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Linear Momentum. Momentum (vector quantity). p = mass X velocity p = m v Unit = kg-m/s. Momentum: Example 1. What is the momentum of a 100.0 kg football player running at 6.0 m/s? p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s. Momentum: Example 2.

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momentum vector quantity
Momentum (vector quantity)
  • p = mass X velocity
  • p = mv
  • Unit = kg-m/s
momentum example 1
Momentum: Example 1

What is the momentum of a 100.0 kg football player running at 6.0 m/s?

p = mv = (100.0 kg)(6.0 m/s) = 600 kg-m/s

momentum example 2
Momentum: Example 2

If Mr. West is running at 2 m/s and has a momentum of 80 kg m/s, what is his mass?

the law of conservation of momentum
The Law of Conservation of Momentum

m1v1 + m2v2 = m1v1’ + m2v2’

momentumbefore = momentumafter

slide6
A 0.015 kg bullet is fired with a velocity of 200 m/s from a 6 kg rifle. What is the recoil velocity of the rifle (consider it’s direction)?
slide7
The Rhino has a mass of 900 kg and hits the Hulk running at 2 m/s. After the collision, the Hulk is moving at 1.3 m/s and Rhino stops. What is the Hulk’s mass?
slide8
A 10,000 kg railroad car moving at 24.0 m/s strikes an identical car that is stationary. They lock together. What will be there common speed afterwards?
slide9
m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 + m2v2 = (m1 + m2)vf

(10,000 kg)(24.0 m/s) + 0 = (20,000 kg)(vf)

vf = +12.0 m/s

(Notice that the speed was cut exactly in half. Why?)

slide10

Two boats meet in the middle of a lake. Boat A has a mass of 150 kg, and boat B a mass of 250 kg. The person in Boat A pushes against Boat B and has a speed of -1.4 m/s.

  • What is the initial momentum of Boat A?
  • Calculate the velocity of Boat B.
rocket ship
Rocket Ship
  • Before moving, p = 0
  • Which moves more, the rocket or the gas?
slide12
A 0.145 kg ball is thrown at +40.2 m/s. The bat has a mass of 0.840 kg. How fast must you swing the bat to return the ball at 20.1 m/s? Assume the bat stops when you hit the ball. (ANS: -10.4 m/s)
  • How fast must you swing the bat to return the ball at 40.2 m/s? (ANS: -13.9 m/s)
types of collisions
Types of Collisions
  • Elastic
    • KE and momentum are conserved
  • Inelastic
    • Momentum is conserved
    • KE is not conserved
      • Lost (heat, sound)
      • Added (chemical reaction)
  • Perfectly inelastic
    • Two objects stick together after the collision
elastic collisions
Elastic Collisions
  • Both KE and momentum are conserved

m1v1i + m2v2i = m1v1f + m2v2f

m1v1i - m1v1f = m2v2f - m2v2i

m1(v1i - v1f) = m2(v2f - v2i)

slide15
½ m1v1i2 + ½ m2v2i2 = ½ m1v1f2 + ½ m2v2f2

m1v1i2 - m1v1f2 = m2v2f2 - m2v2i2

m1(v1i2 - v1f2) = m2(v2f2 - v2i2)

Remember that (a2 – b2) = (a + b)(a – b)

m1(v1i+ v1f) (v1i - v1f) = m2 (v2i+ v2f)(v2f- v2i)

Divide the yellow box equations:

m1(v1i+ v1f) (v1i - v1f) = m2 (v2i+ v2f)(v2f- v2i)

m1(v1i - v1f) m2(v2f - v2i)

v1i+ v1f = v2i+ v2f

elastic collisions example 1
Elastic Collisions: Example 1

A cue ball moving at 2.0 m/s strikes the red ball. What is the speed of both balls after the elastic collision if they have equal mass?

slide17
mvci + mvri = mvcf + mvrf all masses are equal

mvci + 0 = mvcf + mvrf The red ball is still

vci = vcf + vrf Factored out the m

2.0 m/s = vcf + vrf Need another equation

vci+ vcf = vri+ vrf

2.0 m/s + vcf = 0 + vrf` Substitute

vrf` = 2.0 m/s + vcf

2.0 m/s = vcf + 2.0 m/s + vcf

0 = 2vcf or vcf = 0

2.0 m/s = vcf + vrf = 0 + vrf

vrf = 2.0 m/s

elastic collisions example 2
Elastic Collisions: Example 2

Two pool balls of equal mass collide. One is moving to the right at 20 cm/s, and the other to the left at 30 cm/s. Calculate their velocities after they collide elastically.

