Transcript Chapter Six - School District of Haverford Township

Raymond A. Serway
Chris Vuille
Chapter Six
Momentum and Collisions
Link to
ESPN Running with Momentum
Link to a
Conceptual Explanation of
Momentum and Impulse
Lesson One
Collisions
• Conservation of momentum allows complex
collision problems to be solved without
knowing about the forces involved
• Information about the average force can be
derived
Introduction
Momentum
• The linear momentum of an object of mass
m moving with a velocity is defined as the
product of the mass and the velocity
–
– SI Units are kg m / s
– Vector quantity, the direction of the momentum is
the same as the velocity’s
Section 6.1
More About Momentum
• Momentum components
– px = m vx and py = m vy
• Applies to two-dimensional motion
• Momentum is related to kinetic energy
–
Section 6.1
Read and take notes for pages 161-163
Link to
ESPN Relaxing with Impulse
Link to
Introduction to
Momentum (Khan Academy)
Impulse
• In order to change the momentum of an object, a
force must be applied
• The time rate of change of momentum of an object is
equal to the net force acting on it
–
– Gives an alternative statement of Newton’s second law
– Also valid when the forces are not constant
Section 6.1
Impulse cont.
• When a single, constant force acts on the
object, there is an impulse delivered to the
object
–
– is defined as the impulse
– Vector quantity, the direction is the same as the
direction of the force
– SI unit of impulse: kg . m / s
Section 6.1
Impulse-Momentum Theorem
• The theorem states that the impulse acting on
the object is equal to the change in
momentum of the object
–
– If the force is not constant, use the average force
applied
Section 6.1
Average Force in Impulse
• The average force can
be thought of as the
constant force that
would give the same
impulse to the object in
the time interval as the
actual time-varying
force gives in the
interval
Section 6.1
Average Force cont.
• The impulse imparted by a force during the
time interval Δt is equal to the area under the
force-time graph from the beginning to the
end of the time interval
• Or, the impulse is equal to the average force
multiplied by the time interval,
Section 6.1
EXAMPLE 6.1 Teeing Off
A golf ball being struck by a club.
Goal Use the impulse-momentum
theorem to estimate the average force
exerted during an impact.
Problem A golf ball with mass 5.0 10-2
kg is struck with a club as in the figure.
The force on the ball varies from zero
when contact is made up to some
maximum value (when the ball is maximally deformed) and then back to
zero when the ball leaves the club, as in a graph of force vs. time.
Assume that the ball leaves the club face with a velocity of +44 m/s. (a)
Find the magnitude of the impulse due to the collision. (b) Estimate the
duration of the collision and the average force acting on the ball.
Strategy In part (a), use the fact that the impulse is equal to the change in
momentum. The mass and the initial and final speeds are known, so this
change can be computed. In part (b), the average force is just the change
in momentum computed in part (a) divided by an estimate of the duration
of the collision. Guess at the distance the ball travels on the face of the
club (about 2 cm, roughly the same as the radius of the ball). Divide this
distance by the average velocity (half the final velocity) to get an estimate
of the time of contact.
SOLUTION
(a) Find the impulse delivered to the ball.
The problem is essentially one dimensional. Note that vi = 0, and
calculate the change in momentum, which equals the impulse.
I = Δp = pf − pi = (5.0 10-2 kg )(44 m/s) − 0 = +2.2 kg · m/s
(b) Estimate the duration of the collision and the average force acting on
the ball. Estimate the time interval of the collision, Δt, using the
approximate displacement (radius of the ball) and its average speed (half
the maximum speed).
Δt =
Δx
vav
Fav =
=
Δp
Δt
2.00 10-2 m
= 9.1 10-4 s
22 m/s
2.2 kg · m/s
3
=
=
+2.4
10
N
-4
9.1 10 s
LEARN MORE
Remarks This estimate shows just how large such contact forces can be.
A good golfer achieves maximum momentum transfer by shifting weight
from the back foot to the front foot, transmitting the body's momentum
through the shaft and head of the club. This timing, involving a short
movement of the hips, is more effective than a shot powered exclusively
by the arms and shoulders. Following through with the swing ensures that
the motion isn't slowed at the critical instant of impact.
Question What average club speed would double the average force?
31.1 m/s
PRACTICE IT
Use the worked
example above to
help you solve this
problem. A golf ball
with mass 4.90 10-2
kg is struck with a
club as shown in the
figure. The force on
the ball varies from zero when contact is made up to some maximum
value (when the ball is maximally deformed) and then back to zero when
the ball leaves the club, as in the graph of force vs. time in the figure
below. Assume that the ball leaves the club face with a velocity of +50
m/s.
(a) Find the magnitude of the impulse due to the collision.
(2.45 kg · m/s)
(b) Estimate the duration of the collision and the average force acting on
the ball. 0.0008 s and Average Force is 3060N
Read and take notes for pages 165-166
EXAMPLE 6.2 How Good Are the Bumpers?
Goal Find an impulse and estimate
a force in a collision of a moving
object with a stationary object.
Problem In a crash test, a car of
mass 1.50 103 kg collides with a
wall and rebounds as in the figure.
The initial and final velocities of the car are vi = -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.
Strategy This problem is similar to the previous example, except that the
initial and final momenta are both nonzero. Find the momenta and
substitute into the impulse-momentum theorem, solving for Fav.
SOLUTION
(a) Find the impulse delivered to the car.
Calculate the initial and final momenta of the car.
pi = mvi = (1.50 103 kg)(-15.0 m/s) = -2.25 104 kg · m/s
pf = mvf = (1.50 103 kg)(+2.60 m/s) = + 3.90 103 kg · m/s
The impulse is just the difference between the final and initial momenta.
I =pf - pi = +3.90 103 kg · m/s - (-2.25 104 kg · m/s)
I = 2.64 104 kg · m/s
(b) Find the average force exerted on the car.
Apply the impulse-momentum theorem.
fav =
Δp
Δt
=
2.64 104 kg · m/s
0.150 s
= +1.76 105 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 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.
The momentum of the driver relative to
the ground is greater in a head-on collision.
Impulse Applied to Auto Collisions
• The most important factor is the collision
time, or the time it takes the person to come
to a rest
– Increasing this time will reduce the chance of
dying in a car crash
• Ways to increase the time
– Seat belts
– Air bags
Section 6.1
Typical Collision Values
• For a 75 kg person
traveling at 27 m/s
(60.0 mph) and coming
to stop in 0.010 s
• F = -2.0 x 105 N
• a = 280 g
• Almost certainly fatal
Section 6.1
Seat Belts
• Seat belts
– Restrain people so it takes more time for them to
stop
– New time is about 0.15 seconds
– New force is about 9.8 kN
– About one order of magnitude below the values
for an unprotected collision
Section 6.1
Air Bags
• The air bag increases the time of the collision
• It will also absorb some of the energy from the body
• It will spread out the area of contact
– Decreases the pressure
– Helps prevent penetration wounds
Section 6.1
Now you have the ability to answer numbers
1-7
in Unit 1 D Momentum and Collisions on
Webassign