1. Lecture 12 Chapter 9: Momentum & Impulse
Goals:
Chapter 9: Momentum & Impulse
Solve problems with 1D and 2D Collisions
Solve problems having an impulse (Force vs. time)
Chapter 10
Understand the relationship between motion and energy
Define Potential & Kinetic Energy
Develop and exploit conservation of energy principle
Assignment:
HW5 due Wednesday
For Wednesday: Read all of chapter 10 1

2. Inelastic collision in 1-D: Example
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 ? x v V
before
after

3. Inelastic collision in 1-D: Example
What is the momentum of the bullet with speed v ?
Key question: Is x-momentum conserved ?
Before
After v V x
before
after

4. Exercise Momentum 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 elastically back, while the ball hitting box 2 sticks.
Which box ends up moving fastest ?
Box 1
Box 2
same 1 2

5. Exercise Momentum Conservation
The ball hitting box 1 bounces elastically back, while the ball hitting box 2 sticks.
Which box ends up moving fastest ?
Notice the implications from the graphical solution: Box 1’s momentum must be bigger because the length of the summed momentum must be the same.
The longer the green vector the greater the speed
Before
After
Ball 1
Ball 2
Box 1
Box 2
Box 1+Ball 1
Box 2+Ball 2 1 2

6. A perfectly inelastic collision in 2-D
Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). V v1 q
m1 + m2 m1 m2 v2
before
after
If no external force momentum is conserved.
Momentum is a vector so px, py and pz

7. A perfectly inelastic collision in 2-D
If no external force momentum is conserved.
Momentum is a vector so px, py and pz are conseved V v1
m1 + m2 q m1 m2 v2
before
after
x-dir px : m1 v1 = (m1 + m2 ) V cos q
y-dir py : m2 v2 = (m1 + m2 ) V sin q

8. Elastic Collisions Elastic means that the objects do not stick.
There are many more possible outcomes but, if no external force, then momentum will always be conserved
Start with a 1-D problem.
Before
After

9. Billiards Consider the case where one ball is initially at rest. after
before
pa q pb
vcm
Pa f F
The final direction of the red ball will depend on where the balls hit.

10. Conservation of Momentum
Billiards: All that really matters is conservation momentum (and energy Ch. 10 & 11)
Conservation of Momentum
x-dir Px : m vbefore = m vafter cos q + m Vafter cos f
y-dir Py : = m vafter sin q + m Vafter sin f
before
after
pafter q pb
Pafter f F

11. Force and Impulse (A variable force applied for a given time)
Gravity: At small displacements a “constant” force t
Springs often provide a linear force (-k x) towards its equilibrium position (Chapter 10)
Collisions often involve a varying force
F(t): 0  maximum  0
We can plot force vs time for a typical collision. The impulse, J, of the force is a vector defined as the integral of the force during the time of the collision.

12. Force and Impulse (A variable force applied for a given time)
J reflects momentum transfer t ti tf t F
Impulse J = area under this curve !
(Transfer of momentum !)
Impulse has units of Newton-seconds

13. Force and Impulse
Two different collisions can have the same impulse since J depends only on the momentum transfer, NOT the nature of the collision. F t
same area F t t t
t big, F small
t small, F big

14. Average Force and Impulset
Fav F
Fav t t t
t big, Fav small
t small, Fav big

15. Exercise 2 Force & 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 ? F
light
heavy
heavier
lighter
same
can’t tell

16. Boxing: Use Momentum and Impulse to estimate g “force”

17. Back of the envelope calculation
(1) marm~ 7 kg (2) varm~7 m/s (3) Impact time t ~ 0.01 s
 Question: Are these reasonable?
 Impulse J = p ~ marm varm ~ 49 kg m/s
 F ~ J/t ~ 4900 N
(1) mhead ~ 6 kg
 ahead = F / mhead ~ 800 m/s2 ~ 80 g !
Enough to cause unconsciousness ~ 40% of fatal blow
Only a rough estimate!

18. Woodpeckers ahead ~ 600 - 1500 g How do they survive?
During "collision" with a tree 
ahead ~ g
How do they survive?
Jaw muscles act as shock absorbers
Straight head trajectory reduces damaging rotations (rotational motion is very problematic)

19. Home Exercise
The only force acting on a 2.0 kg object moving along the x-axis. Notice that the plot is force vs time.
If the velocity vx is +2.0 m/s at 0 sec, what is vx at 4.0 s ?
Dp = m Dv = Impulse
m Dv = J0,1 + J1,2 + J2,4
m Dv = (-8)1 N s +
½ (-8)1 N s + ½ 16(2) N s
m Dv = 4 N s
Dv = 2 m/s
vx = m/s = 4 m/s

20. Chapter 10: Energy
We need to define an “isolated system” ?
We need to define “conservative force” ?
Recall, chapter 9, force acting for a period of time gives an
impulse or a change (transfer) of momentum
What if a force acting over a distance:
Can we identify another useful quantity?

