6.1 momentum and Impulse
Objectives
1. Compare the momentum of different moving
objects.
2. Compare the momentum of the same object
moving with different velocities.
3. Identify examples of change in the
momentum of an object.
4. Describe changes in momentum in terms of
force and time.
Questions
1. International regulations specify the mass of official
soccer balls. How does the mass of a ball affect the
way it behaves when kicked?
2. How does the velocity of the player’s foot affect the
final velocity of the ball?
3. Is the pins more likely to move rapidly when
the ball travels at a high speed or at a low
speed?
4. If two bowling balls of different masses move
at the same speed, which is more likely to
move the pin?
Linear Momentum
• The language of physics:
– As seen before, words used in our everyday language, such as
work and energy, often have precise definitions in physics,
momentum is another example of such a word.
• Examples of everyday meaning: strength gained by a series of
events:
– The company has had a successful year and hopes to maintain
its momentum by introducing new products.
– The movie loses momentum toward the end.
• Physics definition of momentum: a property of a moving body, is
equal to the product of the body's mass and velocity.
Momentum equation
Momentum = mass • velocity
p=m•v
m: mass in kg
v: velocity in m/s
p: momentum in kg∙m/s
• Momentum is a vector quantity
• The direction of the momentum vector is the same as the
direction of the velocity of the ball. Which is the same as
the direction that an object is moving.
Mass and velocity have equal importance in
momentum
• Momentum is directly proportional to mass and momentum
is directly proportional to velocity
• Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and
moving with a speed of 2.0 m/s. Its momentum is 2.0 kg•m/s.
If the cart was instead loaded with three 0.5-kg bricks, then
the total mass of the loaded cart would be 4.0 kg•m/s. A
doubling of the mass results in a doubling of the momentum.
• Similarly, if the 2.0-kg cart had a velocity of 4.0 m/s (instead of
2.0 m/s), then the cart would have a momentum of 8.0
kg•m/s (instead of 4.0 kg•m/s). A doubling in velocity results
in a doubling of the momentum.
Example 1
•
Express your understanding of the concept and
mathematics of momentum by answering the
following questions 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.
Example 2
2. A car possesses 20 000 units of momentum.
What would be the car's new momentum if ...
a. its velocity was doubled.
b. its velocity was tripled.
c. its mass was doubled (by adding more
passengers and a greater load)
d. both its velocity and mass was doubled.
Determine sum of momentum – example 3
• Four billiard balls, each of mass .5 kg, all are traveling in the
same direction on a billiard table, with speeds 2 m/s, 4 m/s, 8
m/s and 10 m/s. What is the linear momentum of this system?
ptotal = p1 + p2 + p3 + p4
ptotal = (0.5 kg)(2 m/s + 4 m/s + 8 m/s + 10 m/s)
ptotal = 12 kg∙m/s
• If all four balls collide, what is the total momentum after the
collision?
12 kg∙m/s
Example 4 - Sample Problem 6A
• A 2250 kg pickup truck has a velocity of 25
m/s to the east. What is the momentum of
the truck?
Class work
• Page 200 – practice 6A
1. 2500 kg∙m/s
2. a. 120 kg∙m/s NW; b. 94 kg∙m/s NW; c 27 kg∙m/s NW
3. 46 m/s E
• Page 232 #12-13, 41,43, 42*
12. a. 8.35 x 10-21 kg∙m/s up;
c. 750 kg∙m/s SW;
13. 5.2 toward home plate
41. 42.0 m/s toward 2nd base
42. 3.0 kg; 10. m/s
43. a. 0 kg∙m/s;
b. 4.88 kg∙m/s right
d. 1.78 x 1029 kg∙m/s forward
b. 1.0 kg∙m/s upward
A change in Momentum takes force and time
•
•
•
•
demo
p = m∙v
∆p = m∙∆v
∆p = m∙a∙∆t
∆p = F∙∆t
a 
v
t
v  a  t
A change in momentum takes force and time
p  F  t
The equation states that a net external force, F, applied to an object for a
certain time interval, ∆t, will cause a change in the object’s momentum. This
change in the object’s momentum equals to the product of F and the time
interval.
