Download Physics 350 - Los Rios Community College District

Chapter 7
Momentum and Collisions
Momentum
 Newton’s Laws give a description of forces
○ There is a force acting or their isn’t
○ But what about in between
 Constant velocity
 Momentum is a quantity defined by the
velocity of a mass
○ Velocity changes -> Force
 Impulse
 Momentum describes collisions
○ Magnitude gives an indication of how hard it is
to stop
○
Momentum
From Newton’s laws: force must be present to change an object’s
velocity (speed and/or direction)
 Wish to consider effects of collisions and corresponding change in
velocity

Golf ball initially at rest, so
some of the KE of club
transferred to provide motion
of golf ball and its change in
velocity

Method to describe is to use concept of linear momentum
Linear momentum = product of mass
scalar
 velocity
vector
Momentum and Impulse

The linear momentum p of an object of
mass m moving with velocity v is defined
as
p = mv
 Same direction as velocity
m
 Units
○ kg x m / s
v
p
Momentum
p  mv
Vector quantity, the direction of the momentum is the same
as the velocity’s
 Applies to two-dimensional motion as well

p x  m vx and p y  m vy
Size of momentum: depends upon mass
depends upon velocity
Impulse


In order to change the momentum of an object (say,
golf ball), a force must be applied
The time rate of change of momentum of an object is
equal to the net force acting on it

F n et
 p m (v f  v i )


 m a or :  p  F n et  t
t
t
 Gives an alternative statement of Newton’s second law
 (F Δt) is defined as the impulse
 Impulse is a vector quantity, the direction is the same as the
direction of the force
Graphical Interpretation of
Impulse

Usually force is not
constant, but timedependent
im pulse 
 F t
 ti


i
i
 area under F ( t ) curve
If the force is not constant,
use the average force
applied
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
If force is constant: impulse = F t
Example: 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
 This will reduce the chance of dying in a car crash

Ways to increase the time
 Seat belts
 Air bags

The air bag increases the time of the collision and
absorbs some of the energy from the body
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward
you. Both have the same momentum, and you exert the same
force to stop each. How do the time intervals to stop them
compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time.
3. It takes more time to stop the ping-pong ball.
ConcepTest
Suppose a ping-pong ball and a bowling ball are rolling toward
you. Both have the same momentum, and you exert the same
force to stop each. How do the time intervals to stop them
compare?
1. It takes less time to stop the ping-pong ball.
2. Both take the same time. 
3. It takes more time to stop the ping-pong ball.
Note: Because force equals the time rate of change of
momentum, the two balls loose momentum at the same
rate. If both balls initially had the same momenta, it
takes the same amount of time to stop them.
Problem: Teeing Off
A 50-g golf ball at rest is hit by “Big
Bertha” club with 500-g mass.
After the collision, golf leaves
with velocity of 50 m/s.
a) Find impulse imparted to ball
b) Assuming club in contact with
ball for 0.5 ms, find average force
acting on golf ball
Problem: teeing off
1. Use impulse-momentum relation:
Given:
mass:
m=50 g
= 0.050 kg
velocity: v=50 m/s
Find:
impulse=?
Faverage=?
impulse  p  mv f  mvi
 0.050 kg 50 m s   0
 2.50 kg  m s

2. Having found impulse, find the average
force from the definition of impulse:
p 2.50 kg  m s

t
0.5 10 3 s
 5.00 103 N
p  F  t , thus F 

Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball:
Momentum and Impulse
Can describe Newton’s laws more
precisely with momentum

Law I : If no forces act, Fnet = 0 so a = 0 which
implies that velocity is constant p = mv,
therefore momentum is constant
 Momentum constant => conserved quantity


Law II : F = Δp / Δt = m (Δv / Δt) = ma
Law III : Rate of change of momentum
generated by action force is exactly opposite
to the rate of change of momentum by the
reaction force
Conservation of Momentum

Definition: an isolated system is the one that
has no external forces acting on it
Momentum in an isolated system in which a
collision occurs is conserved (regardless of
the nature of the forces between the objects)
 A collision may be the result of physical contact
between two objects
 “Contact” may also arise from the electrostatic
interactions of the electrons in the surface atoms of
the bodies
Conservation of Momentum
The principle of conservation of
momentum states when no external
forces act on a system consisting of two
objects that collide with each other, the
total momentum of the system before the
collision is equal to the total momentum
of the system after the collision
Conservation of Momentum

