Stopping distance after a collision down a ramp

Ep = Ek
mgh = Ek
mgh = ½mv²
v = √2gh

As the collision is elastic, m1u1 + m2u2 = m1v1 + m2v2. It is known that m2 = 2m1.

m√2gh = 2m1v2
v2 = (m√2gh)/2m
v2 = (√2gh)/2

Force body diagram of m2:

̂̂̂̂̂̂Fnet = ma
Fnet = Fn + Fg + Ff
Fnet = Ff
m2a = Ff
a = Ff/m2
a = Ff/2m1
a = uN/2m1
a = 0.5*m2g/2m1
a = m1g/2m1
a = g/2 (negative)

v^2=u^2+2ax
0= v2^2 – gx
gx = v2^2
x = (v2^2)/g
x = 2h/9

The answer is 8/9h. My solution differs from the lecturers at this point:

I do not understand what he was done

http://ift.tt/1lbADbU

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