Calculate the friction coefficient

[AfuBaf]

Homework Statement

Benjamin (4 years, 18 kg incl sledding) is at the top of a hill run which has slope of 20 degrees to the horizontal and the length 15 m. Slope followed by a large horizontal surface. Benjamin embarks straight down the hill, gets speed and just when he reaches the horizontal surface he runs into Elsa, (2.5 years, 16 kg including diaper) coming waddling from the right. They collide, Elsa falls on the sled and the race continues until the two bewildered children remain 14 meters forward and 2 m to the left of the collision point.

Determine the coefficient of friction between the sled and the snow and Elsas speed just before the collision. Suppose the friction coefficient is constant during the movement, and that the collision between Elsa and Benjamin can be considered a shock. Ignore air resistance.

The Attempt at a Solution

M = benjamins weight
m = Elsas weight
V1 = benjamins velocity
V2 = Elsas velocity
V'= Velocity after the collision
s = length of the hill
d = the length they together travel after the collision

So here I was my though:

We can use the principal of energy:

M*g*h = M*V1^2/2 + F*S
and
(M+m)*V'^2/2 = F*d

Momentum Conservation:

we have conservation of momentum:
a = 20 degrees
p = tan (2.14)

x-axis: MV1 * cos (a) - MV2 = (M + m) V'cos (p)

y direction: mV1sin (a) - 0 = - (M + m) V '* sin (p)

Then we know that the total work done by the friction force is equal to the difference in kinetic energy

U=T2 - T1
T2 = 0

-u(M + m)*g*d = 0 - (M + m) V '^ 2/2

But here I get stuck ...
How should i after this (if it's correct) get hold of the friction coefficient?

please help!

 

[Nathanael]

Welcome to physics forums.

You can't use conservation of momentum on the way down the slope because there are external forces acting.

Once you hit the horizontal part of the slope, you can use conservation of momentum but only in the x-direction.