Mathematically:
m 1 v 1i  m 2 v 2 i  m 1 v 1 f  m 2 v 2 f
 Momentum is conserved for the system of objects
 The system includes all the objects interacting with each other
 Assumes only internal forces are acting during the collision
 Can be generalized to any number of objects
Momentum and Impulse

Example:
Conservation of Momentum

If the isolated system consists of two
objects undergoing a collision, the total
momentum of the system is the same
before and after the collision.
 Same concept as energy
m1v1i + m2v2i = m1v1f + m2v2f
Conservation of Momentum

Remember conservation of momentum
applies to the system
 The system includes all the objects
interacting with each other
 Assumes only internal forces are acting
during the collision

You must define the isolated system
 Can be generalized to any number of
objects
ConcepTest 7.1 Rolling in the Rain
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what happens
to the speed of the cart as the rain
collects in it? (assume that the
rain falls vertically into the box)
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
ConcepTest 7.1 Rolling in the Rain
An open cart rolls along a
frictionless track while it is
raining. As it rolls, what happens
to the speed of the cart as the rain
collects in it? (assume that the
rain falls vertically into the box)
1) speeds up
2) maintains constant speed
3) slows down
4) stops immediately
Since the rain falls in vertically, it adds
no momentum to the box, thus the box’s
momentum is conserved. However,
since the mass of the box slowly
increases with the added rain, its
velocity has to decrease.
Follow-up: What happens to the cart when it stops raining?
Collisions
Momentum is conserved in any collision
 We can define two types of collisions to
describe how momentum is transferred
within the system

 Elastic
○ Both momentum and kinetic energy are
conserved
 Inelastic
○ Kinetic energy is not conserved, momentum is
conserved
Collisions
Elastic Collisions


Momentum is conserved
Total Energy is
conserved
m 1v 1i  m 2 v 2 i  m 1v 1f  m 2 v 2 f
1
1
1
1
2
2
2
m 1v 1i  m 2 v 2 i  m 1v 1f  m 2 v 22 f
2
2
2
2
Collisions

Elastic Collisions
 Instead of using the energy quadratic
equation, we can simplify the equation to
v 1i  v 2 i   ( v 1 f  v 2 f )
 The relative velocity of the two objects
before the collision equals the negative of
the relative velocity of the two objects after
the collision
Collisions

Elastic Collision
 Consider the case: both targets moving
 So,
m1v1i + m2v2i = m1v1f + m2v2f
½ m1v21i + ½ m2v22i = ½ m1v21f + ½m2v2
We know m1, m2, v1i, and v2i
Solve for v1f, v2f
Collisions

General Equation
m1 – m2
2m2
v1f = ————— v1i + ————— v2i
m1 +m2
m1 + m2
2m2
m1 – m 2
v2f = ————— v1i + ————— v2i
m1 +m2
m1 + m2
Collisions

Consider the case v2i = 0
we get,
m1 – m2
v1f = ———— v1i
m1 + m2
2m2
v2f = ———— v1i
m1 + m2
Collisions (special cases)
Special situations
Equal masses (m2 = m1)

1.


Exchange velocities
If v2i = 0, v1f = 0 and v2f = v2i
○
body 1 stops and body 2 takes off with speed
v1i
Collisions (special cases)
Massive target (m2 >> m1)
1.
v1f ≈ -v1i and v2f ≈ (2m1 / m2)v1i
body 1 bounces back in opposite direction,
body 2 moves slowly


○
Example: golf ball at cannonball
Collisions (special cases)
Massive projectile (m1 >> m2)
1.
v1f ≈ v1i and v2f ≈ 2v1i
Body 1 continues at same speed, body two at
twice the speed!


○
Example: cannonball at golf ball
Collisions
 Example
Types of Collisions

Momentum is conserved in any collision
what about kinetic energy?

