Download Chapter 8 – Momentum, Impulse, and Collisions

Chapter 8 – Momentum, Impulse, and Collisions
I.
Discuss methods of analyzing situations where interactions between particles take place. The
starting point is again Newton's Second Law.
II.
Impulse
A.

Suppose we have a single particle of mass m moving with a velocity v1 at time t 1 . Suppose forces

act on the particle and cause the velocity of the particle to change to v2 at time t 2 .




dv
,
F  ma  m
dt
multiply both sides by dt and look at the time interval from t 1 to t 2 , then
t2
 
t1

Fdt 


v2

v1



mdv  mv2  mv1 .
 
Remember mv  p  linear momentum. So,

t2
  Fdt  p

t1
2


 p1  p , or
 



J  p2  p1  p , where J is defined as the impulse. In other words,
Impulse = the change in the linear momentum.
This equation is called the Impulse-Momentum Equation. It is a vector equation, which means that
the impulse in some direction (for example, along the x-axis) equals the change in the momentum in
that direction.
Units: Newton sec = kg m/s
B.
In general, when an interaction (collision) occurs, we get a force
that varies with time, e.g., the graph to the right shows an
interaction along the x-axis where the force is present between
times t1 and t2. So, the impulse is
Jx 
F
x
t2
  F dt  area under the curve.
t1
area under
curve
x
Generally, the force is a complex function of time, however, it is
possible to define a constant average force that provides the
same impulse, and therefore it has the same area under the
curve. That is,
Jx  Fav x t 2  t 1   area of rectangle.
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t1
t2
t
area of
rectangle
So, we can also write the impulse-momentum equation as
 

J  Fav t  p ,
or,



Fav t 2  t 1   mv 2  mv 1  .
Along the x-axis, it can be written as
Fav x t 2  t 1   mv x 2  mv x1 
C.
Examples: "following through," catching an egg, hitting a tennis ball, etc.
D.
Example: Suppose a 0.5 kg baseball traveling at 30 m/s is struck by a bat,
and the ball rebounds with a velocity of 70 m/s. The time of contact
between the ball and the bat is 0.04 sec.
1.
What is the value of the impulse the bat exerts on the ball?
2.
What is the average force the bat exerts on the ball?
8-2
III. Interactions or Collisions in One-Dimension
Look at two particles of masses mA and mB traveling along the x-axis with velocities vA1 and vB1
before the collision and velocities vA2 and vB2 after the collision.
vB1
vA1
before the collision
mA
mB
FBA
the collision
vA2
FAB
vB2
after the collision
Apply the impulse-momentum equation to each of the masses: f
for mA :
FBA t  m A v A2  m A v A1
for mB :
 FAB t  m B v B2  m B v B1
Remember that:
FAB  FBA .
So, adding the two equations, we get:
0 = mAvA2 + mBvB2 - mAvA1 - mBvB1
upon rearranging:
mAvA1 + mBvB1 = mAvA2 + mBvB2 ,
or
pA1 + pB1 = pA2 + pB2 ,
or, more generally,
The total linear momentum
before the collision
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=
The total linear momentum
after the collision
What about Energy? The work-energy equation is
Wnon = E = K + U
If the time of contact is relatively short, and it generally is in collisions, then the change in potential
energy is small, and U = 0. Also rewrite Wnon as Q. Then the energy equation for the collision can
be written as
K
1
Q 
K
2
.
IV. General One-dimensional Interaction (Collision)
Linear Momentum is Conserved:
p p
1
Work-energy equation:
K
2
1
Q 
K
2
Q represents the loss or gain in energy during the interaction (collision):
Q < 0 , loss of energy in the collision - inelastic collision
Q > 0 , gain of energy in the collision - inelastic collision
Q = 0 , no loss or gain of energy in the collision – elastic
A.
Perfectly (Totally) Inelastic Collision - objects stick together
If the objects stick together after the collision, then their final velocity must be the same:
vA2 = vB2 = v2
Examples:
1.
A 1 kg block traveling to the right at 4 m/s undergoes a head-on totally inelastic collision with
a 2 kg block traveling to the left at 3 m/s. Find the velocity of the blocks after the collision.
How much energy is lost/gained in the collision? Where did the energy go?
8-4
2.
3.
A 70 g ball traveling at 6 m/s undergoes a totally inelastic collision
with a 260 g block at the end of a rod (ballistic pendulum).
a)
What is the velocity immediately after the collision?
b)
How much energy is lost in the collision?
c)
How high does the pendulum block rise after the collision?
A 0.5 kg block traveling on a smooth surface at 3 m/s
undergoes a totally inelastic collision with a 2 kg block
that is attached to a spring in its unstretched position. If
the spring constant k = 300 N/m, then what is the
maximum distance the spring is compressed?
8-5
k
B.
Inelastic Collision, in general - energy is lost(gained) but the particles do not necessarily stick
together.
Example: A 1 kg particle traveling to the right at 4 m/s undergoes a head-on collision with a 2 kg
particle traveling to the left at 3 m/s. In the collision 20% of the initial energy is lost in the collision.
What are the velocities of the particles after the collision?
8-6
C.
Elastic Collisions: no energy is lost in the collision, Q = 0
Suppose an elastic collision occurs between to blocks. We know that linear momentum is conserved
and, by definition, kinetic energy is also conserved (Q = 0).
Conservation of Momentum:
m A v A1  m B v B1  m A v A2  m B v B2
Conservation of Kinetic Energy:
1
2
Rearranging the equations:
m A v 2A1  v 2A2  mB v B2 1  v B2 2
m A v A1  v A2   mB v B1  v B2 
Dividing the first by the second:
v A1  v A2  v B1  v B2 ,
m A v 2A1  21 m B v B2 1  21 m A v 2A2  21 m B v B2 2




