physics 1501: lecture 17, pg 1 physics 1501: lecture 17 today’s agenda l homework #6: due friday l...
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Physics 1501: Lecture 17, Pg 1
Physics 1501: Lecture 17Physics 1501: Lecture 17
TodayToday’’s Agendas Agenda
Homework #6: due Friday
Midterm I: Friday only
TopicsChapter 9
» Momentum
» Introduce Collisions
Physics 1501: Lecture 17, Pg 2
Chapter 9Chapter 9Linear MomentumLinear Momentum
Newton’s 2nd Law:
FF = maa
dv vp
Units of linear momentum are kg m/s.
pp = mvv(pp is a vector since vv is a
vector).
So px = mvx etc.
Definition: Definition: For a single particle, the momentum pp is defined as:
Physics 1501: Lecture 17, Pg 3
Momentum ConservationMomentum 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 (mechanical) energy is not conserved.
Physics 1501: Lecture 17, Pg 4
Elastic vs. Inelastic CollisionsElastic vs. Inelastic Collisions A collision is said to be elastic when energy as well as
momentum is conserved before and after the collision. Kbefore = Kafter
Carts colliding with a spring in between, billiard balls, etc.
vvi
A collision is said to be inelastic when energy is not conserved before and after the collision, but momentum is conserved. Kbefore Kafter
Car crashes, collisions where objects stick together, etc.
Physics 1501: Lecture 17, Pg 5
Inelastic collision in 1-D: Example 1Inelastic 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 momentum of the bullet with speed v ?What is the initial energy of the system ?What is the final energy of the system ? Is energy conserved ?
v V
before after
x
Physics 1501: Lecture 17, Pg 6
Example 1...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 !Momentum is conserved in the x direction !
Px,before = Px,after
mv = (M+m) V
vV
before after
x
Physics 1501: Lecture 17, Pg 7
Example 1...Example 1... Now consider the energy of the system before and after:
Before:
After:
So
Energy is NOT conservedEnergy is NOT conserved (friction stopped the bullet)
However momentum was conserved, and this was useful.
Physics 1501: Lecture 17, Pg 8
Lecture 17,Lecture 17, ACT 1 ACT 1Inelastic Collision in 1-DInelastic Collision in 1-D
ice
(no friction)
Winter in Storrs
Physics 1501: Lecture 17, Pg 9
Lecture 17,Lecture 17, ACT 1 ACT 1Inelastic Collision in 1-DInelastic Collision in 1-D
vvff = ?
finally
m
v = 0ice
M = 2m
VV00(no friction)
initially
Vf = A) 0 B) Vo/2 C) 2Vo/3 D) 3Vo/2 E) 2Vo
Physics 1501: Lecture 17, Pg 10
Lecture 17, Lecture 17, ACT 2ACT 2Momentum ConservationMomentum 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 fastest ?
(a)(a) Box 1 (b)(b) Box 2 (c)(c) same
1 2
Physics 1501: Lecture 17, Pg 11
Inelastic collision in 2-DInelastic collision in 2-D
Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).
vv1
vv2
VV
before after
m1
m2
m1 + m2
Physics 1501: Lecture 17, Pg 12
Inelastic collision in 2-D...Inelastic collision in 2-D... There are no net external forces acting.
Use momentum conservation for both components.
vv1
vv2
VV = (Vx,Vy)
m1
m2
m1 + m2
Physics 1501: Lecture 17, Pg 13
Inelastic collision in 2-D...Inelastic collision in 2-D...
So we know all about the motion after the collision !
VV = (Vx,Vy)
Vx
Vy
Physics 1501: Lecture 17, Pg 14
Inelastic collision in 2-D...Inelastic collision in 2-D...
We can see the same thing using vectors:
PP
pp1
pp2
PP
pp1
pp2
Physics 1501: Lecture 17, Pg 15
Comment on Energy ConservationComment 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, easier to satisfy than energy conservation.
Physics 1501: Lecture 17, Pg 16
Elastic CollisionsElastic Collisions
Elastic means that energy is conserved as well as momentum.
