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Page 1
Lecture 8, Pg 1
University Physics Mechanics: Lecture 8
Agenda for Today
Linear Momentum and Force
Conservation of Linear Momentum
Impulse
Conservation of Energy and Momentum
Elastic Collisions in One Dimension
Inelastic Collisions in One Dimension
Lecture 8, Pg 2
Linear Momentum
Linear momentum p is defined as the product of an object’s mass m and its velocity v.
Momentum is a vector quantity
the direction of the momentum vector is the same as the velocity vector
The unit of linear momentum is kg.m/s.
Nb. This is not the same as the unit of force (1 N = 1 kg.m/s2).
vmp
Lecture 8, Pg 3
The Momentum Statement of Newton’s Second Law
A force is required to change an object’s momentum and Newton’s second law is related to the change of momentum as follows:
where p is the change in momentum and t is the time interval during which the change in momentum occurs.
tp
tv
mamF
Lecture 8, Pg 4
Conservation of Linear Momentum
In any collision between two or more objects, the vector sum of the momenta before impact equals the vector sum of the momenta after impact.
Ie. momentum is conserved
Law of Conservation of Momentum
For a collision between two objects, the law of conservation of momentum can be written as follows:
'22
'112211 vmvmvmvm
Page 2
Lecture 8, Pg 5
Impulse
An impulse is defined as the product of the force F acting on an object and the time t during which the force acts.
The unit of impulse is N.s.
An impulse causes a change in an object’s momentum:
tFJ
vmvmpttv
mtamtFJ
'
Lecture 8, Pg 6
Discussion Problem 31:Force and Impulse
Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second.
Which box has the most momentum after the force acts ?
(a) heavier (b) lighter (c) same
F F light heavy
Lecture 8, Pg 7
Discussion Problem 31:Solution
F F light heavy
tav
p
FWe know tav Fpso
In this problem F and t are the same for both boxes !
The boxes will have the same final momentum.
Lecture 8, Pg 8
Force and Impulse:Baseball Example
A pitcher pitches the ball ( m = 0.72 kg ) at 145 km/hr (about 90 mph).
The batter makes contact with the ball for 0.0010 s causing the ball to leave the bat going 190 km/hr (about 120 mph).
Find the net average force on the ball, disregarding gravity.
Page 3
Lecture 8, Pg 9
Force and Impulse
F
t
ti tf
f
i
t
tdt
FJ
t
The diagram shows the force versus time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the collision.
Impulse J = area under this curve !
Lecture 8, Pg 10
Force and Impulse
Two different collisions can have the same impulse since J dependsonly on the change in momentum,not the nature of the collision.
ti tft
F
t
F
t
ti tft
same area
t big, F smallt small, F big
Lecture 8, Pg 11
Force and Impulse
ti tft
F
t
F
t
ti tft
t big, F smallt small, F big
soft spring
stiff spring
Lecture 8, Pg 12
Force and Impulse
F
t
ti tf
dtdP
F
t
Using
PPPP
PF
i
t
t f
t
t
t
t
f
i
f
i
f
i
d
dtdtd
dt
J
the impulse becomes:
impulse = change in momentum !
PJ
Page 4
Lecture 8, Pg 13
Conservation of Linear Momentum
Internal forces cancel in pairs, thus the vector sum over all forces can be replaced by a sum over the external forces only:
This has several interesting implications:
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 can not change.
» The total momentum of a system is conserved if there are no external forces acting.
dtd
extP
F
madtdv
mdtmvd
dtd
ext )(P
F
Lecture 8, Pg 14
Conservation of Linear Momentum
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.
dtd
ext
PF 0
dtdP
0extF
Lecture 8, Pg 15
Energy and Momentum Conservation in Collisions
In collisions between two (or more) objects, momentum along a certain direction is conserved when there are noexternal forces acting in this direction.
However, some of the kinetic energy may be converted into other forms of energy, usually thermal energy.
Energy is lost:
» Bending of metal (crashing cars)
» Heat (bomb)
Kinetic energy is not conserved since work is done during the collision !
Lecture 8, Pg 16
Elastic 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.
vi
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.
Page 5
Lecture 8, Pg 17
Comment on Energy Conservation
The total kinetic energy of a system undergoing an inelastic collision is not conserved.
Energy is lost:
» Bending of metal (crashing cars)
» Heat (bomb)
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.
Lecture 8, Pg 18
Elastic Collision in 1-D
v1,i v2,i
before
x
m1m2
v1,fv2,f
after
m1m2
Lecture 8, Pg 19
Should be no problem 2 equations & 2 unknowns !
v1,i v2,i
v1,f v2,f
before
after
x
m1 m2
ffii vmvmvmvm 22112211
2
22
2
11
2
22
2
11 21
21
21
21
ffii vmvmvmvm
Elastic Collision in 1-D
Conserve PX
Conserve Energy
Suppose we know v1,i and v2,i
We need to solve for v1,f and v2,f
Lecture 8, Pg 20
Discussion Problem 32:Elastic Collisions: Equal Masses
A moving ball hits an identical but stationary ball head on. The collision is elastic.
