# Collision with friction

1. Nov 4, 2004

### Mivz18

I'm having trouble solving for the correct answer for this problem:

A block of mass M1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. It collides with block of mass M2 = 1.7 kg which is initially stationary. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1.85 m . What is the coefficient of kinetic friction on the rough surface?

With the help the problem provided, I get that KE = (1/2)m*v^2 which also equals momentum squared over 2 times the two masses added together; this equals the post-collision KE. I also find out that the Force of friction = (friction constant)g(m1 + m2) . Then through the vector dot product you can figure out that delta KE = F*d = (sliding friction coefficient)gd(m1 + m2)cos(180) . I calculate the initial KE as 69.4575 and the post-collision KE as 46.75 leaving delta KE as 22.7073. Then when I solve for the sliding friction coefficient, from the equation
delta KE = (friction coefficient)gd(m1+m2)cos(180).
After my numbers are plugged in and I solve, I get 0.240859 as the answer. However, the online program I'm using says it is incorrect. Am I doing something wrong?

2. Nov 4, 2004

### arildno

It is not meaningful to use (delta KE) here!
What you then actually are saying, is that all of your initial KE is distibuted between post-collision KE, and the work of friction.
This is patently untrue, since some of the initial KE were lost in the collision.

What IS true, is that ALL post-collision KE is dissipated by friction.

So, equate post-collision KE with work of friction!

3. Nov 4, 2004

Thank you!!