Physics 207: Lecture 2 Notes

1
Lecture 14

• Goals
• More Energy Transfer and Energy Conservation
• Define and introduce power (energy per time)
• Introduce Momentum and Impulse
• Compare Force vs time to Force vs distance
• Employ conservation of momentum in 1 D 2D
• Note 2nd Exam, Monday, March 19th, 715 to 845
  PM

2
Energy conservation for a Hookes Law spring

• Associate ½ kx2 with the potential energy of
  the spring

• Ideal Hookes Law springs are conservative so the
  mechanical energy is constant if the spring and
  mass are the system

3
Hookes Law spring in the vertical

• Gravity and perfect Hookes Law spring are
  conservative forces
• New equilibrium length at position where the
  gravitational force equals the spring force.

4
Energy (with spring gravity)
Given m, g, h k, how much does the spring
compress?

• Emech constant (only conservative forces)
• At 1 y1 h v1y 0 At 2 y2 0 v2y ?
  At 3 y3 -x v3 0
• Em1 Ug1 Us1 K1 mgh 0 0
• Em2 Ug2 Us2 K2 0 0 ½ mv2
• Em3 Ug3 Us3 K3 -mgx ½ kx2 0

5
Energy (with spring gravity)
Given m, g, h k, how much does the spring
compress?

• Emech constant (only conservative forces)
• At 1 y1 h v1y 0 At 2 y2 0 v2y ?
  At 3 y3 -x v3 0
• Em1 Ug1 Us1 K1 mgh 0 0
• Em2 Ug2 Us2 K2 0 0 ½ mv2
• Em3 Ug3 Us3 K3 -mgx ½ kx2 0
• Given m, g, h k, how much does the spring
  compress?
• Em1 Em3 mgh -mgx ½ kx2 ? Solve ½ kx2
  mgx - mgh 0

6
Energy (with spring gravity)
1
mass m
2
h
3
0
-x

• When is the childs speed greatest?
• (Hint Consider forces energy)
• (A) At y1 (top of jump)
• (B) Between y1 y2
• (C) At y2 (child first contacts spring)
• (D) Between y2 y3
• (E) At y3 (maximum spring compression)

7
Energy (with spring gravity)
1
2
h
3
kx
mg
0
-x

• When is the childs speed greatest? (D) Between
  y2 y3
• A Calc. soln. Find v vs. spring displacement
  then maximize
• (i.e., take derivative and then set to zero)
• B Physics As long as Fgravity gt Fspring then
  speed is increasing
• Find where Fgravity- Fspring 0 ? -mg
  kxVmax or xVmax -mg / k
• So mgh Ug23 Us23 K23 mg (-mg/k) ½
  k(-mg/k)2 ½ mv2
• ? 2gh 2(-mg2/k) mg2/k v2 ? 2gh mg2/k
  vmax2

8
Work Power

• Two cars go up a hill, a Corvette and a ordinary
  Chevy Malibu. Both cars have the same mass.
• Assuming identical friction, both engines do the
  same amount of work to get up the hill.
• Are the cars essentially the same ?
• NO. The Corvette can get up the hill quicker
• It has a more powerful engine.

9
Work Power

• Power is the rate at which work is done.
• Average Power is,
• Instantaneous Power is,
• If force constant in 1D, W F Dx F (v0 Dt ½
  aDt2)
• and P F v F (v0 aDt)

1 W 1 J / 1s
10
Exercise Work Power

• P dW / dt and W F d (Ff - mg sin q) d
• and d ½ a t2 (constant acceleration)
• So W F ½ a t2 ? P F a t F v
• (A)
• (B)
• (C)

Power
time
Power
Z3
time
Power
time
11
Work Power

• Power is the rate at which work is done.

Example

• A person of mass 80.0 kg walks up to 3rd floor
  (12.0m). If he/she climbs in 20.0 sec what is
  the average power used.
• Pavg F h / t mgh / t 80.0 x 9.80 x 12.0 /
  20.0 W
• P 470. W

12
Ch. 9 Momentum Impulse An alternative
perspective (force vs time)

• Energy, Energy Conservation and Work
• Good approach if evolution with time is not
  needed.
• Energy is Conserved if only conservative (C)
  forces.
• Work relates applied forces (C and NC) along the
  path to energy transfer (in or out).
• Usually employed in situations with long times,
  large distances
• Are there any other relationships between mass
  and velocity that remain fixed in value (i.e. a
  new conservation law)?

13
Newtons 3rd Law

• If object 1 and object 2 are the system then
  any change in the momentum of one is reflected
  by and equal and opposite change in the other.

14
Momentum Conservation

• Momentum conservation (recasts Newtons 2nd Law
  when net external F 0) is an important
  principle (most often useful when forces act over
  a short time)
• It is a vector expression so must consider px, py
  and pz
• if Fx (external) 0 then px is constant
• if Fy (external) 0 then py is constant
• if Fz (external) 0 then pz is constant

15
A collision in 1-D

• 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 final speed
  V.
• Because there is no external force momentum is
  conserved
• Because there is an internal non-conservative
  force energy conservation cannot be used unless
  we know the WNC
• So pxi pxf
• In terms of m, M, and V, what is the momentum of
  the bullet with speed v ?

16
A collision in 1-D

• What is the momentum of the bullet with speed v
  ?
• Key question Is x-momentum conserved ?

p After
p Before
17
A collision in 1-D Energy

• What is the initial energy of the system ?
• What is the final energy of the system ?
• Is energy conserved?
• Examine Ebefore-Eafter

v
No! This is an example of an inelastic collison
18
Explosions A collision in reverse

• A two piece assembly is hanging vertically at
  rest at the end of a 20 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?

After
Before
19
Explosions A collision in reverse

• A two piece assembly is hanging vertically at
  rest at the end of a 20 m long massless string.
  The mass of the two pieces are 60 and 20 kg
  respectively. Suddenly you observe that the 20
  kg mass 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 paththere must be centripetal
  (i.e., radial acceleration)

After
Before
20
Explosions A collision in reverse

• A two piece assembly is hanging vertically at
  rest at the end of a 20 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 2030/ 60 10 m/s
• Tbefore Weight (6020) x 10 N 800 N
• SFy m acy M V2/r T Mg
• T Mg MV2 /r 600 N 60x(10)2/20 N 900 N

After
21
Impulse (A variable force applied for a given
time)

• Collisions often involve a varying force
• F(t) 0 ? maximum ? 0
• We can plot force vs time for a typical
  collision. The impulse, I, of the force is a
  vector defined as the integral of the force
  during the time of the collision.
• The impulse measures momentum transfer

22
Recap

• Read through all of Chapter 9