8.43.
A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless,
horizontal surface of an icy New York sidewalk and has a head-on collision with a larger 30.0-g
marble sliding to the right with a velocity of magnitude 0.200 m/s (see figure). (a) Find the
velocity of each marble (magnitude and direction) after the collision. (since the collision is headon, all the motion is along a line.) (b) Calculate the change in momentum (that is, the momentum
after the collision minus the momentum before the collision for each marble. Compare values
you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy
after the collision minus the kinetic energy before the collision) for each marble. Compare the
values you get for each marble.
Identify: Since the collision is elastic, both momentum conservation and Eq. 8.27
vB2x – vA2x = - (vB1x – vA1x) apply.
Set Up:
Let object A be the 30.0 kg marble and let object B be the 10.0 g marble. Let +x be to
the right.
Execute:
(a) Conservation of momentum gives
(0.0300 kg)(0.200 m/s)  (0.0100 kg)(0.400 m/s)  (0.0300 kg)vA2 x  (0.0100 kg)vB 2 x
3vA2 x  vB 2 x  0.200 m/s
.
. Eq. 8.27 says
vB 2 x  vA2 x  (0.400 m/s  0.200 m/s)  0.600 m/s
. Solving this pair of equations
gives vA2 x  0.100 m/s and vB 2 x  0.500 m/s . The 30.0 g marble is moving to the
left at 0.100 m/s and the 10.0 g marble is moving to the right at 0.500 m/s.
(b) For marble A,
PAx  mAvA2 x  mAvA1x  (0.0300 kg)(0.100 m/s  0.200 m/s)  0.00900 kg  m/s
.
For marble B,
PBx  mBvB 2 x  mBvB1x  (0.0100 kg)(0.500 m/s  [0.400 m/s])  0.00900 kg  m/s
.
The changes in momentum have the same magnitude and opposite sign.
(c) For marble A,
K A  12 mAvA2 2  12 mAvA21  12 (0.0300 kg)([0.100 m/s]2  [0.200 m/s]2 )  4.5 104 J
.
For marble B,
KB  12 mBvB2 2  12 mBvB21  12 (0.0100 kg)([0.500 m/s]2  [0.400 m/s]2 )  4.5 104 J
.
The changes in kinetic energy have the same magnitude and opposite sign.
Evaluate: The results of parts (b) and (c) show that momentum and kinetic energy are
conserved in the collision.