A ball with mass 0.45kg and speed 1.3m/s rolls across a level table and into a box. The box has a mass
0.25kg. The box (with the ball inside) then slides a distance of 0.64m before stopping. What is the
coefficient of kinetic friction that describes the box-table interaction?
So, we basically have an inelastic collision followed by a WET problem.
Working backwards, we can see that the friction force will dissipate the system’s kinetic energy.
K o + W fr = 0 → K o = W fr , and we know that we can write the work done by a constant force as
 
W = F • Δx , so with the friction force as the force performing work upon the
1
2
(mball + mbox )vbb
Ko
v2
2
=
→ µK = bb
box… K o = µK MgΔx → µK =
MgΔx (mball + mbox )gΔx
2gΔx
So, now we need to find the velocity of the ball-box system immediately following the collision.
Even though the collision is inelastic and will not conserve energy, the forces that dissipate energy are
nternal and will conserve momentum.
Since the objects stick together following the collision, they move at the center of mass velocity. So,
2
 m1v1o 
2
2
the post-collision velocity is the same as the center of mass velocity. vbb = VCM = 

 m1 + m2 
2

kgm 
0.59


2
s 

 m1v1o 


 0.7kg 
m
+
m



2
µK = 1
=
= 0.06


m
2gΔx
2  9.8 2  0.64m

s 