Transcript Gravitation and Momentum

Momentum & Impulse
Excerpts from
Chapters 9 and 11
Impulse and Momentum
 Newton’s 2nd Law of motion can be rewritten by
using the definition of acceleration as the change in
velocity over the change in time.
F  ma
v 
F  m 
t 
Impulse and Momentum
 If the change in time is multiplied out of the
denominator, we are left with the following:
Ft  mv
Ft  mv
 The product of force and change in time is
called the impulse (symbol is J).
 Impulse is a vector quantity and is
measured in Newton-seconds (Ns).
Impulse
 If a car hits a haystack or the same car hits a wall,
momentum is decreased by same impulse – the
same products of force and time.
 However, impact force is greater into the wall than
it is into the haystack as the haystack extends
impact time, lessening the impact force.
 Impact time is the time during which momentum is
brought to zero.

  mv
 The product of the mass and the velocity is
called the momentum (symbol  -“rho”) of
an object.
 Momentum is also a vector quantity and is
measured in kgm/s.
 Note that the units for impulse and
momentum appear different, but they are
actually the same unit when simplified.
Momentum
 Momentum can be increased with an
increase in either mass or in velocity or
both.
 Ex: a rolling bowling ball has greater momentum
than a tennis ball rolling at the same speed
because its mass is greater
 Ex: a racecar going forward at 120 mi/hr has
greater momentum than the same size car going
90 mi/hr due to its greater velocity
 If an object is not moving (no matter how big
it is), the momentum is equal to zero.

Ft  mv f  mvi
 The impulse-momentum theorem states
that the impulse on an object is equal to
the object’s final momentum minus the
object’s initial momentum.
 Can also be written as:
Ft   f   i
Example 1
 Tiger Woods hits a 0.050kg golf ball, giving
it a speed of 75m/s. What is the impulse
given to the ball?
Example 1 Answer
m  0.050kg
Ft  mv
v  75m /s
impulse  ?
impulse  0.050  75
impulse  3.75  3.8N  s

Example 2
 Shane hits a stationary 0.12kg hockey puck
with a force that lasts for 1.0x10-2s and
makes the puck shoot across the ice with a
speed of 20.0m/s, scoring a goal for the
team. With what force did Shane hit the
puck?
Example 2 Answer
m  0.12kg
Ft  mv
t  1.0 102 s
v  20.0m /s
F ?
mv
F
t
0.12  20.0
F
2
1.0 10
F  240N
Example 3
 Diana, whose mass is 50.0kg, leaves a ski
jump with a velocity of 21.0m/s. What is
her momentum as she leaves the ski jump?
Conservation of Momentum
 A system is the environment and all of the
objects examined in a problem.
 A closed system is a system in which no mass
is gained or lost.
 An isolated system is a system in which the
net external force is zero… no forces acting
outside of the system have an effect inside
of it.
Conservation of Momentum
 The law of conservation of momentum states that the
sum of momentum of any closed, isolated system does
not change… or that the sum of the momentum of the
objects in that system is constant.
Conservation of Momentum
 Mathematically, we can view this as a BEFORE and
AFTER situation.
 For any two objects A and B:
Ai  Bi  Af  Bf
Types of Collisions
 If two objects bounce apart when they collide, it is called an
elastic collision and can be written:
m1v1i  m2v 2i  m1v1 f  m2v 2 f
 If two objects stick together when they collide, it is called an
inelastic collision and can be written:
m1v1i  m2v 2i  (m1  m2 )v f
Example 1
 Tubby and his twin brother Chubby have a
combined mass of 200.0kg and are zooming
along in a 100.0kg amusement park bumper
car at 10.0m/s. They bump Melinda’s car,
which is sitting still. Melinda has a mass of
25.0kg. After the elastic collision, the twins
continue ahead with a speed of 4.12m/s.
How fast is Melinda’s car bumped across the
floor?
Example 1 Picture
Before Collision
After Collision
T&C
Mel
T&C
Mel
m1  300.0kg
m2  125.0kg
v1i  10.0m /s
v 2i  0m /s
m1  300.0kg
v1 f  4.12m /s
m 2  125.0kg
v2 f  ?



Example 1 Answer
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f  m2v 2 f
m1v1i  m2v 2i  m1v1 f
 v2 f
m2
Example 1 Answer
[m1v1i  m2v 2i  m1v1 f ]
 v2 f
m2
[(300.0 10.0)  (125.0  0)  (300.0  4.12)]
 v2 f
125.0
14.1m /s  v 2 f
Example 2
 If an 800.kg sports car slows to 13.0m/s to
check out an accident scene and the
1200.kg pick-up truck behind him continues
traveling at 25.0m/s, with what velocity
will the two move if they lock bumpers
after a rear-end collision?
Example 2 Picture
Before Collision
After Collision
m1  1200.kg
m2  800.kg
v1i  25.0m /s
v 2i 13.0m /s


(m1  m2 )  2000.kg
vf  ?
Example 2 Answer
m1v1i  m2v 2i  (m1  m2 )v f
(m1v1i  m2v 2i )
 vf
(m1  m2 )
Example 2 Answer
(m1v1i  m2v 2i )
 vf
(m1  m2 )
[(1200  25.0)  (800 13.0)]
 vf
(1200  800)
20.2 m/s forward  v f
Journal #
 What do we mean when we ask people to “conserve
water”?
 What do you think it will mean if we say that
momentum is conserved?
Conservation Lab
 Objective:
 Prove that the law of conservation of momentum
is true.
 Law of Conservation of Momentum:
 The sum of the momentum of the objects prior
to a collision is equal to the sum of the
momentum of the objects after a collision.
Journal #
 Explain how an airbag protects you by
making you come to a stop differently
than hitting steering wheel.
Try to use the words impulse, force, and time in your answer.
Journal #
 Impulse is a force applied over an
interval of time. In this question, the
impulse of hitting the steering wheel and
hitting the airbag are the same amount
because they both cause you to stop.
However, the airbag applies a smaller
force to your body over a larger time,
therefore keeping you safer.
Journal #

Order these objects from the most momentum to the
least.
A.
B.
C.
D.
A bullet shot from a rifle
An elephant standing still
A bowling ball rolling
A fly buzzing by your ear