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```Estimate the amount of work the
engine performed on a 1200-kg car as
it accelerated at 1.2 m/s2 over a 150meter distance.
W  KE  Fd
m = 1200 kg
a = 1.2 m/s2
d = 150 m
F = ma
W = ∆KE=Fcosqd
W = ma×d
W = (1200 kg)(1.2 m⁄s^2 )(cos0o)(150 m)
W = 2.2x10^5 J
To complicate matters further, consider
the fact that only about 15% of the
energy extracted from gasoline actually
propels the car. The rest gets exhausted
as heat and unburnt fuel.

Linear momentum of an object is equal
to the product of its mass and velocity
SI Units of momentum are kgm/s
 A lowercase p is used for the momentum
of an individual particle

• The total linear momentum of a system
is the vector sum of the momenta of
the individual particles
• The total linear momentum of a

system, expressed by an uppercase P,
is used for the total momentum of the
individual particles of the system
Given: mcar = 1500 kg
mtruck = 40 000 kg
vcar = 25 m/s
vtruck = 1 m/s
a) pcar=mcarvcar= (1500 kg)(25 m/s)
= 3.75 x 104 kgm/s
b) ptruck = mtruckvtruck=(40000kg)(1m/s)
=4.00 x 104 kgm/s
The much slower truck has more
momentum than the faster car because
the truck has much greater mass.
Given:
m1 = m2 = 1500 kg
v1 = 25.0 m/s
v2 = 15.0 m/s
p1 = m1 v1 = (1500kg)(25.0m/s) = 3.75 x 104kgm/s
p2 = m2v2 = (1500kg)(15.0m/s) = 2.25 x 104kgm/s
Continued on next slide…
From vector addition, the magnitude of the
total momentum is
And the direction is given by
1. Determine the momentum of a ...
a. 60-kg halfback moving eastward at 9 m/s.
b. 1000-kg car moving northward at 20 m/s.
c. 40-kg freshman moving southward at 2 m/s.
2. A car possesses 20 000 units of momentum. What
would be the car's new momentum if ...
a. its velocity were doubled.
b. its velocity were tripled.
c. its mass were doubled (by adding more
d. both its velocity were doubled and its mass were
doubled.
1. a) p = m*v = 60 kg*9 m/s
p = 540 kg•m/s, east
b) p = m*v = 1000 kg*20 m/s
p = 20 000 kg•m/s, north
c) p = m*v = 40 kg*2 m/s
p = 80 kg•m/s, south
2. a) p = 40 000 units (doubling the velocity will
double the momentum)
b) p = 60 000 units (tripling the velocity will
triple the momentum)
c) p = 40 000 units (doubling the mass will
double the momentum)
d) p = 80 000 units (doubling the velocity will
double the momentum and doubling the
mass will also double the momentum; the
combined result is that the momentum is

Newton’s second law, F=ma, can also
be expressed in terms of momentum.

If the mass is constant, then
• If an object’s momentum changes, then
a force must have acted upon it.

Solve the problems on “Chapter 4”
Momentum and Energy”
› Momentum p. 27 Worksheet

Impulse is the change in momentum.
• When a moving object stops, its impulse
depends only on its change in momentum.
This can be accomplished by a large force
acting for a short time, or a smaller force
acting for a longer time.

geLSYXUE

Predict what will happen when two
identical balls are dropped from the
same height onto the same type of
surface
Solve the problems on
“Chapter 4” Momentum and Energy”
- Impulse-Momentum p.31
• Use the Rally Coach Method to work
through the Examples 3, 4, 5, and 6.
A hockey player applies an average
force of 80.0 N to a 0.25 kg hockey puck
for a time of 0.10 seconds. Determine the
impulse experienced by the hockey
puck.
Given:
F=80.0 N
m=0.25 kg
t=0.10 s I = Ft
When bunting, a baseball player uses the bat
to change both the speed and direction of
the baseball.
 (a) How will the magnitude of the momentum
of the baseball before and after the bunt
change? (Describe in words, not numbers)
 (b) The baseball has a mass of 0.16 kg; its
speeds before and after the bunt are 15m/s
and 10 m/s respectively; the bunt lasts 0.025
s. What is the change in momentum of the
baseball?
 (c) What is the average force on the ball by
the bat?

(b) Choose the direction of motion before
the bunt as positive.
v = 10 m/s, vo = 15 m/s.
p = mv – mvo
= (0.16 kg)(10 m/s) – (0.16 kg)(15
m/s)
=  4.0 kgm/s in direction opposite v0
(c)
A boy catches—with bare hands and his
arms rigidly extended—a 0.16-kg
baseball coming directly toward him at
a speed of 25 m/s. He emits an audible
“Ouch!” because the ball stings his
hands. He learns quickly to move his
hands with the ball as he catches it.
 If the contact time of the collision is
increased from 3.5 ms to 8.5 ms in this
way, how do the magnitudes of the
average impulse forces compare?

Given: m=0.16 kg, v0=25 m/s, v=0m/s,
t1=3.5ms, t2=8.5ms
 Favg t = mv  mvo = mvo, 
So the magnitude is

A karate student tries not to follow through
in order to break a board. How can the
abrupt stop of the hand (with no followthrough) generate so much force?
 Assume that the hand has a mass of 0.35 kg
and that the speeds of the hand just before
and just after hitting the board are and 0,
respectively. What is the average force
exerted by the fist on the board if (a) the fist
follows through, so the contact time is 3.0
ms, and (b) the fist stops abruptly, so the
contact time is only 0.30 ms?


By stopping, the contact time is short. From
the impulse momentum theorem (Favg t =
p = mv  mvo), a shorter contact time will
result in a greater force if all other factors
(m, vo, v) remain the same.
```