University Physics I - Classical Mechanics, 2019

82 CHAPTER 4. KINETIC ENERGY
4.4 Examples
4.4.1 Collision graph revisited
Look again at the collision graph from example 3.5.1 from the point of view of the kinetic energy
of the two carts.
(a) What is the initial kinetic energy of the system?
(b) How much of this is in the center of mass motion, and how much of is convertible?
(c) Does the convertible kinetic energy go to zero at some point during the collision? If so, when?
Is it fully recovered after the collision is over?
(d) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution?
Solution
(a) From the solution to example 3.5.1 we know that
v 1i = −1 m s
v 1f =1 m s
v 2i =0.5 m s
v 2f = −0.5 m s
and m 1 =1kgandm 2 = 2 kg. So the initial kinetic energy is
K sys,i = 1 2 m 1v 2 1i + 1 2 m 2v 2 2i =0.5J+0.25 J = 0.75 J (4.17)
(b) To calculate K cm = 1 2 (m 1 + m 2 )v 2 cm , we need v cm, which in this case is equal to
v cm = m 1v 1i + m 2 v 2i −1+2× 0.5
= =0
m 1 + m 2 3
so K cm = 0, which means all the kinetic energy is convertible. We can also calculate that directly:
K conv,i = 1 2 μv2 12,i = 1 ( ) 1 × 2
(
2 1+2 kg × 0.5 m s − (−1) m ) 2 1.5 2
= J=0.75 J (4.18)
s 3
(c)Ifwelookatfigure3.5, we can see that the carts do not pass through each other, so their
relative velocity must be zero at some point, and with that, the convertible energy. In fact, the
figure makes it quite clear that both v 1 and v 2 are zero at t = 5 s, so at that point also v 12 =0,
and the convertible energy K conv = 0. (And so is the total K sys = 0 at that time, since K cm =0
throughout.)