Question

EXAMPLE 6.4A Truck Versus a Compact

GOAL Apply conservation of momentum to a one-dimensional inelastic
collision.

PROBLEM A pickup truck with mass 1.80 103 kg is traveling eastbound
at +15.0 m/s, while a compact car with mass 9.00 102 kg is
traveling westbound at −15.0 m/s. (See figure.) The vehicles
collide head-on, becoming entangled. (a) Find the speed
of the entangled vehicles after the collision.
(b) Find the change in the velocity of each
vehicle. (c) Find the change in the kinetic
energy of the system consisting of both vehicles.

STRATEGY The total momentum of the vehicles before the collision,
pi, equals the total momentum of the vehicles after the collision,
pf , if we ignore friction and assume the two vehicles form an
isolated system. (This is called the "impulse approximation.")
Solve the momentum conservation equation for the final velocity of
the entangled vehicles. Once the velocities are in hand, the other
parts can be solved by substitution.

SOLUTION

(A) Find the final speed after collision.

Let

m1

and

v1i

represent the mass and initial velocity of the pickup truck,
while

m2

and

v2i

pertain to the compact. Apply conservation of momentum.

pi = pf

m1v1i + m2v2i = (m1 + m2)vf

Substitute the values and solve for the final velocity, vf.

(1.80 ✕ 103 kg)(15.0 m/s) + (9.00 ✕ 102 kg)(−15.0 m/s)

= (1.80 ✕ 103 kg + 9.00 ✕ 102 kg)vf

vf = +5.00 m/s

(B) Find the change in velocity for each vehicle.

Change in velocity of the pickup truck.

Δv1 = vf − v1i = 5.00 m/s − 15.0 m/s = −10.0 m/s

Change in velocity of the compact car.

Δv2 = vf − v2i = 5.00 m/s − (−15.0 m/s) = 20.0 m/s

(C) Find the change in kinetic energy of the system.

Calculate the initial kinetic energy of the system.

KEi =

1

2

m1v1i2 +

1

2

m2v2i2

=

1

2

(1.80 ✕ 103 kg)(15.0 m/s)2

+

1

2

(9.00 ✕ 102 kg)(−15.0 m/s)2

= 3.04 ✕ 105 J

Calculate the final kinetic energy of the system and the change in kinetic energy, ΔKE.

KEf =

1

2

(m1 + m2)vf2

=

1

2

(1.80 ✕ 103 kg + 9.00 ✕ 102 kg)(5.00 m/s)2

= 3.38 ✕ 104 J

ΔKE = KEf − KEi = −2.70 105 J

LEARN MORE

REMARKS During the collision, the system lost almost 90% of its
kinetic energy. The change in velocity of the pickup truck was only
10.0 m/s, compared to twice that for the compact car. This example
underscores perhaps the most important safety feature of any car:
its mass. Injury is caused by a change in velocity, and the more
massive vehicle undergoes a smaller velocity change in a typical
accident.

QUESTION If the mass of both vehicles were doubled, how would the
final velocity and the change in kinetic energy be affected?
(Select all that apply.)

The change in kinetic energy would be 2 times as great.The final
velocities would each be 1/2 times as large in magnitude.The change
in kinetic energy would be half as great.The final velocities would
each be 2 times as large in magnitude.The change in kinetic energy
would be unchanged.The final velocities would each have the same
magnitude as before.

PRACTICE IT

Use the worked example above to help you solve this problem. An
pickup truck with mass 1.85 103 kg is traveling eastbound at +14.5
m/s, while a compact car with mass 9.34 102 kg is traveling
westbound at -14.5 m/s. (See figure.) The vehicles collide head-on,
becoming entangled.

(a) Find the speed of the entangled vehicles after the
collision.

m/s

(b) Find the change in the velocity of each vehicle.

Δvtruck = m/s

Δvcar = m/s

(c) Find the change in the kinetic energy of the system consisting
of both vehicles.

J

EXERCISEHINTS: GETTING
STARTED | I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise.
Suppose the same two vehicles are both traveling eastward, the
compact car leading the pickup truck. The driver of the compact car
slams on the brakes suddenly, slowing the vehicle to 5.91 m/s. If
the pickup truck traveling at 17.6 m/s crashes into the compact
car, find the following.

(a) the speed of the system right after the collision, assuming the
two vehicles become entangled

m/s

(b) the change in velocity for both vehicles

Δvtruck = m/s

Δvcar = m/s

(c) the change in kinetic energy of the system, from the instant
before impact (when the compact car is traveling at 5.91 m/s) to
the instant right after the collision

ΔKE = J

Answer #1

a)

m_{c} = mass of compact car = 934 kg

m_{p} = mass of pickup = 1850 kg

V_{ci} = velocity of compact car before collision = 5.91
m/s

V_{pi} = velocity of pickup before collision = 17.6
m/s

V = velocity of combination after the collision

using conservation of momentum

m_{c} v_{ci} + m_{p} v_{pi} =
(m_{c} + m_{p}) V

934 (5.91) + 1850 (17.6) = (934 + 1850) V

V = 13.7 m/s

b)

V_{pickup} = 13.7 - 17.6 = - 3.9 m/s

V_{car} = 13.7 - 5.91 = 7.8 m/s

c)

change in KE = (0.5) (m_{c} v^{2}_{ci} +
m_{p} v^{2}_{pi} - (m_{c} +
m_{p}) V^{2} )

change in KE = (0.5) ((934) (5.91)^{2} + (1850)
(17.6)^{2}- (934 + 1850) (13.7)^{2} ) = 4.2 x
10^{4} J

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