Use the worked example above to help you solve this problem. In a crash test, a car of mass 1.41 x 10³ kg collides with a wall and rebounds as shown in the figure. The initial and final velocities of the car are v; = -15.5 m/s and vf = 2.35 m/s, respectively. If the collision lasts for 0.155 s, find the following. (a) the impulse delivered to the car due to the collision kg. m/s (b) the size of and direction of the average force exerted on the car (Indicate the direction with the sign of your answer.) N

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EXAMPLE 6.2
GOAL Find an impulse
and estimate a force in a
collision of a moving object
with a stationary object.
How Good Are the Bumpers?
PROBLEM In a crash
test, a car of mass
-15.0 m/s
+2.60 m/s
SOLUTION
(A) Find the impulse delivered to the car.
Calculate the initial and final momenta
of the car.
The impulse is just the difference
between the final and initial momenta.
Before
1.50 x 10³ kg collides with
a wall and rebounds as in
the figure. The initial and
final velocities of the car
are v₁ = -15.0 m/s and vf
= 2.60 m/s, respectively.
If the collision lasts for
0.150 s, find (a) the
impulse delivered to the
car due to the collision and (b) the size and direction of the average force exerted on the car.
a
(a) This car's momentum changes as a result of its collision
with the wall. (b) In a crash test (an inelastic collision),
much of the car's initial kinetic energy is transformed into
the energy it took to damage the vehicle.
(B) Find the average force exerted on the car.
Apply the impulse-momentum
After
STRATEGY This problem is similar to the a golf club striking a stationary golf ball, except that the initial
and final momenta are both nonzero. Find the momenta and substitute into the impulse-momentum
theorem, solving for Fav.
P₁ = mv; =
=
ⒸVolvo Car Corporation
F =
НЬ
4257
= -2.25 x 104 kg. m/s
Pf = mvf = (1.50 × 10³ kg)(+2.60 m/s)
= +0.390 x 104 kg. m/s
Ap
I = Pf - Pi
+0.390 x 104 kg · m/s - (-2.25 x 104 kg • m/s)
I = 2.64 x 104 kg. m/s
(1.50 × 10³ kg)(-15.0 m/s)
2.64 x 104 kg. m/s
= +1.76 x 105 N
Transcribed Image Text:EXAMPLE 6.2 GOAL Find an impulse and estimate a force in a collision of a moving object with a stationary object. How Good Are the Bumpers? PROBLEM In a crash test, a car of mass -15.0 m/s +2.60 m/s SOLUTION (A) Find the impulse delivered to the car. Calculate the initial and final momenta of the car. The impulse is just the difference between the final and initial momenta. Before 1.50 x 10³ kg collides with a wall and rebounds as in the figure. The initial and final velocities of the car are v₁ = -15.0 m/s and vf = 2.60 m/s, respectively. If the collision lasts for 0.150 s, find (a) the impulse delivered to the car due to the collision and (b) the size and direction of the average force exerted on the car. a (a) This car's momentum changes as a result of its collision with the wall. (b) In a crash test (an inelastic collision), much of the car's initial kinetic energy is transformed into the energy it took to damage the vehicle. (B) Find the average force exerted on the car. Apply the impulse-momentum After STRATEGY This problem is similar to the a golf club striking a stationary golf ball, except that the initial and final momenta are both nonzero. Find the momenta and substitute into the impulse-momentum theorem, solving for Fav. P₁ = mv; = = ⒸVolvo Car Corporation F = НЬ 4257 = -2.25 x 104 kg. m/s Pf = mvf = (1.50 × 10³ kg)(+2.60 m/s) = +0.390 x 104 kg. m/s Ap I = Pf - Pi +0.390 x 104 kg · m/s - (-2.25 x 104 kg • m/s) I = 2.64 x 104 kg. m/s (1.50 × 10³ kg)(-15.0 m/s) 2.64 x 104 kg. m/s = +1.76 x 105 N
REMARKS When the car doesn't rebound off the wall, the average force exerted on the car is smaller
than the value just calculated. With a final momentum of zero, the car undergoes a smaller change in
momentum.
QUESTION When a person is involved in a car accident, why is the likelihood of injury greater in a
head-on collision as opposed to being hit from behind? Answer using the concepts of relative velocity,
momentum, and average force. (Select all that apply.)
The momentum of the driver relative to the ground is greater in a head-on collision.
The average force on the driver is greater in the head-on collision.
The change in momentum is greater in the head-on collision.
The collapse of the crumple zone in the front of the car occurs only in the head-on collision.
The velocity of the driver relative to the ground is greater in a head-on collision.
ооо
PRACTICE IT
Use the worked example above to help you solve this problem. In a crash test, a car of mass
1.41 x 10³ kg collides with a wall and rebounds as shown in the figure. The initial and final velocities of
the car are v; = -15.5 m/s and vf = 2.35 m/s, respectively. If the collision lasts for 0.155 s, find the
following.
(a) the impulse delivered to the car due to the collision
kg. m/s
(b) the size of and direction of the average force exerted on the car (Indicate the direction with
the sign of your answer.)
N
EXERCISE
HINTS: GETTING STARTED I I'M STUCK!
Use the values from PRACTICE IT to help you work this exercise. Suppose the car doesn't rebound off the
wall, but the time interval of the collision remains at 0.155 s. In this case, the final velocity of the car is
zero. Find the average force exerted on the car. (Indicate the direction with the sign of your answer.)
N
Transcribed Image Text:REMARKS When the car doesn't rebound off the wall, the average force exerted on the car is smaller than the value just calculated. With a final momentum of zero, the car undergoes a smaller change in momentum. QUESTION When a person is involved in a car accident, why is the likelihood of injury greater in a head-on collision as opposed to being hit from behind? Answer using the concepts of relative velocity, momentum, and average force. (Select all that apply.) The momentum of the driver relative to the ground is greater in a head-on collision. The average force on the driver is greater in the head-on collision. The change in momentum is greater in the head-on collision. The collapse of the crumple zone in the front of the car occurs only in the head-on collision. The velocity of the driver relative to the ground is greater in a head-on collision. ооо PRACTICE IT Use the worked example above to help you solve this problem. In a crash test, a car of mass 1.41 x 10³ kg collides with a wall and rebounds as shown in the figure. The initial and final velocities of the car are v; = -15.5 m/s and vf = 2.35 m/s, respectively. If the collision lasts for 0.155 s, find the following. (a) the impulse delivered to the car due to the collision kg. m/s (b) the size of and direction of the average force exerted on the car (Indicate the direction with the sign of your answer.) N EXERCISE HINTS: GETTING STARTED I I'M STUCK! Use the values from PRACTICE IT to help you work this exercise. Suppose the car doesn't rebound off the wall, but the time interval of the collision remains at 0.155 s. In this case, the final velocity of the car is zero. Find the average force exerted on the car. (Indicate the direction with the sign of your answer.) N
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