1. Momentum, Energy, and Collisions
Lecture 12
Momentum, Energy, and Collisions

2. Announcements EXAM: Thursday, October 17 (chapter 6-9)
Equation sheet to be posted
Practice exam to be posted

3. Linear Momentum
Impulse
With no net external force:

4. Center of Mass For two objects:
The center of mass of a system is the point where the system can be balanced in a uniform gravitational field.
For two objects:
The center of mass is closer to the more massive object.
Think of it as the “average location of the mass”

5. Center of Mass a) higher b) lower c) at the same place
The disk shown below in (1) clearly has its center of mass at the center.
Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ?
a) higher
b) lower
c) at the same place
d) there is no definable CM in this case
Answer: a
(1)
X CM
(2)

6. Center of Mass CM a) higher b) lower c) at the same place
The disk shown below in (1) clearly has its center of mass at the center.
Suppose the disk is cut in half and the pieces arranged as shown in (2).
Where is the center of mass of (2) as compared to (1) ?
a) higher
b) lower
c) at the same place
d) there is no definable CM in this case
The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line.
(1)
X CM
(2) CM

7. Gun Control
When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?
a) it is much sharper than the gun
b) it is smaller and can penetrate your body
c) it has more kinetic energy than the gun
d) it goes a longer distance and gains speed
e) it has more momentum than the gun

8. Gun Control
When a bullet is fired from a gun, the bullet and the gun have equal and opposite momenta. If this is true, then why is the bullet deadly (whereas it is safe to hold the gun while it is fired)?
a) it is much sharper than the gun
b) it is smaller and can penetrate your body
c) it has more kinetic energy than the gun
d) it goes a longer distance and gains speed
e) it has more momentum than the gun
Even though it is true that the magnitudes of the momenta of the gun and the bullet are equal, the bullet is less massive and so it has a much higher velocity. Because KE is related to v2, the bullet has considerably more KE and therefore can do more damage on impact.

9. Rolling in the Rain
An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately

10. Rolling in the Rain
An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)
a) speeds up
b) maintains constant speed
c) slows down
d) stops immediately
Because the rain falls in vertically, it adds no momentum to the box, thus the box’s momentum is conserved. However, because the mass of the box slowly increases with the added rain, its velocity has to decrease.

11. Two objects collide... and stick
mass m
No external forces... so momentum of system is conserved
initial
px = mv0
mv0 = (2m)vf
vf = v0 / 2
final
px = (2m)vf
A completely inelastic collision: no “bounce back”

12. Inelastic collision: What about energy?
mass m
vf = v0 / 2
initial
final
Kinetic energy is lost!
KEfinal = 1/2 KEinitial

13. This is an example of an “inelastic collision”
Collisions
This is an example of an “inelastic collision”
Collision: two objects striking one another
Elastic collision ⇔ “things bounce back”
⇔ energy is conserved
Inelastic collision: less than perfectly bouncy
⇔ Kinetic energy is lost
Time of collision is short enough that external forces may be ignored so momentum is conserved
Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss

14. Elastic vs. Inelastic Completely inelastic collision:
colliding objects stick together, maximal loss of kinetic energy
Inelastic collision:
momentum is conserved but kinetic energy is not
Elastic collision:
momentum and kinetic energy is conserved.

15. Completely Inelastic Collisions in One Dimension
Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses:
Completely inelastic only
(objects stick together, so have same final velocity)
Momentum Conservation:
KEfinal < KEinitial

16. Crash Cars I
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt?
Answer: e

17. Crash Cars I
a) I
b) II
c) I and II
d) II and III
e) all three
If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt?
In case I, the solid wall clearly stops the car.
In cases II and III, because ptot = 0 before the collision, then ptot must also be zero after the collision, which means that the car comes to a halt in all three cases.

18. Crash Cars II
If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)?
a) I
b) II
c) III
d) II and III
e) all three
Answer: c

19. Crash Cars II
If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)?
a) I
b) II
c) III
d) II and III
e) all three
The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv2 and the car in case III has the largest velocity.

20. Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.
vf = m v0 / (m+M)
momentum conservation
in inelastic collision
KE = 1/2 (mv0)2 / (m+M)
energy conservation
afterwards
PE = (m+M) g h
hmax = (mv0)2 / [2 g (m+M)2]

21. Velocity of the ballistic pendulum
approximation
Pellet Mass (m): 2 g
Pendulum Mass (M): 3.81 kg
Wire length (L): 4.00 m

22. Inelastic Collisions in 2 Dimensions
For collisions in two dimensions, conservation of momentum is applied separately along each axis:
Energy is not a vector equation: there is only 1 conservation of energy equation
Momentum is a vector equation: there is 1 conservation of momentum equation per dimension

23. In elastic collisions, both kinetic energy and momentum are conserved.
One-dimensional elastic collision:

24. Elastic Collisions in 1-dimension
For special case of v2i = 0
We have two equations:
conservation of momentum
conservation of energy
and two unknowns (the final speeds).
solving for the final speeds:
Note: relative speed is conserved for head-on (1-D) elastic collision

