Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

30 CHAPTER 2
where E A is the energy lost by block A and φ is the binding energy of block B. Ina
collision of this type, where energy is expended to free one of the colliding bodies,
kinetic energy is not conserved and the collision is therefore not elastic, that is, it
is inelastic.
W EXAMPLE 2.13
A stationary block B, whose mass is 2 kg, is held by an elastic cord whose elastic
constant is 10 N/m and whose ultimate strength is 5 N. Another block A, whose
mass is 10 kg, is moving with a velocity of 2 m/s on a frictionless surface. If block A
strikes block B, with what velocity will block B move after the collision?
Solution
From Example 2.12, it is seen that the energy lost by block A in this collision is
11 1
J. The energy expended in breaking the rubber band may be calculated from
9
the product of the force needed to break the elastic cord and the distance that the
elastic cord stretches before breaking. In the case of a spring, rubber band, or any
other substance that is elastically deformed, the deforming force is opposed by a
restoring force whose magnitude is proportional to the deformation. That is,
f = k × r, (2.44)
where f is the force needed to deform the elastic body by an amount r, and k is
the “spring constant” or the force per unit deformation. Since Eq. (2.44) shows
that the force is not constant but rather is proportional to the deformation of the
rubber band, the work done in stretching the rubber band must be computed by the
application of calculus. The infinitesimal work, dW, done in stretching the rubber
band through a distance dr is
dW = f dr,
and the total work done in stretching the rubber band from r = 0tor is given by
W =
r
0
f dr
Substituting Eq. (2.44) for f , we have
W =
r
0
kr dr (2.45)
and solving Eq. (2.45) shows the work done in stretching the rubber band to be
W =
kr 2
2
. (2.46)
Since in this example k is equal to 10 N/m, the ultimate strength of the elastic cord,