Inelastic collisions
 Kinetic energy is not conserved
K Ei  K E
f
 lo s t e n e r g y
○ Some of the kinetic energy is converted into other types of energy
such as heat, sound, work to permanently deform an object
 Perfectly inelastic collisions occur when the objects stick
together
○ Not all of the KE is necessarily lost

Actual collisions
 Most collisions fall between elastic and perfectly inelastic
collisions
Collisions

Perfectly Inelastic
 Objects stick together
○ Not all of the KE is
necessarily lost
 Most real life collisions
fall between elastic and
perfectly inelastic
collisions
Collisions

Perfectly Inelastic
Collisions
 Since both masses
have the same
velocity, conservation
of momentum
becomes,
m1v1i + m2v2i = (m1 + m2)vf
Collisions

Example of perfectly inelastic collision
Momentum is
conserved, kinetic
energy is not
conserved and
converted into
heat, sound, and
destruction of
smaller car
Collisions

Application –
Perfectly Inelastic
Collision
 Ballistic Pendulum
○ Initially used to
measure speeds of
bullets before
electronic timing
devices were
developed.
Sketches for Collision
Problems


Draw “before” and
“after” sketches
Label each object
 include the direction
of velocity
 keep track of
subscripts
Sketches for Perfectly
Inelastic Collisions



The objects stick
together
Include all the
velocity directions
The “after” collision
combines the
masses
Perfectly Inelastic
Collisions:


When two objects stick
together after the
collision, they have
undergone a perfectly
inelastic collision
Suppose, for example,
v2i=0. Conservation of
momentum becomes
m 1v1i  m 2 v 2 i  ( m 1  m 2 ) v f
m 1v1i  0  ( m 1  m 2 ) v f
E.g., if m1  1000 kg , m 2  1500 kg :
(1000 kg )( 50 m s )  0  ( 2500 kg ) v f ,
5  10 4 kg  m s
vf 
 20 m s.
3
2 .5  10 kg
Perfectly Inelastic
Collisions:

What amount of KE lost
during collision?
1
1
m1v12i  m 2 v 22i
2
2
1
 (1000 kg )( 50 m s ) 2  1 .25  10 6 J
2
KE before 
1
( m1  m 2 ) v 2f
2
1
 ( 2500 kg )( 20 m s ) 2  0 .50  10 6 J
2
KE after 
 K E lo st  0 .7 5  1 0 6 J
lost in heat/”gluing”/sound/…
Collisions

Example – Perfectly Inelastic Collision
 Two objects, one with mass of 5kg and the
other 3kg, approach each other with
velocities of 2m/s from opposite directions.
After they collide, they stick together. What
is the initial and final kinetic energy?
M = 5 kg
M = 3 kg
Glancing Collisions

For a general collision of two objects in
three-dimensional space, the
conservation of momentum principle
implies that the total momentum of the
system in each direction is conserved

m 1 v1ix  m 2 v 2 ix  m 1 v1 fx  m 2 v 2 fx and
m 1 v1iy  m 2 v 2 iy  m 1 v1 fy  m 2 v 2 fy
 Use subscripts for identifying the object,
initial and final, and components
Glancing Collisions



The “after” velocities have x and y components
Momentum is conserved in the x direction and in
the y direction
Apply separately to each direction
Two-dimensional Collisions

For a general collision of two objects in threedimensional space, the conservation of
momentum principle
m 1 v 1i  m 2 v 2 i  m 1 v 1 f  m 2 v 2 f
… implies that the total momentum of the system in
each direction is conserved
m 1 v1ix  m 2 v 2 ix  m 1 v1 fx  m 2 v 2 fx and
m 1 v1iy  m 2 v 2 iy  m 1 v1 fy  m 2 v 2 fy
 Use subscripts for identifying the object, initial and
final, and components
Glancing Collisions

Two dimensional collisions
 Momentum is still conserved
 Apply vector components to conservation of
momentum equation

Applications
 Billiard balls
 Particle physics
Center of Mass
In (a), the diver’s motion is pure translation; in (b) it is translation plus rotation.
There is one point that moves in the same path a particle would take if subjected to
the same force as the diver. This point is called the center of mass (CM).
Center of Mass
The general motion of an object can be considered
as the sum of the translational motion of the CM, plus
rotational, vibrational, or other forms of motion about
the CM.
Center of Mass
For two particles, the center of mass lies closer to the one with the most mass
where M is the total mass.
Center of Mass
Example