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Physics 111: Lecture 14, Pg 1
Physics 111: Lecture 14
Today’s Agenda
Momentum Conservation
Inelastic collisions in one dimension
Inelastic collisions in two dimensions
Explosions
Comment on energy conservation
Ballistic pendulum
Physics 111: Lecture 14, Pg 2
Center of Mass Motion: Review
We have the following law for CM motion:
This has several interesting implications:
It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: We can use it to relate F and A like we are used to doing.
It tells us that if FEXT = 0, the total momentum of the system does not change.The total momentum of a system is conserved if there are
no external forces acting.
F P AEXT CMddt
M
Physics 111: Lecture 14, Pg 3
(2)
Lecture 14, Act 1Center of Mass Motion
Two pucks of equal mass are being pulled at different points with equal forces. Which experiences the bigger acceleration?
FM
M
A1
T
T
A2
(a) A1 A2 (b) A1 A2 (c) A1 = A2
(1)
Pucks
Physics 111: Lecture 14, Pg 4
We have just shown that MA = FEXT
Acceleration depends only on external force, not on where it is applied!
Expect that A1 and A2 will be the same since F1 = F2 = T = F / 2
The answer is (c) A1 = A2.
So the final CM velocities should be the same!
Lecture 14, Act 1Solution
Physics 111: Lecture 14, Pg 5
The final velocity of the CM of each puck is the same! Notice, however, that the motion of the particles in each of
the pucks is different (one is spinning).
V
V
This one has morekinetic energy(rotation)
Lecture 14, Act 1Solution
Physics 111: Lecture 14, Pg 6
Momentum Conservation
The concept of momentum conservation is one of the most fundamental principles in physics.
This is a component (vector) equation.We can apply it to any direction in which there is no
external force applied. You will see that we often have momentum conservation
even when energy is not conserved.
F PEXT
ddt
ddtP 0 FEXT 0
Physics 111: Lecture 14, Pg 7
Elastic vs. Inelastic Collisions
A collision is said to be elastic when kinetic energy as well as momentum is conserved before and after the collision. Kbefore = Kafter
Carts colliding with a spring in between, billiard balls, etc.
vi
A collision is said to be inelastic when kinetic energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter
Car crashes, collisions where objects stick together, etc.
Physics 111: Lecture 14, Pg 8
Inelastic collision in 1-D: Example 1
A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :What is the initial speed of the bullet v?What is the initial energy of the system?What is the final energy of the system?Is energy conserved?
vV
before after
x
Physics 111: Lecture 14, Pg 9
Example 1...
Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction!
Px, i = Px, f
mv = (M+m)V
vV
initial final
v M mm
V
x
Physics 111: Lecture 14, Pg 10
Example 1...
Now consider the kinetic energy of the system before and after:
Before:
After:
So
E mv m M mm
V M mm
M m VB
12
12
12
22
2 2
E M m VA 12
2
E mM m
EA B
Kinetic energy is NOT conserved! (friction stopped the bullet)However, momentum was conserved, and this was useful.
v M mm
V
Physics 111: Lecture 14, Pg 11
Inelastic Collision in 1-D: Example 2M
m
M + m
V v = 0
v = ?
ice(no friction)
Physics 111: Lecture 14, Pg 12
Example 2...
Before the collision:
Use conservation of momentum to find v after the collision.
After the collision:
Conservation of momentum:
P Vi M m ( )0 P vf M m ( )
P Pi f
M M mV v ( )
Vv)mM(
M
vector equation
Air track
Physics 111: Lecture 14, Pg 13
Example 2...
Now consider the K.E. of the system before and after:
Before:
After:
So
E MV M M mM
v M mM
M m vBUS
12
12
12
22
2 2
E M m vA 12
2
E MM m
EA B
Kinetic energy is NOT conserved in an inelastic collision!
v M mm
V
Physics 111: Lecture 14, Pg 14
Lecture 14, Act 2Momentum Conservation
Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface.
The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.Which box ends up moving faster?
(a) Box 1 (b) Box 2 (c) same
1 2
Physics 111: Lecture 14, Pg 15
Lecture 14, Act 2Momentum Conservation
Since the total external force in the x-direction is zero, momentum is conserved along the x-axis.
In both cases the initial momentum is the same (mv of ball). In case 1 the ball has negative momentum after the collision,
hence the box must have more positive momentum if the total is to be conserved.
The speed of the box in case 1 is biggest!
1 2
x
V1 V2
Physics 111: Lecture 14, Pg 16
Lecture 14, Act 2Momentum Conservation
1 2
xV1 V2
mvinit = MV1 - mvfin
V1 = (mvinit + mvfin) / M
mvinit = (M+m)V2
V2 = mvinit / (M+m)
V1 numerator is bigger and its denominator is smaller than that of V2.