25. Limiting cases of elastic collisions
note: relative speed conserved

26. Limiting cases
note: relative speed conserved

27. Limiting cases
note: relative speed conserved

28. Toy Pendulum Incompatible!
Could two balls recoil and conserve both momentum and energy?
Incompatible!

29. Elastic Collisions I 2 1 at rest v a) situation 1 b) situation 2
Consider two elastic collisions:
1) a golf ball with speed v hits a stationary bowling ball head-on.
2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
a) situation 1
b) situation 2
c) both the same
Answer: b v 2 1
at rest

30. Elastic Collisions I v 1 v 2 a) situation 1 b) situation 2
Consider two elastic collisions:
1) a golf ball with speed v hits a stationary bowling ball head-on.
2) a bowling ball with speed v hits a stationary golf ball head-on.
In which case does the golf ball have the greater speed after the collision?
a) situation 1
b) situation 2
c) both the same v 1
Remember that the magnitude of the relative velocity has to be equal before and after the collision!
In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v. v 2 2v
In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v.

31. A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?

32. A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.
So lets try a frame where one body (the elephant) is at rest!
What is the speed of the ball relative to the elephant?
In the frame, what is the final speed of the ball?
Back in the frame of the ground:

33. A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?
Our simplest formulas for speed after an elastic collision relied on one body being initially at rest.
So lets try a frame where one body (the elephant) is at rest!
What is the speed of the ball relative to the elephant?
NOTE: Formulas for 1-D elastic scattering with non-zero initial velocities are given in end-of-chapter problem 88.
In the frame, what is the final speed of the ball?
Back in the frame of the ground:

34. Elastic Collisions in 2-D
Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object

35. CEBAF at JLab Precision hadronic microscopy
CEBAF at Jefferson Lab
recirculation through continuous-wave superconducting RF linacs
simultaneous beam delivery to 3 experimental halls with large complementary spectrometers
cold RF = stable, clean, quiet
up to 200 microAmps per hall, E ~ GeV, >80% polarization
Let me briefly introduce CEBAF
This photograph shows the characteristic “racetrack” shape of two cold RF linacs connected by recirculating arcs.
There are three experimental halls with complementary standard spectrometer equipment. Beam can be simultaneously delivered to all 3.
The machine operates in continuous wave mode, which maximizes the useful luminosity.
The cold RF is key in producing a stable, quiet and clean beam.
With up to 200 microAmps to a hall, an energy range of ~ GeV, and beam polarization routinely over 80%, there has never been an instrument like this for precision hadron microscopy.
This facility has over 1200 outside users, performing a wide range of studies.
An ideal machine for precision hadronic microscopy!
3 experimental Halls
First experiments begun in 1994
User community 1200 members strong

37. v1 v0 = 3.5x105 m/s 37o 53o initial v2 final
A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.5x105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37o to the original direction of motion, the other recoils at 53o to that same axis. Find the final speeds of the two protons.
v0 = 3.5x105 m/s
initial
37o
53o v2 v1
final

38. v1 v0 = 3.5x105 m/s 37o 53o initial v2 final
A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.5x105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37o to the original direction of motion, the other recoils at 53o to that same axis. Find the final speeds of the two protons.
Momentum conservation:
v0 = 3.5x105 m/s
initial
37o
53o v2 v1
final
if we’d been given only 1 angle, would have needed conservation of energy also!

39. Rotational Kinematics

40. Angular Position

41. θ > 0
θ < 0
Angular Position
Degrees and revolutions:

42. Arc Length s = rθ Arc length s, from angle measured in radians:
- What is the relationship between the circumference of a circle and its diameter?
C / D = π
C = 2 π r
- Arc length for a full rotation (360o) of a radius=1m circle?
s = 2 π (1 m) = 2 π meters
1 complete revolution = 2 π radians
1 rad = 360o / (2π) = 57.3o

43. Angular Velocity

44. Instantaneous Angular Velocity
Period = How long it takes to go 1 full revolution
Period T:
SI unit: second, s

45. Linear and Angular Velocity

46. Greater translation for same rotation

47. Bonnie and Klyde II
Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go- round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for both of them
Answer: b ω
Bonnie
Klyde

48. Bonnie and Klyde II
Bonnie sits on the outer rim of a merry-go- round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity?
a) Klyde
b) Bonnie
c) both the same
d) linear velocity is zero for both of them
Answer: b
Their linear speeds v will be different because v = r ω and Bonnie is located farther out (larger radius r) than Klyde. ω
Klyde
Bonnie

49. Angular Acceleration

50. Instantaneous Angular Acceleration

51. Rotational Kinematics, Constant Acceleration
If the angular acceleration is constant:
If the acceleration is constant:
v = v0 + at

52. Analogies between linear and rotational kinematics:

53. ½ Angular Displacement I ¾ ¼ ½ a) θ b) θ c) θ d) 2 θ e) 4 θ
An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t?
Answer: b

54. ½ Angular Displacement I ¾ ¼ ½ a) θ b) θ c) θ d) 2 θ e) 4 θ
An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in the time t, through what angle did it rotate in the time t ?
The angular displacement is θ = αt 2 (starting from rest), and there is a quadratic dependence on time. Therefore, in half the time, the object has rotated through one-quarter the angle.

55. Which child experiences a greater acceleration?
(assume constant angular speed)

56. Larger r: - larger v for same ω - larger ac for same ω
ac is required for circular motion.
An object may have at as well, which implies angular acceleration

57. Angular acceleration and total linear acceleration