1. Lecture 13 Goals: Chapter 9 Employ conservation of momentum in 2D
Examine forces over time (aka Impulse)
Chapter 10
Understand the relationship between motion and energy
Assignments:
HW5, due tomorrow, HW6 due Tuesday, Oct. 19th
For Wednesday, Read all of Chapter 10 1

2. Exercise: Momentum is a Vector (!) quantity
Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart
System….block and cart only (assume Earth has infinite mass)
Px is conserved once the block leaves the slide
Py is not conserved (if “system” is block and cart only)
2 kg
5.0 m
30°
What is the final velocity of the cart?
10 kg
7.5 m

3. Exercise Momentum is a Vector (!) quantity
1) ai = g sin 30°
= 5 m/s2
2) 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
x-direction: No net force so Px is conserved
y-direction: vy of the cart + block will be zero and we can ignore vy of the block when it lands in the cart. i j N
5.0 m mg
30°
30°
Initial Final
Px: MVx + mvx = (M+m) V’x
M 0 + mvx = (M+m) V’x
V’x = m vx / (M + m)
= 2 (8.7)/ 12 m/s
V’x = 1.4 m/s
7.5 m y x

4. Home Exercise Inelastic Collision in 1-D with numbers
Do not try this at home!
ice
(no friction)
Before: A 4000 kg bus, twice the mass of the car, moving at 30 m/s impacts the car at rest.
What is the final speed after impact if they move together?

5. Home exercise Inelastic Collision in 1-D
v = 0
ice
M = 2m V0
(no friction)
initially
vf =?
finally
2Vo/3 = 20 m/s

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. Billiards: Without external forces, conservation of momentum (and energy Ch. 10 & 11)
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. Explosions: A collision in reverse
A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. The time of the “explosion” is short compared to the swing of the string.
Does the tension in the string increase or decrease after the explosion?
Before
After

12. Explosions: A collision in reverse
A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s.
Decipher the physics:
1. The green ball recoils in the –x direction (3rd Law) and, because there is no net force in the x-direction the x-momentum is conserved.
2. The motion of the green ball is constrained to a circular path…there must be centripetal (i.e., radial acceleration)
Before
After

13. Explosions: A collision in reverse
A two piece assembly is hanging vertically at rest at the end of a 3.0 m long massless string. The mass of the two pieces are 60 & 20 kg respectively. Suddenly you observe that the 20 kg mass is suddenly ejected horizontally at 30 m/s.
Cons. of x-momentum
Px before= Px after = 0 = - M V + m v
V = m v / M = 20*30/ 60 = 10 m/s
Tbefore = Weight = (60+20) x 10 N = 800 N
SFy = m ay (radial) = M V2/r = T – Mg
T = Mg + MV2 /r = 600 N + 60x(10)2/3 N = 2900 N
Before
After

14. 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.

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

16. 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

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

18. Exercise 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

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

20. 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
 Favg ~ J/t ~ 4900 N
(1) mhead ~ 7 kg
 ahead = F / mhead ~ 700 m/s2 ~ 70 g !
Enough to cause unconsciousness ~ 35% of a fatal blow
Only a rough estimate!

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

22. Discussion 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

23. Newton’s Laws rearranged (Ch. 10)
From motion in the y-dir, Fy = m ay and let ay 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
and eliminating Dt yields
2 ay Dy= (vy2 - vy02 ) which is nothing new
or if ay= -g
-mg Dy= ½ m (vy2 - vy02 )

24. -mg (yf – yi) = ½ m ( vyf2 -vyi2 )
“Energy” defined
-mg Dy= ½ m (vy2 - vy02 )
-mg (yf – yi) = ½ m ( vyf2 -vyi2 )
A relationship between
y- displacement and change in the y-speed squared
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”

25. 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)

26. Lecture 13 Assignment: HW6 up soon
For Wednesday: Read all of chapter 10 1