# Block sliding down wedge (w/ friction), wedge on frictionless floor

1. Jun 21, 2010

### vitus

1. The problem statement, all variables and given/known data

A large Wedge with mass 10 kg rests on a horizontal frictionless surface (think right triangle on a unit circle, theta <90 deg.). A block with a mass of 5 kg starts at rest (near the top) and slides down the inclined surface of the wedge, which is rough (does this imply friction?). At one instant the vertical component of the block's velocity is 3.0 m/s and the horizontal component is 6 m/s. At that instant the velocity of the wedge is:

2. Relevant equations

None provided, here's what I know:

Mass of the block = Mb = 5 kg
Mass of the Wdge = Mw = 10 kg
theta = tan^-1(3/6) ~ 26.56 deg. (from x-axis)
Velocity of the block = (3^2 + 6^2)^(1/2) ~ 6.7m/s @ 206.56 deg.

Here's what I think is possibly useful:

conservation of energy:
KEb + PEb + KEb + KEw = Me (total mechanical energy)
(Kinetic energy of the block, Potential energy of the block, etc for the wedge)

Sum of Forces:
SUM(Fx) = 0
(Sum of forces in the X-axis is zero)
SUM(Fy) = 0

Conservation of momentum:
MbVbi + MwVwi = MbVbf + MwVwf
(sum of the initial mass x velocity of objects = sum of final mass x velocity of objects)

3. The attempt at a solution

I've taken several approaches
1) Reference Frames

Vb/e = Vb/w + Bw/e
(Velocity of the block w/ respect to earth is the sum of Vblock w/ res. to wedge + wedge to earth)

6 m/s = (gravity - friction here) + Bw/e
6 m/s = (Mb(g)cos(theta) - Fnus) + Bw/e
6 m/s = (5gcos(26.56) - (5gcos(26.56)us) + Bw/e
6 m/s = (1 - us) + Bw/e

except us is neither stated nor computable...

2) Conservation of Momentum
MbVbi + MwVwi = MbVbf + MwVwf
5(0) + 10(0) = 5(6.7) + 10(Vwf)
except everything starts at rest, yet end up moving so momentum isn't conserved here

3) Conservation of Energy
KEb + PEb + KEb + KEw = Me
0.5Mb(Vbo)^2 + Mbgho + 0.5Mw(Vwo)^2 = 0.5Mb(Vbf)^2 + Mbghf + 0.5Mw(Vwf)^2

Here there is no inital KE so it becomes
Mbgho = 0.5Mb(Vbf)^2 + Mbghf + 0.5Mw(Vwf)^2
(5)g(ho) = 0.5(5)(6.7)^2 + (5)g(hf) + 0.5(10)(Vwf)^2
g(ho) = 0.5(Vbf)^2 + g(hf) + 0.5(10/5)(Vwf)^2
Lets choose the instant where the block hits h=0, so this removes g(hf) from the equation:
g(ho) = 0.5(Vbf)^2 + 0.5(10/5)(Vwf)^2
(g(ho) / 0.5(Vbf)^2)^(1/2) = Vwf

great, except for two catches. First what is the original height (ho) and what about the energy lost to friction (does the prof. mean there's frinction when he says rough?)

Just from basic reasoning, if the block is going 6m/s left as it slides and the ratio of masses is 1:2 then the wedge will go 3 m/s right, but I can't seem to get that answer...

btw, I already took the test this question was on and he gave me 4/5 points for the work showed and the explanation I gave above (even though the work didn't really fit with my explanation). I'm just curious as to how one would solve this problem.

2. Jun 21, 2010

### Staff: Mentor

- Yes, a rough surface does imply friction. Thus mechanical energy is not conserved.
- Momentum can be conserved in one direction, yet not conserved in another. (That's the key to solving this with barely any work at all.)

3. Jun 21, 2010

### vitus

I see so it becomes
MbVbix + MwVwix = -MbVbfx + MwVwfx
which, since everything is at rest initially, becomes:
0 = -MbVbfx + MwVwfx
MbVbfx = MwVwfx
(5 kg)(6 m/s) = (10 kg)(Vwfx)

0.5(6 m/s) = 3 m/s

Seems rather obvious now that you point it out. Thanks.