Transcript conservation of momentum in two dimensions

Momentum, Impulse 1-14, 20-24, 36, 37, 45-50
text: p.233 1-3 p.237 5-10 p.238 2-11 p.269
9,10
Conservation of Momentum 15-18, 26-35, 38-40,
51, 56*, 58*, 59, 60, 61*, 76, 84
Text: p.243
1-8 p.244 1-10 p.269 11-13
Collisions 62-74, 76-83 Text: p.251 10-14
p.253 1-6, 7* p.269 14, 15
Momentum (vector) 43, 52-55, 57, 75 Text:
p.257 1-5 p.258 1-4 p.269 16-18, 21
Rockets 25
Booklet Questions are in Momentum section,
p.32-40.
MOMENTUM AND COLLISIONS
Chapter 5, p. 230-271
In Newton’s time it was known that momentum of
objects was conserved in collisions. Momentum is
defined as:


vector!!!!
pmv
units of Ns
Newton stated his second law of motion in
terms of changing momentum.


dp
F
dt
If the force is constant then:


p
F

-impulse (Ns)
- change in
momentum
t

F t  p
Another obvious equation for impulse assumes
the mass remains constant (not always true).


 p  m v
In fact the force acting on an object is often not
constant. The force in the preceding equations
then must be thought of as an average force over
the time interval.
In many interactions this net force is very large
and this allows one to ignore small forces like
gravity and credit the force to the main agent
causing the impulse.
Take note at this time that the area under a
F vs. t graph would also yield the impulse.
a)What is the impulse given to a 3 kg ball
that moves towards a wall with a velocity of
12 m/s [E] and bounces off the wall with a
velocity of 10 m/s [W]?
b) If the ball was in contact with the wall for
0.0056 s then what was the average force
exerted by the wall on the ball?
a) 66 Ns [W]
b) 11 786 N [W]
LAW OF CONSERVATION OF LINEAR
MOMENTUM
If the net force acting on a system of interacting
objects is zero, then the linear momentum of the
system before the interaction equals the linear
momentum of the system after the interaction.
This law is more fundamental than the
conservation of energy and is considered the
most important law of mechanics.
Consider a system where two objects exert
equal and opposite forces on each other
(from Newton’s Third Law).


F AB   F BA


F AB t   F BA t


mB  v B   m A  v A


 pB   p A


 p A   pB  0


mB  v B   m A  v A




mA v A1  mB vB1  mA v A2  mB vB 2
The mechanical energy of a system is
conserved only when conservative forces
alone act on the system. Momentum is
conserved regardless of the nature of the
internal forces.
There is only one Ft within an interaction
but there is more than one possible Fd.
Some Fd terms serve to remove kinetic
energy from the system under study.
Studying the complete system would
conserve energy but this can be difficult to
do.
Calculate the recoil velocity of an unconstrained
rifle of mass 5 kg after it shoots a 50 g bullet at a
speed of 300 m/s. with respect to Earth.
assume the bullet is shot in the positive direction




mb vb1  mr vr1  mb vb 2  mr vr 2

m
0  0  (0.05kg)( 300 )  (5kg) vr 2
s

m
vr 2  3.00
s
A loaded railway car of mass 6000 kg is rolling
to the right at 2 m/s when it collides and couples
with an empty car of mass 3000 kg , rolling to
the left on the same track at 3 m/s. What is the
velocity of the pair of cars after the collision?
right is the positive direction




ma va1  mb vb1  ma va 2  mb vb 2
the cars will have the same final velocity




ma va1  mb vb1  ma v2  mb v2


ma va1  mb vb1 
 v2
ma  mb
m
m
(6000kg)( 2 )  (3000kg)( 3 ) 
s
s v
2
6000  3000

m
v2  0.333
s
A 1 kg ball moving with a velocity of 2 m/s to
the right collides straight-on with a stationary 2
kg ball. After the collision the 2 kg ball moves
off to the right with a velocity of 1.2 m/s. What
is the velocity of the 1 kg ball after the
collision?
right is the positive direction

