Transcript Momentum, impulse, and collisions - wbm

Momentum,
impulse, and
collisions
Chapter 8
Sections 1-5
Momentum

The linear momentum of an object is
defined as


p  mv

Momentum has a magnitude of mv and
the same direction as v
Impulse

When a particle is acted on by a constant
force, the impulse of the force is defined
as
 
J  Ft
Impulse-momentum theorem
v  v0  at
mv  mv0  mat

For a constant force
mv  mv0  Ft
Ft  mv  mv0
Ft  mv


J  p
Example

A 45-g golf ball initially at rest is given a
speed of 25.0 m/s when a club strikes. If
the club and ball are in contact for 2.00
ms, what average force acts on the ball?

562 N
On your own

A 0.160-kg hockey puck is moving on an
icy, frictionless, horizontal surface. At
t = 0 the puck is moving to the right at
3.00 m/s. Calculate the velocity of the
puck after a force of 12.0 N directed to the
left has been applied for 0.050 s.

0.75 m/s to the left
Conservation of momentum

If the vector sum of the external forces on
a system is zero, the total momentum of
the system is conserved.
pbefore  pafter
p1  p2
Example

A runaway 14,000 kg railroad car is rolling
horizontally at 4 m/s toward a switchyard.
As it passes by a grain elevator, 2000 kg
of grain are suddenly dropped into the car.
How long does it take the car to cover the
500-m distance from the elevator to the
switchyard? Neglect friction and air drag.

143 s
On your own

During repair of the Hubble Space
Telescope, an astronaut replaces a damaged
solar panel. Pushing the detached panel
away into space, she is propelled in the
opposite direction. The astronaut’s mass is
60 kg and the panel’s mass is 80 kg. The
astronaut is at rest relative to the spaceship
when she shoves away the panel, and she
shoves it at 0.3 m/s relative to the spaceship.
What is her subsequent velocity relative to
the spaceship?

-0.4 m/s
Components of momentum

We can separate momentum into
components, just like we can with force or
velocity
px  mvx
p y  mv y
Example
30°
A
90°
B
Example

A 2.0-kg ball, A, is moving at a velocity
of 5.0 m/s. It collides with a stationary
ball, B, also of mass 2.0 kg. After the
collision, ball A moves off in a direction
of 30° to the left of its original direction.
Ball B moves off in a direction of 90° to
the right of ball A’s final direction. Find
the speeds of the balls after the
collision.

vA=4.3 m/s, vB=2.5 m/s
On your own
30°
A
B
45°
On your own
A hockey puck B rests on a smooth ice
surface and is struck by a second puck,
A, which was originally traveling at 40.0
m/s and which is deflected 30.0° from
its original direction. Puck B acquires a
velocity at a 45.0° angle to the original
direction of A. The pucks have the
same mass.
 Compute the speed of each puck after
the collision.


vA=29.3 m/s
vB=20.7 m/s
Elastic collisions
If all the forces acting during a collision are
conservative, then no mechanical energy
is lost or gained in the collision.
 When the total kinetic energy after the
collision is the same as it was before the
collision, then the collision was elastic.
 Collisions between billiard balls, marbles,
or other similar objects are nearly elastic.

Inelastic collisions
Collisions between automobiles are one
example of inelastic collisions.
 The kinetic energy is not conserved,
because some energy goes into crumpling
the cars.
 Collisions in which the two objects stick
together afterwards are inelastic.

Example

In a feat of public marksmanship, you fire
a bullet of mass mb into a hanging wood
block of mass mw. The block, with the
bullet embedded, swings upward. Noting
the height, h, reached at the top of the
swing, you inform the crowd of the bullet’s
speed. How fast was the bullet traveling
before it hit the block?

((m1+m2)/m1)*sqrt(2gh)
On your own

In Dallas, the morning after a winter ice
storm, a 1400-kg automobile going west at
35.0 km/h collides with a 2800-kg truck
going south at 50.0 km/h. If they become
coupled on collision, what are the
magnitude and direction of their velocity
after colliding?

35.3 km/h @ 19.3 degrees west of south
Elastic collisions

For elastic collisions, both p and K are
conserved.
1
2
mAv A1  mB vB1  mAv A2  mB vB 2
1
1
1
1
2
2
2
m Av A 2  mB vB 2  m Av A1  mB vB21
2
2
2
2
Elastic collisions
1
mB vB 2  mB vB1  mAv A1  mAv A2
1
mB vB 2  vB1  mA v A1  v A2




Elastic collisions
2
1
1
1
1
2
2
2
mB vB 2  mB vB1  m Av A1  m Av A2 2
2
2
2
2

2


mB vB2 2  vB21  mA v A21  v A2 2
2






mB vB 2  vB1 vB 2  vB1  mA v A1  v A2 v A1  v A2

Elastic collisions
2÷1





mB vB 2  vB1 vB 2  vB1 mA v A1  v A2 v A1  v A2

mB vB 2  vB1
mA v A1  v A2


vB 2  vB1  v A1  v A2



Be careful!
The last equation is only true for elastic
collisions.
 It is a special case, not the general case.
 For all collisions, momentum is conserved,
but kinetic energy is only conserved for
elastic collisions.
 Do not assume that a collision is elastic
unless you are told that it is.

On your own


A neutron of mass mn and speed vn1 collides
elastically with a carbon nucleus of mass mc
initially at rest. What are the final velocities of
both particles?
Be careful! - The velocity equation developed
for elastic collisions isn’t enough to solve the
problem.
 It
has 2 unknowns, so you need another equation.
 Use the conservation of momentum equation.


Vn2=-((mc-mn)/(mn+mc))*vn1
Vc2=((2mn)/(mn+mc))*vn1
Center of mass
The center of mass of a system of
particles is a weighted average of the
position of the particles
 If we have several particles with masses
m1, m2, etc. and coordinates (x1 , y1), (x2 ,
y2), etc.
 The center of mass of the system is
defined as the point with the following
coordinates.

Center of mass
mx

m x  m x  m x  ...
X

m  m  m  ...
m
i i
1 1
2 2
1
2
3 3
i
3
i
i
my

m y  m y  m y  ...
Y

m  m  m  ...
m
i
1 1
2
1
2
2
3 3
i
3
i
i
i
Example
Find the center of mass of the Earth-Sun
system.
 The distance between their centers is
1.49 x 1011 m.
 The mass of the Sun is 1.99 x 1030 kg.
 The mass of the Earth is 5.98 x 1024 kg.

Example, continued
mx

m x  m x  m x  ...
X

m  m  m  ...
m
i i
1 1
2 2
1

1.99 10
X

3 3
2
i
3

i
i

kg 0 m  5.98 1024 kg 1.49 1011 m
1.99 1030 kg  5.98 1024 kg
30
X  4.48 105 m

The radius of the sun is 6.95 x 108 m

Velocity of the center of mass
Vx 
 mi vix
i iy
Vy
i
 mi
i
i
i
i

V

m v

m

 mi vi
i
M
Where M = the total mass
Momentum of the center of mass

Total momentum equals the total mass
times the velocity of the center of mass.



P  MV   mi vi
i
Acceleration of the Center of Mass

A

 mi ai
i
M
Newton’s second law

All internal forces between the particles
cancel out


 Fext  MA
See page 196
The center of mass of the wrench moves
in a straight line.
 The center of mass of the shell fragments
follows the parabolic trajectory of the intact
shell.
