*1
mv^2=40
v^2=20
v=sqrt(20)ms^-1

unless if i missed something, sqrt(20)=sqrt(20)

[LinusX]

I have no idea what happened there...
Thanks for all the help xZero 

[Shark 774]

Way to steal my thread...