This gives us more constraints.We can solve more complicated problems !!Billiards (2-D collision).The colliding objects
have separate motionsafter the collision as well as before.
Start with a simpler 1-D problem.
Before After
Physics 1501: Lecture 17, Pg 17
Elastic Collision in 1-DElastic Collision in 1-D
v1,b v2,b
before
x
m1m2
v1,av2,a
afterm1
m2
Physics 1501: Lecture 17, Pg 18
Elastic Collision in 1-DElastic Collision in 1-D
v1,b v2,b
v1,av2,a
before
after
x
m1 m2
Conserve PX
m1v1,b + m2v2,b = m1v1,a + m2v2,a
Conserve Energy
1/2 m1v21,b + 1/2 m2v2
2,b = 1/2 m1v21,a + 1/2 m2v2
2,a
Suppose we know v1,b and v2,b
We need to solve for v1,a and v2,a
Should be no problem 2 equations & 2 unknowns !
m1v1,b + m2v2,b = m1v1,a + m2v2,a
Physics 1501: Lecture 17, Pg 19
Elastic Collision in 1-DElastic Collision in 1-D
After some moderately tedious algebra, we can derive the following equations for the final velocities,
Physics 1501: Lecture 17, Pg 20
Example - Elastic CollisionExample - Elastic Collision
Suppose I have 2 identical bumper cars. One is motionless and the other is approaching it with velocity v1. If they collide elastically, what is the final velocity of each car ?
Note that this means,
m1 = m2 = m
v2B = 0
Physics 1501: Lecture 17, Pg 21
Example - Elastic CollisionExample - Elastic Collision
Let’s start with the equations for conserving momentum and energy, mv1,B = mv1,A + mv2,A (1)
1/2 mv21,B = 1/2 mv2
1,A + 1/2 mv22,A (2)
Initially, v2 = 0. After the collision, v1 = 0.
To satisfy equation (1), v2A = v1B.
v1A=0 , v2A=0 or both Both: contradicts (2)
If v2A=0 : v1A= v1B from (1) or (2) no collision !
Physics 1501: Lecture 17, Pg 22
Lecture 17,Lecture 17, ACT 3ACT 3Elastic CollisionsElastic Collisions
I have a line of 3 bumper cars all touching. A fourth car smashes into the others from behind. Is it possible to satisfy both conservation of energy and momentum if two cars are moving after the collision?All masses are identical, elastic collision.
A) Yes B) No
Before
After?
Physics 1501: Lecture 17, Pg 23
Example of 2-D elastic collisions:Example of 2-D elastic collisions:BilliardsBilliards
If all we know is the initial velocity of the cue ball, we don’t have enough information to solve for the exact paths after the collision. But we can learn some useful things...
Physics 1501: Lecture 17, Pg 24
BilliardsBilliards
Consider the case where one ball is initially at rest.
ppa
ppb
FF PPa
before afterthe final direction of the
red ball will depend on
where the balls hit.
vvcm
Physics 1501: Lecture 17, Pg 25
We know momentum is conserved: ppb = ppa + PPa
We also know that energy is conserved:
Comparing these two equations tells us that:
BilliardsBilliards
pb2 = (ppa + PPa )2 = pa
2 + Pa2 + 2 ppa PPa
m2
P
m2
p
m2
p 2a
2a
2 b
ppa PPa = 0
and must therefore be orthogonal!
Or … one momentum must be zero.
ppa
ppb
PPa
Ppp 2a
2a
2 b
Physics 1501: Lecture 17, Pg 26
BilliardsBilliards
The final directions are separated by 90o.
ppa
ppb
FF PPa
before after
vvcm
Active Figure
Physics 1501: Lecture 17, Pg 27
Lecture 17 –Lecture 17 – ACT 4ACT 4Pool SharkPool Shark
Can I sink the red ball without scratching ?Ignore spin and friction.
A) Yes B) No C) More info needed
Physics 1501: Lecture 17, Pg 28
BilliardsBilliards..
More generally, we can sink the red ball without sinking the white ball – fortunately.
Physics 1501: Lecture 17, Pg 29
BilliardsBilliards
However, we can also scratch. All we know is that the angle between the balls is 90o.