+20 cm/s

-30 cm/s

slide19
mvwi + mvbi = mvwf + mvbf

vwi + vbi = vwf + vbf

20 cm/s + (-30 cm/s) = vwf + vbf

-10 cm/s = vwf + vbf Need another equation

vwi+ vwf = vbi+ vbf KE is conserved

vwf = vbi+ vbf - vwi

vwf = -30 cm/s+ vbf – 20 cm/s

vwf = -50 cm/s+ vbfNow substitute

-10 cm/s = -50 cm/s+ vbf + vbf

40 cm/s = 2vbf

vbf = +20 cm/s

slide20
vwf = -50 cm/s+ vbf

vwf = -50 cm/s+ (+20 cm/s) = -30 cm/s

+20 cm/s

-30 cm/s

-30 cm/s

+20 cm/s

Note how they exchanged velocities (because they have equal mass)

elastic collisions example 3
Elastic Collisions: Example 3

A proton of mass 1.01 amu moving at 3.60 X 104 m/s elastically collides head on with a still Helium nucleus (4.00 amu). What are the velocities of the particles after the collision?

vpf = -2.15 X 104 m/s

vhf = 1.45 X 104 m/s

slide22
vpf = vpi(mp – mh) = 3.60 X 104 m/s(1.01 amu-4,00 amu)

(mp + mh) (1.01 amu + 4.00 amu)

vpf = -2.15 X 104 m/s

vpi+ vpf = vhf

vhf = 3.60 X 104 m/s + (-2.15 X 104m/s)

vhf = 1.45 X 104 m/s

perfectly inelastic collision ex 1
Perfectly Inelastic Collision: Ex 1

An 1800-kg Cadillac is stopped at a traffic light. It is struck in the rear by a 900-kg Cube moving at 20.0 m/s. The cars become entangled. Calculate their velocity after the collision.

KE is not conserved

m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 + m2v2 = (m1 + m2)vf

0 + (900 kg)(20.0 m/s) = (1800kg + 900kg)vf

vf= +6.67 m/s

slide24
How much KE was lost as a result of the collision?

KEi = ½ (900 kg)(20 m/s)2 = 180,000 J

KEf = ½ (2700 kg)(6.67 m/s)2 = 60,000 J

DKE = 60,000 J – 180,000 J = -120,000J

perfectly inelastic collision ex 2
Perfectly Inelastic Collision: Ex 2

Two balls of mud collide head on and stick together. The first ball of mud had a mass of 0.500 kg and was moving at +4.00 m/s. The second had a mass of 0.250 kg and was moving at –3.00 m/s. Find the velocity of the ball after the collision.

m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 + m2v2 = (m1 + m2)vf

(0.500 kg)(4.00 m/s)+(0.250 kg)(-3.00 m/s) = (0.750 kg)(vf)

vf = +1.67 m/s

perfectly inelastic collision ex 3
Perfectly Inelastic Collision: Ex 3

A 95.0 kg running back running at 3.50 m/s collides with a 120.0 kg linebacker running in the opposite direction at 4.00 m/s. They grapple and stick together

  • Calculate the velocity after the collision. (-0.69 m/s)
  • Calculate the kinetic energy of the system before the collision. (1542 J)
  • Calculate the kinetic energy after the collision. (51 J)
  • What percent of the kinetic energy was lost in the collision? (97%)
ballistic pendulum
Ballistic Pendulum

Used to determine velocities

Perfectly Inelastic collisions (KE converts to PE)

perfectly inelastic collision ex 328
Perfectly Inelastic Collision: Ex 3

A 5.00 gram bullet is fired into a 1.00 kg block of wood. The wood-bullet system rises 5 cm. Calculate the initial velocity of the bullet.

slide29
m1v1 + m2v2 = m1v1’ + m2v2’

m1v1 + m2v2 = (m1 + m2)vf

0 +(0.005 kg)(vi) = (1.00 kg + 0.005 kg)(vf)

(0.005 kg)(vi) = (1.005 kg)(vf)

We need a second equation (two unknowns)

½ mv21 + mgy1 = ½ mv22 + mgy2

½(1.005 kg)(vf2) +0 =0 + (1.005 kg)(9.8 m/s2)(0.05 m)

vf = 0.990 m/s

(0.005 kg)(vi) = (1.005 kg)(0.990 m/s)

vi = 199 m/s

Moment of impact

slide30

A 7.00 g bullet is fired into a 0.950 kg ballistic pendulum. The bob rises to a height of 22.0 cm. Calculate the initial speed of the bullet.

(284 m/s)

slide31

An 8.00 g bullet is fired at 350 m/s into a ballistic pendulum bob of 2.00 kg. Calculate how high the coupled bullet and bob will rise.

slide32

A 10.0 g bullet is fired into a 1.20 kg ballistic pendulum at a speed of 320 m/s. The string on the pendulum is 150 cm.