v A2
m
 0.400
s
ROCKET PROPULSION
The operation of a rocket depends upon the law
of conservation of momentum as applied to a
system of particles where the system is the
rocket plus its ejected fuel.
This type of motion still involves forces since
forces are at the core of momentum changes.
However in this case both parts of the system
are moving as opposed to a car’s movement
where the road is stationary.
Suppose at some time t, the momentum of a
rocket is (mR + m)v. At some short time later,
t, the rocket ejects some fuel of mass m and
the rocket’s speed increases to v + v. The gases
pushed against the rocket to be ejected thereby
increasing the momentum of the rocket. If the
fuel is ejected with a velocity ve relative to the
rocket then the actual speed of the fuel is v-ve.
 (mR  m)v  mR (v  v)  m(v - ve )
mR v  ve m
mR dv  ve dmR
v2
v
1
dv  ve 
mR 2
mR 1
this is better said
as this
this equation can
give us thrust
dmR
mR
mR1
v2  v1  ve ln(
)
mR 2
basic
expression of
rocket
propulsion
The following question applies to rocket
propulsion. It asks if it is better to eject all of
the fuel at one time or is it better to let it out
slowly. The answer is based on the
momentum questions already completed
not on the calculus of the past few pages. It
is a long solution. There is one simplification
to this question that effects the “reality” of
the question. Can you pick it out and
explain why it is a problem?
A competition is held between two teams of
physics students, each team made up of three
members, and each member having a mass of 60
kg. The teams take turns climbing onto a cart
(120 kg) and jumping off. They want to see
whose cart will be moved fastest by propelling it
with a jump. Each member will jump off the cart
by running eastwards at 10 m/s relative to the
cart. The first team decides that its members
will all jump of together while the second team
will have each member jump off individually.
Calculate the final velocity of each cart.
The first team:

(assume east is positive)



mc vc1  3ms vs1  mc vc 2  3ms vs 2


v s 2  v c 2  10


0  0  mc vc 2  3ms (vc 2  10)


0  0  mc vc 2  3ms vc 2  3ms10

0  vc 2 (mc  3ms )  3ms 10

3ms 10
vc 2  
mc  3ms
m
3(60kg)(10 )

s
vc 2  
120kg  3(60kg)

m
vc 2  6.00 [ E ]
s

m
vs 2  4.00 [ E ]
s
The second team:
east is positive
x,y,z are the students
after the first student has jumped





mc vc 0  3ms vs 0  mc vc1  2ms v yz1  ms vx1



v c1  v yz1


vx1  vc1  10


0  mc vc1  2ms vc1  ms (vc1  10)

0  vc1 (mc  2ms  ms )  10ms

10ms
vc1  
mc  2ms  ms
m
10 (60kg)

s
vc1  
120kg  120  60


m
m
vc1  2.00 [ E ] vx1  8.00 [ E ]
s
s
after the second student has jumped






mc vc0  3ms vs 0  mc vc 2  ms vz 2  ms v y 2  ms vx1



v y 2  vc 2  10
v c 2  vz 2





0  mc vc 2  ms vc 2  ms (vc 2  10)  ms vx1


0  vc 2 (mc  2ms )  10ms  ms vx1

 10ms  ms vx1
vc 2 
(mc  2ms )

m
m
)(60kg)  60(8 )
  (10
s
s
vc 2 
(120kg  120)


m
m
vc 2  4.50 [ E ] v y 2  5.50 [ E ]
s
s
after the third student has jumped






mc vc0  3ms vs 0  mc vc3  ms vz 3  ms v y 2  ms vx1


vz 3  vc3  10


m
m
vc3  7.80 [ E ] vz 3  2.20 [ E ]
s
s
Members jumping separately gave the cart a
greater velocity than jumping all at once. This is
part of the theory of rocket propulsion.
Elastic and Inelastic Collisions
An elastic collision takes place when the total kinetic
energy after the collision equals the total kinetic energy
before the collision.
An inelastic collision takes place when the total kinetic
energy after the collision is different from the total kinetic
energy before the collision(mostly lose sometimes gain).
A completely inelastic collision occurs when two objects
colliding stick together after contact. This brings about a
maximum decrease of Ek
In any collision there is a loss of Ek during the
collision as the “collision spring” compresses. In
an elastic collision all of the Ek returns as the
“collision spring” decompresses. In an inelastic
collision the “collision spring” doesn’t
decompress at all.
MOMENTUM IS CONSERVED IN ALL
COLLISIONS!!
Analysis of Elastic Collisions
-one-dimensional only




mA v A1  mB vB1  mA v A2  mB vB 2




 (mB vB 2  mB vB1 )  m A v A2  mA v A1

(1)



 mB (vB 2  vB1 )  mA (v A2  v A1 )
2
m Av A1
2
2
mAv A1

2
mB vB1
2
2
 mB vB1
2
 (mB vB 2


2
 mB vB1 )
2
(2)  mB (vB 2
2
 vB1 )
2
m Av A2
2
2
m Av A 2



2
mB v B 2
2
2
 mB vB 2
2
m Av A 2
2
 mAv A1
2
m A (v A 2
2
 v A1 )
divide equation (2) by equation (1)




vB 2  vB1  v A2  v A1
Use the above equation and equation one to
solve for the final velocities. If we assume the
initial velocity of object B is zero, then . . .
v A2
m A  mB
2m A
(
)v A1 vB 2  (
)v A1
m A  mB
m A  mB
A 3 kg ball moving right at 5 m/s collides
elastically with a stationary 2 kg ball. What are
the final velocities of each ball?
right is +
m m
v A2  (
v A2

v A2
A
B
)v A1
m A  mB
3kg  2kg
m
(
)( 5 )
3 2
s
m
 1.00 [ R]
s

v B2
m
 6.00 [ R]
s
A 4 kg ball moving to the right at 5 m/s collides
with a 2 kg ball moving left at 4 m/s. If the
collision is elastic then what are the final
velocities of the balls.
To use our derived equations the second object’s
velocity must be zero. This can be accomplished
by using a moving frame of reference.