V1 > V2
Physics 111: Lecture 14, Pg 17
Inelastic collision in 2-D
Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).
v1
v2
V
before after
m1
m2
m1 + m2
Physics 111: Lecture 14, Pg 18
Inelastic collision in 2-D...
There are no net external forces acting.Use momentum conservation for both components.
v1
v2
V = (Vx,Vy)
m1
m2
m1 + m2
P Px i x f, , m v m m Vx1 1 1 2
V mm m
vx
1
1 21
P Py i y f, , m v m m Vy2 2 1 2
V mm m
vy
2
1 22
Physics 111: Lecture 14, Pg 19
Inelastic collision in 2-D...
So we know all about the motion after the collision!
V = (Vx,Vy)
Vx
Vy
V m
m mvx
1
1 21
V m
m mvy
2
1 22
1
2
11
22
x
y
pp
vmvm
VV
tan
Physics 111: Lecture 14, Pg 20
Inelastic collision in 2-D...
We can see the same thing using vectors:
tan pp
2
1
P
p1
p2
P
p1p2
Physics 111: Lecture 14, Pg 21
Explosion (inelastic un-collision)
Before the explosion:M
m1 m2
v1 v2
After the explosion:
Physics 111: Lecture 14, Pg 22
Explosion...
No external forces, so P is conserved.
Initially: P = 0
Finally: P = m1v1 + m2v2 = 0
m1v1 = - m2v2 M
m1 m2
v1 v2
RocketBottle
Physics 111: Lecture 14, Pg 23
Lecture 14, Act 3Center of Mass
A bomb explodes into 3 identical pieces. Which of the following configurations of velocities is possible?
(a) 1 (b) 2 (c) both
m mv V
v
m
m mv v
v
m
(1) (2)
Physics 111: Lecture 14, Pg 24
Lecture 14, Act 3Center of Mass
m mv v
v
m
(1)
No external forces, so P must be conserved. Initially: P = 0 In explosion (1) there is nothing to balance the upward
momentum of the top piece so Pfinal 0.
mvmv
mv
Physics 111: Lecture 14, Pg 25
Lecture 14, Act 3Center of Mass
No external forces, so P must be conserved. All the momenta cancel out.
Pfinal = 0.
(2)
m mv v
v
mmv
mv
mv
Physics 111: Lecture 14, Pg 26
Comment on Energy Conservation
We have seen that the total kinetic energy of a system undergoing an inelastic collision is not conserved.Energy is lost:
» Heat (bomb)» Bending of metal (crashing cars)
Kinetic energy is not conserved since work is done during the collision!
Momentum along a certain direction is conserved when there are no external forces acting in this direction.In general, momentum conservation is easier to satisfy
than energy conservation.
Physics 111: Lecture 14, Pg 27
Ballistic Pendulum
H
L LL L
m
M
A projectile of mass m moving horizontally with speed v strikes a stationary mass M suspended by strings of length L. Subsequently, m + M rise to a height of H.
Given H, what is the initial speed v of the projectile?
M + mv
V
V=0
Physics 111: Lecture 14, Pg 28
Ballistic Pendulum...
Two stage process:
1. m collides with M, inelastically. Both M and m then move together with a velocity V (before having risen significantly).
2. M and m rise a height H, conserving K+U energy E. (no non-conservative forces acting after collision)
Physics 111: Lecture 14, Pg 29
Ballistic Pendulum... Stage 1: Momentum is conserved
in x-direction: mv m M V ( ) V mm M
v
Stage 2: K+U Energy is conserved
( )E EI F12
2( ) ( )m M V m M gH V gH2 2
Eliminating V gives: gH2mM1v
Physics 111: Lecture 14, Pg 30
Ballistic Pendulum Demo
H
L LL L
m
M
In the demo we measure forward displacement d, not H:
M + mv
d
L
d H
L-H L d L H
H L L d
2 2 2
2 2
Physics 111: Lecture 14, Pg 31
Ballistic Pendulum Demo...
L
d H
L-H
L2d
L2d1LL
Ld1LL
dLLH22
2
2
22
dL
1for
gH2mM1v
Lgd
mM1v
for d << L
Let’s see who can throw fast...
Ballisticpendulum
Physics 111: Lecture 14, Pg 32
Recap of today’s lecture
Inelastic collisions in one dimension (Text: 8-6)
Inelastic collisions in two dimensions (Text: 8-6)
Explosions
Comment on energy conservation
Ballistic pendulum (Ex. 8-14)
Look at textbook problems Chapter 8: # 31, 57, 59, 73, 77