Impulse-Momentum Theorem
F  t  p
F   t  mv
f
 mv i
In physics, the quantity F•∆t is known as impulse, represented by
letter J.
F: force in Newton
J = F∙∆t
t: time in second
J: impulse in N∙s
•Impulse is a vector quantity, its direction is the same as the net
force F.
Impulse
= Change in Momentum
Impulse = Change in momentum
F  t  p
• The impulse experienced by the object is the cause of the
change in momentum.
• The impulse experience by the object equals the change in
momentum of the object.
• There are two way to determine impulse:
– F·t
– m∙∆v
• There are two way to determine change in momentum:
– F·t
– m∙∆v
Example 1
•
A 0.10-kilogram model rocket’s engine is
designed to deliver an impulse of 6.0
newton-seconds. If the rocket engine burns
for 0.75 second, what average force does it
produce?
Example 2
•
In the diagram, a 60.-kilogram rollerskater
exerts a 10.-newton force on a 30.-kilogram
rollerskater for 0.20 second. What is the
magnitude of the impulse applied to the 30.kilogram rollerskater?
Example 3
• If the halfback experienced a force of 800 N
for 0.9 seconds, then we could say that the
720 N•s
impulse was _______________
• This impulse would cause a momentum
720 kg•m/s
change of ____________
• In a collision, the impulse experienced by an
object is always equal to the momentum
change.
Note: the unit N•s = kg•m/s
Example 4
•
A 0.50-kg cart (#1) is pulled with a 1.0-N force for 1 second;
another 0.50 kg cart (#2) is pulled with a 2.0 N-force for 0.50
seconds.
a.
Which cart (#1 or #2) has the greatest acceleration? Explain.
b. Which cart (#1 or #2) has the greatest impulse? Explain.
c.
Which cart (#1 or #2) has the greatest change in
momentum? Explain.
Example 5
•
Two cars of equal mass are traveling down Lake Avenue with
equal velocities. They both come to a stop over different
lengths of time. The ticker tape patterns for each car are
shown on the diagram below.
a. Which car (A or B) experiences the greatest acceleration?
Explain.
b. Which car (A or B) experiences the greatest change in
momentum? Explain.
c. Which car (A or B) experiences the greatest impulse? Explain.
Example 6
•
Calculate the change of momentum for a 1.0-kilogram cart
for the following situations. Which situation has the
greatest change of momentum?
a.
b.
c.
d.
accelerating it from rest to 3.0 m/s
accelerating it from 2.0 m/s to 4.0 m/s
applying a net force of 5.0 N for 2.0 s
applying a net force of 10.0 N for 0.5 s
Example 7
• A 0.15-kilogram baseball moving at 20. meters
per second is stopped by a catcher in 0.010
second. What is the average force stopping
the ball?
Example 8
• A 2.0-kilogram laboratory cart is sliding across a
horizontal frictionless surface at a constant
velocity of 4.0 meters per second east. What will
be the cart’s velocity after a 6.0-newton
westward force acts on it for 2.0 seconds?
Example 9
•
A 1,000-kilogram car traveling due east at 15 meters per
second is hit from behind and receives a forward impulse of
8,000 newton-seconds. Determine the magnitude of the car’s
change in momentum due to this impulse.
Example 10 - Sample problem 6B
• A 1400 kg car moving westward with a velocity of 15 m/s
collides with a utility pole and is brought to rest in 0.30 s. Find
the magnitude of the force exerted on the car during the
collision.
Class work
• Page 211 – practice 6B
1.
2.
3.
4.
380 N right
1100 N up
16.0 kg∙m/s S
a. 9 m/s right;
b. 15 m/s left
• Page 232 #14-15, 47, 55, 56*
14.
15.
47.
55.
56.
160 N right
18 N
400 N
a. 0.83 m/s right; b. 1.2 m/s left
a. 9.9 m/s down; b. 1800 N up
Real-World Applications
• The effect of collision time upon the amount
of force an object experiences
• The effect of rebounding upon the velocity
change and hence the amount of force an
object experiences.
