TI-M: Re: RE: physics
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TI-M: Re: RE: physics
I'll bite....
Ki is initial kinetic energy for the block at the top of the ramp, so 0.
Kf is final kinetic energy for the block right after it leaves the ramp.
Kfh is the final velocity on the horizontal surface, so 0.
Kih is the initial kinetic energy on the horizontal surface.
vf is the final velocity of the block after it leaves the ramp.
vi is the initial velocity on the horizontal surface.
m is the mass of the first block.
M is the mass of the second block.
N is the normal force.
h is the height.
g is gravity.
d is distance traveled on the horizontal surface (what we want to solve
for).
W is work
Wg is work done by gravity
Wn is work done by normal force (0, since normal force is perpendicular to
the change in the position)
Wf is work done by friction (0 for the ramp, -(mu)*N*d for the horizontal
surface)
Up is positive y
Right is positive x
First, for the block sliding down the ramp:
(Work-kinetic energy theorem)
Kf-Ki = Sigma(W) with Ki = 0
= Wg + Wn + Wf
= Wg
Kf = Wg
Wg = m*g*h
Kf = 1/2*m*vf^2
1/2*m*vf^2 = m*g*h
vf = sqrt(2*g*h)
Now, because this is assumed to be a completely inelastic collision and
momentum is conserved:
m*vf = (m+M)*vi
m*(sqrt(2*g*h)) = (m+M)*vi
vi = (m*sqrt(2*g*h))/(m+M)
So, we have the initial velocity of the system and we use the Work-kinetic
energy theorem again:
Kfh-Kih = sigma(W) with Kfh = 0
-Kih = Wg + Wn + Wf, Wg = 0, Wn = 0
-Kih = -(mu)*N*d, with N = m*g
1/2*(m+M)*vi^2 = (mu)*(m+M)*g*d
d = vi^2/(2*g*(mu))
= (m*sqrt(2*g*h)/(m+M))^2 * (2*g*(mu))
= (m^2*h)/((mu)(m+M)^2)
And plugging in the values for m, h, (mu), and M, we get.....
231481 meters
I think that's right :)
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