which says that the sum of the velocities before the collision equals the sum after the collision.
Rearranging again:
v A1  v B1  v A2  v B2  ,
which says the relative velocity before the collision equals the negative of the relative velocity
after the collision.
Example: A 1 kg particle traveling to the right at 4 m/s undergoes an elastic collision with a 2 kg
particle traveling to the left at 3 m/s. What are the velocities of the particles after the collision?
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D.
Recoil - the exact opposite of a totally inelastic collision
Example: A 3 kg object explodes into two separate objects. One of the objects is 1 kg and travels to
the right at 4 m/s. The 2 kg part recoils and travels in the opposite direction. What is the velocity of
the 2 kg part after the explosion? How much energy was gained (lost) in the explosion? From
where did the energy come?
Example: The Rocket (gravitational forces are ignored)
m
v
dm
u
velocity of the rocket:
thrust = vex
v = vo + vex ln
m - dm
mo
m
dm
= force provided by the rocket engines to accelerate the rocket
dt
8-8
v + dv
Example: Compare the final velocity of a single stage rocket to the final velocity of a two stage
rocket. Both have the same amount of fuel and have the same exhaust velocity of the gases.
Single stage: total mass = 13,000 kg, mass of fuel = 9750 kg (75%)
Two stage: total mass = (12,000 + 1,000) kg, mass of fuel = (9000 + 750) kg
V.
Collisions in Two (and more) Dimensions
Linear momentum is conserved along all axes and can be written vectorially as:
p p


1
2
When applied to a two dimensional situation:
Linear momentum is conserved in the x-direction:
p
x1

p
x2
Linear momentum is conserved in the y-direction:
p
y1

p
y2
Energy Equation:
K
8-9
1
Q 
K
2
Example: A 1 kg block travels at 4 m/s in the +x
direction toward the origin and collides with a 2 kg
block traveling at 3 m/s toward the origin in the fourth
quadrant. If the collision is totally inelastic, then how
fast do they travel after the collision and in what
direction? How much energy was lost?
y
4 m/s
mA =
1 kg
37o
x
mB = 2 kg
3 m/s
Example: An elastic collision occurs between two identical masses. One mass is at rest and the
second is traveling toward the first at 2 m/s. After the collision, the second mass travels at an angle
of 53o to the original direction of travel of the second mass. Find the velocities of the two masses
after the collision and the direction of travel of the mass that was originally at rest.
8-10
VI. Recoil in Two (and more) Dimensions
Example: A 10 kg block traveling at constant velocity of 1.5 m/s breaks up into three pieces after a
firecracker explodes in the block. A 5 kg piece continues in the original direction at 4 m/s. A 3 kg
piece travels in a direction perpendicular to the original direction at 6 m/s. How fast and in what
direction does the third piece travel?
VII. Systems of Particles
A.
Center of Mass
We have been dealing with particles where the size and shape were unimportant. We now take into
account the size and shape. The first idea to examine is the center of mass of the object.
If you apply Newton’s Second Law to all of the particles that make up an object, then a point on the
object can be found that undergoes the same type of motion as a particle would when the same
forces act on the particle. This point is defined as the center of mass of the object.
Example: Think about an odd-shaped object as a projectile.
8-11
B.
C.
Qualitatively, the center of mass represents:
1.
the balance point of the system
2.
the point around which the object freely rotates
Quantitatively, the center of mass can be found as follows:
1.
Point masses – given many individual particles of masses m1, m2,…, etc., located at positions (x1,
y1 , z1), (x2 , y2 , z2), …, etc.
Definition: the x-coordinate of the center of mass is
x cm 
m1 x 1  m 2 x 2  

