What's the trick of solving collisions with friction?

Question

A block of mass m lying on a rough horizontal surface is given an initial velocity of vo. After traveling a distance d, it makes a head-on collision with a block of mass 2m. How far does the second block move before coming to rest?( Assume that the coefficient of friction, uk, is the same for both blocks).

I use V2=Vo2 +2ax for m2 but keep getting half the right answer.

answer: 2Vo2 -4d
9ug 9
What am I doing wrong... how do you solve this one.

Answer 1

OK, first off you need more information. Does the first block stop when it hits the second? Or does it continue? The first thing you should probably do is draw a free body diagram and figure out what forces are acting on the blocks. The magnitude of the normal force is equal to the magnitude of the gravitational force and the frictional force is equal to the coefficient of friction (called "mu") times the normal force. That should be all the info you need to solve the problem, let me know if you need me to explain more.

Answer 2

The deacceleration rates for the two blocks are the same (even though the force on the second is double that of the first), namely:

a = g * mu

Suppose V is the velocity of the first block at the time of the collision. The first block slows from Vo to V during the time interval t = (Vo-V)/a and d = (1/2)at^2
-->
Vo^2 - 2ad = V^2

Similarly letting d2 be the distance required for the second block to stop from an initial velocity of v2:

v2^2 = 2a(d2)

Assuming the collision is elastic, the relation between v2 and V is:

v2 = 2m*V/(m+2m) = (2/3)V

Putting the above results together:

d2 = v2^2/2a = (4/9)(1/2a)V^2
= (4/9)(1/2a)[Vo^2 - 2ad]
= (2/9)(Vo^2/a) - (4/9)d