Example 7-6 A Head-On Collision
In a scene from an action movie, a 1.50 * 103-kg car moving north at 35.0 m>s collides head-on with a 7.50 * 103-kg truck
moving south at 25.0 m>s. The car and truck stick together after the collision. (a) How fast and in what direction is the
wreckage traveling just after the collision? (b) How much mechanical energy is lost in the collision?
Set Up
Since the two vehicles stick together,
this is a completely inelastic collision.
We’ll use Equation 7-18 to find the
vf of the wreckage. We’ll
final velocity s
then compare the final kinetic energy
of the wreckage to the combined
kinetic energies of the car and truck
before the collision; the difference is
the amount of mechanical energy that’s
lost in the collision.
Momentum conservation in a
completely inelastic collision:
mcar s
vcar, i + mtruck s
vtruck, i
sf = 1mcar + mtruck 2v
before
mtruck = 7.50 × 103 kg
vtruck,i = 25.0 m/s
(7-18)
Kinetic energy:
1
K = mv 2
2
truck
mcar = 1.50 × 103 kg
(6-8)
car
after
vcar,i = 35.0 m/s
car + truck
velocity = ?
Solve
(a) The collision is along a straight
line, which we call the x axis. We take
the positive x axis to be to the north,
in the direction of the car’s motion
before the collision. We use the x
­component of Equation 7-18 to solve
for the final velocity.
Equation for conservation of x momentum
in a completely inelastic collision:
mcar vcar, ix + mtruck vtruck, ix = (mcar + mtruck)vfx
+x
before
vtruck,ix = –25.0 m/s
Solve for final velocity of wreckage:
vfx =
mcar vcar, ix + mtruck vtruck, ix
mcar + mtruck
With our choice of x axis, we have
vcar, ix = +35.0 m>s
vtruck, ix = -25.0 m>s
Substitute values:
vfx =
11.50 * 103 kg2 1 +35.0 m>s2
c
d
+ 17.50 * 103 kg2 1 -25.0 m>s2
1.50 * 103 kg + 7.50 * 103 kg
= -15.0 m>s
The wreckage moves at 15.0 m>s to the
south (in the negative x direction).
vcar,ix = +35.0 m/s
after
vfx = –15.0 m/s
(b) Calculate the kinetic energies before
and after the collision and compare.
Total kinetic energy before the collision:
1
1
K car, i + K truck, i = m car v 2car, i + mtruck v 2truck, i
2
2
1
3
= 11.50 * 10 kg2 135.0 m>s2 2
2
1
+ 17.50 * 103 kg2 125.0 m>s2 2
2
= 9.19 * 105 J + 2.34 * 106 J
= 3.26 * 106 J
Total kinetic energy after the collision:
1
K car, f + K truck, f = 1mcar + mtruck 2v 2f
2
1
= 11.50 * 103 kg + 7.50 * 103 kg2 115.0 m>s2 2
2
= 1.01 * 106 J
The amount of mechanical energy lost in the collision is the difference between the initial and final energies:
(Kcar, i + Ktruck, i) 2 (Kcar, f + Ktruck, f)
= 3.26 * 106 J 2 1.01 * 106 J
= 2.25 * 106 J
Reflect
It makes sense that the wreckage
moves in the direction of the truck’s
initial motion. Before the collision, the
magnitude of the southbound truck’s
momentum was much greater than
that of the northbound car, so the total
momentum was to the south.
Our result in part (b) shows
that more than two-thirds of the
total mechanical energy is lost in the
collision. This lost energy goes into
doing work to deform the two vehicles,
Momentum of car before the collision:
which makes for a very impressive
collision scene in the movie.
Total momentum before the collision:
Pix = pcar, ix + ptruck, ix
= +5.25 * 104 kgm>s
+ 1 -1.88 * 105 kgm>s2
= -1.35 * 105 kgm>s
pcar,ix = mcar vcar, ix
= 11.50 * 103 kg2 1 +35.0 m>s2
+x
before
ptruck,ix
= –1.88 × 105 kg·m/s
4
= +5.25 * 10 kgm>s
Momentum of truck before the collision:
ptruck, ix = mtruck vtruck, ix
pcar,ix
= +5.25 × 104 kg·m/s
3
= 17.50 * 10 kg2 1 -25.0 m>s2
= -1.88 * 105 kgm>s
after
+x
pfx
= –1.35 × 105 kg·m/s