21. Energy
Fy = m ay and let the force be constant
y(t) = y0 + vy0 Dt + ½ ay Dt2 
Dy = y(t)-y0= vy0 Dt + ½ ay Dt2
vy (t) = vy0 + ay Dt  Dt = (vy - vy0) / ay
Eliminate Dt and regroup
So Dy = vy0 (vy- vy0) / ay+ ½ ay (vy2 - 2vy vy0+vy02 ) / ay2
Dy = (vy0vy- vy02) / ay+ ½ (vy2 - 2vy vy0+vy02 ) / ay
Dy = ( vy02) / ay+ ½ (vy vy02 ) / ay
Dy = ½ (vy2 - vy02 ) / ay

22. Dy = ½ (vy2 - vy02 ) / ay may Dy = ½ m (vy2 - vy02 ) ay= -g Energy
And now
Dy = ½ (vy2 - vy02 ) / ay
can be rewritten as:
may Dy = ½ m (vy2 - vy02 )
And if the object is falling under the influence of gravity
then
ay= -g

23. -mg (yf – yi) = ½ m ( vyf2 -vyi2 )
Energy
-mg Dy= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between y-displacement and change in the
y-speed
Rearranging to give initial on the left and final on the right
½ m vyi2 + mgyi = ½ m vyf2 + mgyf
We now define mgy as the “gravitational potential energy”

24. Energy ½ m vi2 + mgyi = ½ m vf2 + mgyf
Notice that if we only consider gravity as the external force then
the x and z velocities remain constant
To ½ m vyi2 + mgyi = ½ m vyf2 + mgyf
Add ½ m vxi2 + ½ m vzi2 and ½ m vxf2 + ½ m vzf2
½ m vi2 + mgyi = ½ m vf2 + mgyf
where vi2 = vxi2 +vyi2 + vzi2
½ m v2 terms are defined to be kinetic energies
(A scalar quantity of motion)

25. Emech is called “mechanical energy”
If only “conservative” forces are present, the total energy
(sum of potential, U, and kinetic energies, K) of a system is conserved
Emech = K + U
Emech = K + U = constant
K and U may change, but Emech = K + U remains a fixed value.
Emech is called “mechanical energy”

26. Example of a conservative system: The simple pendulum.
Suppose we release a mass m from rest a distance h1 above its lowest possible point.
What is the maximum speed of the mass and where does this happen ?
To what height h2 does it rise on the other side ? v h1 h2 m

27. Example: The simple pendulum.
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum. y
y=h1
y=0

28. Example: The simple pendulum.
What is the maximum speed of the mass and where does this happen ?
E = K + U = constant and so K is maximum when U is a minimum
E = mgh1 at top
E = mgh1 = ½ mv2 at bottom of the swing y
y=h1 h1
y=0 v

29. Example: The simple pendulum.
To what height h2 does it rise on the other side?
E = K + U = constant and so when U is maximum again (when K = 0) it will be at its highest point.
E = mgh1 = mgh2 or h1 = h2 y
y=h1=h2
y=0

30. Cart Exercise Revisited: How does this help?
ai = g sin 30°
= 5 m/s2
d = 5 m / sin 30°
= ½ ai Dt2
10 m = 2.5 m/s2 Dt2
2s = Dt
v = ai Dt = 10 m/s
vx= v cos 30°
= 8.7 m/s
1st Part: Find v at bottom of incline i j N
Emech is conserved
Ki+ Ui = Kf Uf
0 + mgyi = ½ mv2 + 0
(2gy)½ = v = (100) ½ m/s
vx= v cos 30°
= 8.7 m/s
One step, no FBG needed
5.0 m mg
30°
7.5 m y x

31. Home exercise
A block is shot up a frictionless 40° slope with initial velocity v. It reaches a height h before sliding back down. The same block is shot with the same velocity up a frictionless 20° slope.
On this slope, the block reaches height 2h h
h/2
Greater than h, but we can’t predict an exact value.
Less than h, but we can’t predict an exact value.

32. Exercise 3: U, K, E & Path
A ball of mass m, initially at rest, is released and follows three difference paths. All surfaces are frictionless
Ball is dropped
Ball slides down a straight incline
Ball slides down a curved incline
After traveling a vertical distance h, how do the speeds compare? h 1 3 2
(A) 1 > 2 > (B) 3 > 2 > 1 (C) 3 = 2 = 1 (D) Can’t tell