Download Chapter 9

Chapter 9: Linear Momentum and Collisions
Reading assignment:
Homework 9.1
Chapter 9.1 to 9.7
(due Wednesday, Oct. 17):
QQ2, AE1, AE5, 5, 9, 10, 19, 27, 28, 29, 32, 33
Homework 9.2
(due Thursday, Oct. 18):
11, 14, 37, 38, 65
• Momentum


p  mv
• Momentum is conserved – even in collisions with energy
loss.
• Collisions
• Center of mass
• Impulse
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Chapter 9: Linear Momentum and
Collisions
The linear momentum of a particle of mass m and velocity
v is defined as


p  mv
The linear momentum is a vector quantity.
It’s direction is along v.
The components of the momentum of a particle:
px  m  vx
py  m  vy
pz  m  vz
Conservation of linear momentum
p   p  constant
or:
 p  p
i
f
p1,i  p2,i  p1, f  p2, f
Black board example 9.1
(similar to blocks and spring HW problem)
You (100kg) and your skinny friend (50.0 kg) stand face-toface on a frictionless, frozen pond. You push off each other.
You move backwards with a speed of 5.00 m/s.
1. What is the total
momentum of the youand-your-friend system?
A. 0 kgm/s
B. 250 kgm/s
C. 500 kgm/s
D. 750 kgm/s
E. -500 kgm/s
2.
A.
B.
C.
D.
E.
What is your
momentum after you
pushed off?
0 kgm/s
250 kgm/s
500 kgm/s
750 kgm/s
-500 kgm/s
3. What is your friends
speed after you pushed off?
A. 0 m/s
B. 5 m/s
C. 10 m/s
D. -5 m/s
E. -10m/s
4. How much energy (work) did you and your
friend expend?
Demo: How are rocket ships (in space) able to change their velocity?
Elastic and inelastic collisions in one dimension
Momentum is conserved in any collision, elastic and
inelastic.
Mechanical Energy is only conserved in elastic collisions.
Perfectly inelastic collision: After colliding, particles stick
together. There is a loss of kinetic energy (deformation).
Elastic collision: Particles bounce off each other without loss of
kinetic energy.
Inelastic collision: Particles collide with some loss of kinetic
energy, but don’t stick together.
Perfectly inelastic collision of two particles
(Particles stick together)
 
pi  p f



m1v1i  m2v2i  (m1  m2 )v f
Notice that p and v are
vectors and, thus have a
direction (+/-)
Ki  Eloss  K f
1
1
1
2
2
2
m1v1i  m2v2i  (m1  m2 )v f  Eloss
2
2
2
There is a loss
in energy, Eloss
Perfectly elastic collision of two particles
(Particles bounce off each other without loss of energy.
Momentum is conserved:




m1v1i  m2 v2i  m1v1 f  m2v2 f
Energy is conserved:
1
1
1
1
2
2
2
2
m1v1i  m2 v2i  m1v1 f  m2 v2 f
2
2
2
2
By plugging one equation into the
other, we can also derive:
v1i  v1 f  v2i  v2 f
Black board example 9.2
Two carts collide elastically on a frictionless track. The first
cart (m1 = 1kg) has a velocity in the positive x-direction of
2 m/s; the other cart (m = 0.5 kg) has velocity in the
negative x-direction of 5 m/s.
(a) Find the speed of both carts after the collision.
(b) Now, what is the speed if the collision is perfectly
inelastic?
(c) How much energy is lost in the inelastic collision?
Black board example 9.3 and demo
Determining the speed of a bullet
A bullet (m = 0.01kg) is fired into a block (0.1 kg) sitting at the edge of a table. The
block (with the embedded bullet) flies off the table (h = 1.2 m) and lands on the
floor 2 m away from the edge of the table.
a.) What was the speed of the bullet?
b.) What was the energy loss in the bullet-block collision?
vb = ?
h = 1.2 m
x=2m
Two-dimensional collisions (Two particles)
Conservation of momentum:
pi  p f
m1v1i  m2v2i  m1v1 f  m2v2 f
Split into components:
p x ,i  p x , f
m1v1ix  m2v2ix  m1v1 fx  m2v2 fx
p y ,i  p y , f
m1v1iy  m2v2iy  m1v1 fy  m2v2 fy
If the collision is elastic, we can also use conservation of energy.
Black board example 9.4
Accident investigation. Two
automobiles of equal mass approach
an intersection. One vehicle is
traveling towards the east with 29 mi/h
(13.0 m/s) and the other is traveling
13.0 m/s
north with unknown speed. The
vehicles collide in the intersection and
stick together, leaving skid marks at an
angle of 55º north of east. The second
driver claims he was driving below the
speed limit of 35 mi/h (15.6 m/s).
a) Is he telling the truth?
??? m/s
b) What is the speed of the “combined vehicles” right after the
collision?
c) How long are the skid marks (mk = 0.5)?
Motion of a System of Particles.
Newton’s second law for a System of Particles
The center of mass of a system of particles (combined mass M)
moves like one equivalent particle of mass M would move under
the influence of an external force.


Fnet  MaCM
Fnet , x  MaCM , x
Fnet , y  MaCM , y
Fnet , z  MaCM , z
Center of mass
Center of mass for many particles:

rCM 

 mi ri
i
M
Black board example 9.5
Where is the center of mass
of this arrangement of
particles.
(m3 = 2 kg; m1 = m2 = 1 kg)?
Velocity of the center of mass:

vCM 

 mi vi
i
M
Acceleration of the center of mass:

aCM 

 miai
i
M
A rocket is shot up in the air and explodes.
Describe the motion of the center of mass before and after
the explosion.
A method for finding the center of mass of any object.
- Hang object from two
or more points.
- Draw extension of
suspension line.
- Center of mass is at
intercept of these lines.
Impulse (change in momentum)

 

A change in momentum is called “impulse”: J  p  p f  pi
For a constant (average) force:
During a collision, a force F acts on
an object, thus causing a change in
momentum of the object:
 
p  J  Favg  t
tf


p  J   F (t )dt
ti
Think of hitting a soccer ball: A force F acting over a time t
causes a change p in the momentum (velocity) of the ball.
Black board example 9.5
A soccer player hits a ball (mass m = 440 g) coming at him with
a velocity of 20 m/s. After it was hit, the ball travels in the
opposite direction with a velocity of 30 m/s.
1.
What impulse acts on the ball while it is in contact with the foot?
2.
The impact time is 0.1s. What average force is the acting on the ball?
3.
How much work was done by the foot? (Assume an elastic collision.)
1A. 0
2A. 0
3A. 0
1B. 20 kg‧m/s
2B. 200 N
3B. 110 J
1C. 22 kg‧m/s
2C. 220 N
3C. 220 J
1D. 30 kg‧m/s
2D. 300 N
3D. 300 J
1E. 33 kg‧m/s
2E. 330 N
3E. 330 J