m
If v B1  0 then v A1  9
[ R]
s
This is a subtract 4 m/s [L] shift.

v A2
v A2
m A  mB
(
)v A1
m A  mB
v A2
4kg  2kg
m
(
)( 9 )
42
s
m
 3.00 [ R]
s

v B2
m
 12.00 [ R]
s
These velocities are in the shifted frame of
reference (subtract 4 m/s [L]). To return to the
original frame of reference add 4 m/s [L].

v A2

m
m
 1.00 [ R] vB 2  8.00 [ R]
s
s
Questions on the following slide illustrate
questions that can be asked if one is unsure if
the collision is elastic or not.
A 5 kg ball travelling 4 m/s right collides with a
3 kg ball travelling 3 m/s left. After the collision
the first ball is travelling 1 m/s right.
a) What is the final velocity of the second ball?
b) Is this collision elastic or inelastic?
c) If the collision was completely inelastic then
what would the final velocity of the two objects
be?
homework
p. 359 1-4
extra p. 370
23,24,25a
ENERGY CHANGES WITHIN AN
ELASTIC LINEAR COLLISION
An elastic linear collision between two objects
has three definite stages. Let’s say object A
collides with a stationary object B. While they
are in contact one can say the following.
Initially A is moving closer to B. (vA> vB)
At some point A moves at the same velocity as
B. (vA= vB) This is minimum separation.
Finally A is moving farther away from B. (vA <
vB)
Since A and B are in contact with each other
they must be undergoing compression and elongation, in other words they are acting as springs.
Let’s see where this understanding leads.
A 3 kg ball moving 5 m/s [R] collides elastically
with a stationary 2 kg ball. The balls have a
radius of 10 cm. The balls have a spring constant
of 1250 N/m.
a) What are the final velocities of the balls?
b) How much kinetic energy is lost at min. sep.?
c) How close were the centers of the two balls at
min. sep.?
Right is positive
v A2
m A  mB
2m A
(
)v A1 vB 2  (
)v A1
m A  mB
m A  mB
3kg  2kg m
v A2  (
)(5 )
3 2
s
2(3kg) m
vB 2  (
)(5 )
3 2
s
m
v A2  1.00
s
m
vB 2  6.00
s
At minimum separation the two objects have the
same velocity. To solve for this velocity we use
the conservation of momentum equation. For
one side of this equation we can use the initial
velocities given or the final velocities calculated.
Both yield the same answer.
right is positive




mA v A1  mB vB1  mA v A2  mB vB 2
m
(3kg)( 5 )  0  (3kg  2kg)v2
s

m
v 2  3.00 [ R]
s
The kinetic energy lost at minimum
separation turns into elastic potential energy
due to compression of the balls.
2
mAv A1
2

2
mB vB1
2
2
kx1
2
2
mAv A2 mB vB 2 kx2




2
2
2
2
2
m2
m2
m2
(3kg)(5 )
(3kg)(3 ) (2kg)(3 )
2
s 00 
s 
s  kx
2
2
2
2
2
kx
 15 J
2
This is the energy lost at
minimum separation.
Solving for the distortion of the balls . . .
N 2
(1250 ) x
m
 15 J
2
x  0.1549m
Since the balls undergo compression the
negative answer is correct and the balls are
4.51 cm apart at minimum separation.
A 4 kg cart moving 3 m/s [R] collides elastically
with a 2 kg cart moving 4 m/s [L]. One of the
carts has a 30 cm spring with a spring constant
of 2000 N/m in front of it.
a) What are the final velocities of the balls?
b) How much kinetic energy is lost at min. sep.?
c) How close are the two carts at min. sep.?
CONSERVATION OF MOMENTUM IN
TWO DIMENSIONS
Up to this point we have limited ourselves
to one-dimensional situations for analysis
of momentum. Since momentum is a vector
two and three dimensional situations occur
(rather plentifully) however our analysis
will extend only to two dimensions.
Two-dimensional collisions are difficult to
analyze because of collision geometry. In
one dimension all collisions are head on but
this is not the case in two dimensions.
Glancing contact, more direct contact or
elasticity differences lead to very different
results. As a result we simply use the vector
equation of conservation of momentum as
the basis of our analysis involving diagrams
and geometry.
mma mmm