Describe the motion of both balls just after the collision.
V
(a) (b) (c)
2
V
2
V
2
VV 2
V
Page 6
Lecture 8, Pg 21
Discussion Problem 32:Solution
The first ball will transfer all of its energy and momentum to the second ball (the only way both energy and momentum can be conserved when the masses are equal).
After the collision, the second ball is moving and the first is stationary.
Looks just like before the collision since the balls are identical !
V
Lecture 8, Pg 22
Discussion Problem 32:Solution
The first ball will transfer all of its energy and momentum to the second ball (the only way both energy and momentum can be conserved when the masses are equal).
After the collision, the second ball is moving and the first is stationary.
Looks just like the balls missed !
V
Lecture 8, Pg 23
Discussion Problem 33:Elastic Collisions: Massive Target/Projectile
Consider the two elastic collisions shown below. In 1 a golf ball moving with speed V hits a stationary bowling ball head on, in 2a bowling ball moving with the same speed V hits a stationary golf ball.
In which case does the golf ball have the greatest speedafter the collision ?
(a) 1 (b) 2 (c) same
V
1V
2
Lecture 8, Pg 24
Discussion Problem 33:Solution
The speed of approach of two objects before an elastic collision is the same as the speed of recession after colliding.
Since the bowling ball is much heavier than the golf ball, its speed will be changed very little in either collision.
V
1V
2
Page 7
Lecture 8, Pg 25
Discussion Problem 33:Solution
V
1
In case 1 the bowling ball will almost remain at rest, and thegolf ball will bounce back with speed close to V.
V 2
2V
In case 2 the bowling ball will keep going with speed close toV, hence the golf ball will rebound with speed close to 2V.
Lecture 8, Pg 26
Elastic Collision in 1-D
The blocks shown below slide without friction. (a) Is the collision elastic? (b) What is the velocity v of the 1.6 kg block after the collision? (c) What is the velocity v of the 2.4 kg block after the collision? (d) Suppose the initial velocity of the 2.4 kg block is the reverse of that shown. Can the velocity v of the 1.6 kg block after the collision be in the direction shown?
1.6 kg 2.4 kg
5.5 m/s 2.5 m/s
before collision
1.6 kg 2.4 kg
v 4.9 m/s
after collision
Answers:(a) v1f = 1.9 m/s(b) Ki = 31.7 J
Kf = 31.7 Jcollision is elastic
(c) v1f = -5.6 m/s1.6kg block moves to the left
Lecture 8, Pg 27
Inelastic Collision in 1-D: Example 1
M+m
v = 0
v = ?
ice(no friction)
M = 2800 kg
m = 560 kg
V = 40 mph
Lecture 8, Pg 28
Inelastic Collision in 1-D: Example 2
A block of mass 2.50 kg is initially at rest on a frictionless horizontal surface. A bullet of mass 5.00 g 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 1.20 m/s.
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
Page 8
Lecture 8, Pg 29
Discussion Problem 34:Inelastic Collisions in 1D
A box sliding on a frictionless surface collides and sticks to a second identical box which is initially at rest.
What is the ratio of initial to final kinetic energy of the system.
(a) 1
(b)
(c) 2
2
Lecture 8, Pg 30
Discussion Problem 34:Solution
No external forces in the x direction, so Px is constant.
mvPi v
m m
2v
2mPf m m v / 2
x
Lecture 8, Pg 31
Discussion Problem 34:Solution
Compute kinetic energies:
vm m
m m v / 2
2
i mv21
K
if Kv
mK21
22
21
2
2f
i
K
K
Lecture 8, Pg 32
Discussion Problem 34:Another Solution
We can write
m m
m m
m
Pm
mvmvK
221
21 22
2
P is the same before and after the collision.
The mass of the moving object has doubled, hence thekinetic energy must be half.
2f
i
KK
Page 9
Lecture 8, Pg 33
Discussion Problem 35:Elastic vs. Inelastic Collisions
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) Box 1 (b) Box 2 (c) same
1 2
Lecture 8, Pg 34
Discussion Problem 35:Solution
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
Lecture 8, Pg 35
Explosion (Inelastic Un-Collision)
Before the explosion:M
m1 m2
v1 v2
After the explosion:
Lecture 8, Pg 36
Explosion...
No external forces, so P is conserved.
Initially: P = 0
Finally: P = m1v1 + m2v2 = 0
m1v1 = - m2v2 M
m1 m2
v1 v2