m1  m 2  
m x  m x
M
m
i
i
i
i
.
i
The y- and z-coordinates of the center of mass, ycm and zcm, are similarly defined.
Example: Find the center of mass of the system of particles
shown.
y
y'
3 kg
5 cm
2 kg
8-12
10 cm
x
2 kg
2.
Uniform Symmetric Objects
Center of Mass located at the Center of Symmetry
3.
Composite symmetric objects
70 cm
20 cm
50 cm
40 cm
y
4.
Irregular and nonuniform objects
dm
y
x cm 
 xdm  1 xdm

 dm M
and, similarly for ycm and zcm
8-13
x
x
What is dm?
1-D: dm = dx
2-D: dm = dA
3-D: dm = dV
y
Examples:
1.
Find the center of mass of a rod of length L with a
linear density  = Ax, where A is a constant.
8-14
x
y
2.
Find the center of mass of a uniform flat, triangular plate
whose height is h and base is b.
h
b
VIII.
A.
Motion of a System of Particles
Look at the center of mass in one dimension for N particles:
x cm 
m x
i i
, or rearranged
M
Mxcm = mixi . Now take the derivative with respect to time:
M
dx cm

dt
m
This expression can be written as:
i
dx i
.
dt
Mv cm 
m v
i
i
and
vcm =
m v
i
i
M
The first expression says that the total linear momentum of a system of particles is equal to the
linear momentum of the center of mass of the system of particles. The second expression says that
during the interaction (collision), the velocity of the center of mass remains constant. So, for a body
made up of many particles, the momentum of the body can be found by just finding the momentum
of the center of mass of the body.
8-15
x
B.
If we take the time derivative again, we get: M
Ma cm 
This can be rewritten as:
But
m a  F
i i
i
m a
i i
dv cm

dt
m
i
dv i
.
dt
.
which is the sum of all the forces acting on all the particles in the system. This
sum represents the sum of both the external and internal forces acting on the particles. However,
the internal forces cancel (Newton’s Third Law), and therefore, the sum is just the sum of the
external forces. That is,
F  F
i
ext
 Ma cm .
This equation says that when forces act on a system of particles, even though the particles may have
very complex motions, the motion of the center of mass of the system of particles is the same as the
motion that a single particle would have if it were located at the center of mass.
Example: plunger, projectile, fireworks, etc.
C.
Conservation of Momentum
Rewrite Newton’s Second Law for a system of particles:

Suppose




dp
dv



F  Ma cm  M cm 
, where p  Mv cm .
dt
dt


dp


F0
; then the momentum of the system of particles, p  Mv cm  constant . That
dt
is, if there is no net, external forces acting on the system of particle, then the momentum of the center
of mass of the system of particles remains constant, and thus, the velocity of the center of mass
does not change.
8-16
Examples:
1.
A 10 kg block moves to the right on a smooth horizontal surface at 2 m/s. A small firecracker
inside the block explodes and the block breaks up into two fragments. One fragment is 7 kg
and continues to move in the original direction but at a speed of 8 m/s.
a.
Find the velocity of the second 3 kg fragment.
b.
What is the velocity of the center of mass of the system before the firecracker explodes?
c.
What is the velocity of the center of mass of the system after the firecracker explodes?
Note that the velocity of the center of mass remains constant. No net external forces acted on
the system of particles.
2.
A 10 kg block is at rest on a smooth horizontal surface. A 4 kg blob of clay traveling at 5 m/s
hits and sticks to the block. How fast is the block moving after being hit? How fast is the
center of mass moving before and after the collision?
8-17