Example 7-2 Conservation of Momentum: A Collision on the Ice
Gordie, a 100-kg hockey player, is initially moving to the right at 5.00 m>s directly toward Mario, a stationary 80.0-kg
player. After the two players collide head-on, Mario is moving to the right at 3.75 m>s. (a) In what direction and at what
speed is Gordie moving after the collision? (b) What was the change in Gordie’s momentum in the collision? What was the
change in Mario’s momentum?
Set Up
The system that we’re considering is made up
of the two players. The vertical forces on each
player (the normal force exerted by the ice
and the gravitational force) cancel each other,
so there is no net ­external force in the vertical
direction. The friction forces between the
players and the ice are small compared to the
forces that Gordie and Mario exert on each
other. So we can treat the total momentum of
the system as conserved during the collision.
We’ll use this to find Gordie’s final velocity
and the changes in momentum of each player.
We choose the positive x direction to be to
the right, as shown.
Solve
(a) Write the equation of momentum
­conservation, using the subscript i for
values just before the collision and
subscript f for values just after.
Momentum conservation:
sG + p
sM
(7-14)
Ps = p
(G for Gordie, M for Mario)
has the same value just before
and just after the collision
s = mv
s
p
Mario (at rest)
mM = 80.0 kg
Gordie
mG = 100 kg
vGi = 5.00 m/s
before the collision
(7-5)
Just before the collision, the total
momentum is
vMi = 0
+x
Gordie
sGi + p
sMi = mG s
Psi = p
vGi + mM s
vMi
Just after the collision, the total
momentum is
vGf = ?
Mario
vMf = 3.75 m/s
+x
after the collision
sGf + p
sMf = mG s
Psf = p
vGf + mMs
vMf
Momentum is conserved:
Psf = Psi
mG s
vGf + mM s
vMf = mG s
vGi + mM s
vMi
Note that s
vMi = 0 since Mario is originally at rest. We need to find s
vGf
(Gordie’s velocity just after the collision).
Since the motion is entirely along the
x axis, we only need the x component of the
­momentum conservation equation. Solve for
Gordie’s final x velocity, vGfx; then substitute
the values of the players’ masses, Gordie’s velocity before the collision, and Mario’s velocity after the collision.
mG vGfx + mM vMfx = mG vGix + mM vMix
Mario is originally at rest, so vMix = 0 and
mG vGfx + mM vMfx = mG vGix
mG vGfx = mG vGix 2 mM vMfx
mM vMfx
vGfx = vGix mG
Players’ masses: mG = 100 kg, mM = 80.0 kg
Gordie’s initial x velocity: vGix = +5.00 m>s
Mario’s final x velocity: vMfx = +3.75 m>s
vGfx = 1 +5.00 m>s2 -
180.0 kg2 1 +3.75 m>s2
100 kg
= +5.00 m>s - 3.00 m>s
= +2.00 m>s
Gordie ends up moving at 2.00 m>s to the right (in the positive
x direction).
(b) The change in each player’s momentum
equals his momentum after the collision
minus his momentum before the collision.
Reflect
Gordie: DpGx = mG vGfx 2 mG vGix
= (100 kg)(+ 2.00 m>s) - (100 kg)(+ 5.00 m>s)
= 200 kg # m>s - 500 kg # m>s
= -300 kg # m>s
Mario: DpMx = mM vMfx 2 mM vMix
= (80.0 kg)(+3.75 m>s) - (80.0 kg)(0 m>s)
= +300 kg # m>s
Our answer to part (a) tells us that after the collision Gordie is still moving in the positive x direction, but with reduced
speed: He has lost x momentum, while Mario (who was originally at rest) has gained x momentum. In fact, as part (b)
shows, the amount of x momentum that Gordie loses (300 kg # m>s) is exactly the same as the amount of x momentum
that Mario gains. Thus we can think of the collision between the two players as a transfer of momentum between Gordie
and Mario. This is like a financial transaction in which one person gives money to another. No momentum is lost in the
collision; it simply changes hands from one player to the other.
Notice that in this problem, we needed to know Mario’s final x velocity vMfx in order to find Gordie’s final x
velocity vGfx. That’s because the statement that momentum is conserved in the x direction gave us only one equation
which relates vMfx and vGfx. Hence we were able to solve for only one unknown quantity. If we didn’t know either of
the players’ final velocities, we would have needed additional information—in particular, the duration of the collision
and the magnitude of the force that one player exerts on the other during the collision—which we unfortunately do
not have. The law of conservation of momentum is a great tool, but by itself it can’t tell you everything!
Although we used conservation of momentum in this example, note that neither player’s individual momentum
is conserved: Gordie’s momentum after the collision is different than before the collision, and likewise for Mario’s
momentum. It’s only the total momentum of Gordie and Mario’s system that is conserved. In general, the momentum
of any particular object within a system will not be conserved.