The relationship between
Collision Time and Force of Impact
m∙∆v = F∙∆t
•The greater the time over which the collision occurs, the smaller the force acting
upon the object.
•To minimize the effect of the force on an object involved in a collision, the time
must be increased.
•To maximize the effect of the force on an object involved in a collision, the time
must be decreased.
Reduce force by increase time
• Air bags are used in automobiles because they are able to
minimize the affect of the force on an object involved in a
collision. Air bags accomplish this by extending the time
required to stop the momentum of the driver and passenger.
• Padded dashboards also reduces force by increase time.
• A boxer rides the punch in order to extend the time of impact
of the glove with their head.
• Nylon ropes are used in the sport of rock-climbing because of
its ability to stretch. The rock climber can appreciate
minimizing the effect of the force through the use of a longer
time of impact .
The Effect of Rebounding
• Bouncing off each other is known as rebounding. Rebounding
involves a change in the direction of an object; rebounding
situations are characterized by
– a large velocity change
F∙∆t = m∙∆v
– a large momentum change.
– a large impulse
• The importance of rebounding is critical to the outcome
of automobile accidents.
• Automobiles are made with crumple zones. Crumple
zones minimize the affect of the force in an automobile
collision in two ways.
– By crumpling, the car is less likely to rebound upon
impact, thus minimizing the momentum change and
the impulse.
– The crumpling of the car lengthens the time over
which the car's momentum is changed; by increasing
the time of the collision, the force of the collision is
greatly reduced.
Increase velocity by increasing time:
F∙∆t = m∙∆v
• In racket and bat sports, hitters are often encouraged
to follow-through when striking a ball.
• In this situation, both the force applied (as hard as you can) and the
mass (the mass of the ball) are constant. By following through, the
hitter increases the time, the result is increasing the ball’s velocity.
examples
•
A constant force can act on an object for different
lengths of time. As the length of time increases,
1. the impulse imparted to the object
a) decreases
b) increases
c) remains the same
2. The momentum of the object
a) decreases
b) increases
c) remains the same
Example 3
• A cannonball shot from a long-barrel cannon travels
faster than one shot from a short-barrel cannon
because the cannonball receives a greater
a. force.
b. impulse.
A cannonball shot from a cannon receive
the same force regardless of the length of
c. both A and B
its barrel. A long-barrel will take longer
d. neither A nor B
time (t) for the cannonball to travel.
Since J = F∆t = m∆v, the longer the time,
the bigger the impulse, the faster it will
travel.
Example 4
•
A student drops two eggs of equal mass simultaneously from
the same height. Egg A lands on the tile floor and
breaks. Egg B lands intact, without bouncing, on a foam pad
lying on the floor. Compared to the magnitude of the impulse
on egg A as it lands, the magnitude of the impulse on egg B as
it lands is _________________ (less, greater, the same) as
egg A.
Stopping times and distances depend on the
impulse-momentum theorem
F∙∆t = m∙∆v
•
Two identical trucks travels with the same velocity. One is
empty with mass m, and the other is loaded with mass 2m.
1. Why is loaded truck’s stopping time twice as much as the
empty truck’s when acted on by the same force? Loaded is 2x unloaded
2. How do the stopping distances compare? Loaded is 2x unloaded
Example 5
• If the maximum coefficient of kinetic friction
between a car and a road is 0.50, what is the
minimum stopping distance for car moving at
29 m/s? (about 65 mi/hr)
Example 6 - Sample problem 6C
• A 2250 car traveling to the west slows down
uniformly from 20.0 m/s to 5.00 m/s.
1. How long does it take the car to decelerate if
the force on the car is 8450 N to the east?
2. How far does the car travel during the
deceleration?
Class work
• Page 213 practice 6C
1. 5.33 s; 53.3 m E
2. a. 14 m/s N;
b. 42 m/s N;
3. a. 1.22 x 104N E; b. 53.3 m W
• Page 232 #16, 47
16. 0.010 s; 0.13 m
47. 400 N
c. 8.0 s