9781118230725.pdf

9-4 COLLISION AND IMPULSE 229

Impulse:The impulse is then

(Answer)

which means the impulse magnitude is

The angle of is given by

(Answer)

which a calculator evaluates as 75.4. Recall that the physi-

cally correct result of an inverse tangent might be the

displayed answer plus 180. We can tell which is correct here

by drawing the components of (Fig. 9-11c). We find that u

is actually 75.4 180 255.4, which we can write as

u 105 . (Answer)

(b) The collision lasts for 14 ms. What is the magnitude of

the average force on the driver during the collision?

KEY IDEA

From Eq. 9-35 (J Favgt), the magnitude Favgof the aver-

age force is the ratio of the impulse magnitude Jto the dura-

tiontof the collision.

Calculations: We have

. (Answer)
UsingFmawithm80 kg, you can show that the magni-
tude of the driver’s average acceleration during the collision
is about 3.22 103 m/s^2  329 g,which is fatal.
Surviving:Mechanical engineers attempt to reduce the
chances of a fatality by designing and building racetrack
walls with more “give,” so that a collision lasts longer. For
example, if the collision here lasted 10 times longer and the
other data remained the same, the magnitudes of the aver-
age force and average acceleration would be 10 times less
and probably survivable.

2.583 105 N2.6 105 N

Favg

J

t



3616 kg m/s

0.014 s

J

:

utan^1

Jy

Jx

,

J

:

J 2 Jx^2 Jy^2 3616 kg m/s3600 kg m/s.

J

:

(910iˆ3500 jˆ) kg m/s,

Sample Problem 9.04 Two-dimensional impulse, race car–wall collision

Figure 9-11ais an overhead view of
the path taken by a race car driver as his car collides with the
racetrack wall. Just before the collision, he is traveling at
speedvi70 m/s along a straight line at 30from the wall.
Just after the collision, he is traveling at speed vf50 m/s
along a straight line at 10from the wall. His mass mis 80 kg.

(a) What is the impulse on the driver due to the collision?

KEY IDEAS

We can treat the driver as a particle-like body and thus apply
the physics of this module. However, we cannot calculate
directly from Eq. 9-30 because we do not know anything about
the force on the driver during the collision. That is, we do
not have a function of or a plot for it and thus cannot
integrate to find. However, we canfind from the change in
the driver’s linear momentum via Eq. 9-32.

Calculations:Figure 9-11bshows the driver’s momentum p:i

(J

:

:p :pf:pi)

J

:

J

: F

:

(t)

F

:

(t)

J

:

J

:

Race car–wall collision.

Wall

x

y

30 °

10 °

30 °

Path

(a)

x

y

10 °

(b)

pi

pf –105°

x

y

(c)

Jy

Jx

J

The impulse on the car

is equal to the change

in the momentum.

The collision

changes the

momentum.

Figure 9-11(a) Overhead
view of the path taken by a
race car and its driver as the
car slams into the racetrack
wall. (b) The initial momen-
tum and final momentum
of the driver. (c) The
impulse on the driver
during the collision.

J

:

:pf

p:i

Additional examples, video, and practice available at WileyPLUS

before the collision (at angle 30from the positive xdirection)
and his momentum after the collision (at angle 10). From
Eqs. 9-32 and 9-22 , we can write

(9-41)

We could evaluate the right side of this equation directly on
a vector-capable calculator because we know mis 80 kg,
is 50 m/s at  10 , and is 70 m/s at 30. Instead, here we
evaluate Eq. 9-41 in component form.

x component:Along the xaxis we have

Jxm(vfxvix)

(80 kg)[(50 m/s) cos( 10 )(70 m/s) cos 30]

910 kg m/s.

y component:Along the yaxis,

Jym(vfyviy)

(80 kg)[(50 m/s) sin( 10 )(70 m/s) sin 30]

3495 kg m/s3500 kg m/s.

:vi

v:f

J

:

p:fp:imv:fmv:im(v:fv:i).

(p:mv:)

p:f