Help on physics problem finding the coefficient of friction after collision given mass, speed and distance.?

Question

A block of mass M1 = 2.8 kg moves with velocity v1 = 7.4 m/s on a frictionless surface. It collides with block of mass M2 = 1.6 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?

Picture to go along with problem: http://i284.photobucket.com/albums/ll28/bathtub2008/showmepl-3.gif

μk = ______

Answer 1

Use the conservation of momentum to determine the velocity of the blocks after sticking together.

M1V1 + M2V2 = (M1 + M2)V

where

M1 = 2.8 kg.

V1 = 7.4 m/sec

M2 = 1.6 kg.

V2 = 0

V = velocity of blocks after the collision

Substituting appropriate values,

(2.8)(7.4) + (1.6)(0) = (2.8 + 1.6)V

Solving for V,

V = 4.71 m/sec.

The next working formula is

Vf^2 - V^2 = 2as

where

Vf = final velocity = 0 (when the blocks will stop)

V = 4.71 m/sec (as calculated above)

a = acceleration through the rough surface

s = distance travelled by blocks before stopping = 1.85 m

Again, substituting appropriate values and solving for "a",

0 - 4.71^2 = 2(a)(1.85)

a = -4.71^2/(2 * 1.85)

a = - 6 m/sec^2

NOTE -- the negative sign attached to the acceleration indicates that the blocks were slowing down as they were skidding along the rough surface.

From Newton's 2nd Law of Motion,

f = ma

By definition,

-f = (μk)(m)(g)

where

f = frictional force opposing the motion of the blocks

m = M1 + M2 = 4.4 kg.

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

μk = coefficient of friction

NOTE, as well, the negative sign attached to the frictional force. It merely implies that the dircection of the frictional force is opposite that of the blocks' motion.

Equating the two "f's"

ma = μk(mg)

(4.4)(-6) = -μk(4.4)(9.8)

Solving for "μk"

μk = 6/9.8

μk = 0.612