Block Collision: Find Distance for Elastic & Inelastic Collisions

[DavidAp]

Block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.1 m and then collides with stationary block 2, which has mass m2 = 5m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction µk is 0.35 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?

Answers
(a) 0.984 m
(b) 0.246 m

Relevant Equations
m1u1 + m2u2 = m1v1 + m2v2
vf^2 = vi^2 + 2as (where s is the Δx)
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This is what I attempted.
I started the problem by working on part (b) first because it seemed that finding the distance traveled through and inelastic collision would be easier due to the shared final velocity. I know that,
h = 3.1m
m2 = 5m1
μk = 0.35.
so I use the formula m1u1 + m2u2 = (m1 + m2)v. However, since block two has no initial velocity the equation becomes m1u1 = (m1 + m2)v. Replacing m1u1 with mgh I get,
v(m1 + m2) = mgh
v = m1gh/(m1 + m2)
v = m1(9.8)(3.1)/(5m1+m1)
v = 30.38m1/6m1
v = 5.06

From there I use the kinematic equation
vf^2 = vi^2 + 2as
knowing vf = 0 and, by searching the web I found that F = ma = -μkmg making, a = -μkg I plug in my knowns into the equation.
s = -vi^2/-μkg
s = -(5.06^2)/-0.35(9.8)
s = 7.46

Now thank goodness I took geometry because 7.46 = 0.246 does not pass the reflexive property! A joke of course but this brings me to my question: what did I do wrong? Obviously something horrible which will dishearten physics lovers everywhere but please, show pity. I'm having difficulty understanding this.

Can somebody explain to me what I did wrong and what I should have done to get the right answer instead and why (especially why! Baby words please!). I will very much appreciate your help. Thank you for taking the time to review my question.

 

[Doc Al]

This is what I attempted.
I started the problem by working on part (b) first because it seemed that finding the distance traveled through and inelastic collision would be easier due to the shared final velocity. I know that,
h = 3.1m
m2 = 5m1
μk = 0.35.
so I use the formula m1u1 + m2u2 = (m1 + m2)v. However, since block two has no initial velocity the equation becomes m1u1 = (m1 + m2)v. Replacing m1u1 with mgh I get,

There's your problem. m1u1 is a momentum; mgh is an energy. You cannot set one equal to another.

Use mgh to find m1's speed at the bottom, then you can calculate m1u1.