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