  • Calculate the height that the pendulum rose. (35.1 cm)
  • Calculate the angle the pendulum rises. (40.5o)
collisions in 2 or 3 dimensions glancing collisions
Collisions in 2 or 3 Dimensions (Glancing collisions)
  • A moving marble collides with a stationary marble.
  • Which of the following situations is/are not possible after:
collisions in 2 or 3 dimensions
Collisions in 2 or 3 Dimensions

Problem Solving:

  • Resolve all vectors into their x and y components
  • Spix = Spfx Spiy = Spfy
2 d collisions example 1
2-D Collisions: Example 1

At an intersection, a 1500-kg car travels east at 25 m/s. It collides with a 2500-kg van traveling north at 20 m/s. If the vehicles stick together afterwards, calculate the magnitude and velocity of the cars after the collision.

slide37
Let’s first work with the initial components

Spix = (1500 kg)(25 m/s)

Spix = 37,500 kg-m/s

Spiy = (2500 kg)(20 m/s)

Spiy = 50,000 kg-m/s

After the collision:

v

vy = v sinq

q

vx = v cosq

slide38
After the collision:

Spix = Spfx

37,500 kg-m/s = mvx

37,500 kg-m/s = (1500 kg + 2500 kg) vx

37,500 kg-m/s = (4000 kg) vx

vx = 9.375 m/s

Spiy = Spfy

50,000 kg-m/s = mvy

50,000 kg-m/s = (1500 kg + 2500 kg)vy

50,000 kg-m/s = (4000 kg) vy

vy= 12.5 m/s

slide39
Now solve for v

v2 = vx2+vy2

v = 16 m/s

Divide the two equations

tanq = vy = 12.5

vx 9.38

q = 53o

2 d collisions example 140
2-D Collisions: Example 1

A pool ball moving at +3.00 m/s strikes another pool ball (same mass) that is initially at rest. After the collision, the two balls move off at 45.0o angles. Calculate the speeds of the two balls.

vred

q = +450

q = -450

vgreen

slide41
x coordinate

mgvg + mrvr = mgvg + mrvr

3m + 0 = mgvg + mrvr

3 = vg + vr

3 = vgcos(45o) + vrcos(-45o)

3 = 0.707vg + 0.707vr

4.24 = vg + vr

slide42
y coordinate

0 = mgvg + mrvr

0 = mgvg + mrvr

0 = vg + vr

0 = vgsin(45o) + vrsin(-45o)

0 = 0.707vg - 0.707vr

vg = vr

Two equations, two unknowns

4.24 = vg + vr vg = vr

2.12 m/s

slide43

Two zombie heads of unequal mass sit on an icerink. The first head (m1 = 4.00 kg) is propelled toward the stationary second head (m2 = 2.00 kg) at a velocity of v1= 4.00 m/s. After the collision, both heads move off at 35o to the x-axis. Calculate their final speeds.

2.44 m/s, 4.88 m/s

slide44

A 1000 kg car travelling south at 15.0 m/s is hit by a 1500 kg van travelling west at 20.0 m/s. They stick together.

  • Calculate the final speed of the cars. (13.4 m/s)
  • Calculate the direction that the cars will travel in. (26.6o South of West)
  • Can you express your answer in “i and j” notation?
slide45

Two different pucks are placed on an air hockey table. A 15.0 g red puck is pushed to the right at 1.00 m/s. A 20.0 g green puck is pushed to the left at 1.20 m/s. After the collision, the green puck travels at 1.10 m/s at an angle of 40.0o south of the horizontal.

  • Calculate the x and y components of the green puck’s velocity after the collision. (-0.843 m/s, -0.707 m/s)
  • Calculate the x and y components of the green puck’s momentum after the collision. (-0.0169 kg m/s, -0.0141 kg m/s)
  • Find the speed and direction of the red puck. (1.08 m/s, 60.7o north of east)
  • Can you express your answer to (c) in “i and j” notation? ((0.527 i + 0.940j) m/s)
center of mass
Center of Mass

Center of mass – one point on a particle that follows the same path.

center of mass48
Center of Mass
  • Stand perpendicular to the wall. Both feet must be against the baseboard. Now lift your outer leg.
  • Stand with your back against the wall. Be sure your heels touch the baseboard. Now try to pick up an eraser without bending your knees.
center of mass49
Center of Mass

3. Get down on your elbows and knees on the floor. Place your elbows right in front of your knees and then move your hands behind you back. Try to move an eraser with your nose.

general motion
General Motion
  • Translational Motion
    • all points of an object follow the same path
    • Sliding a book across a table
  • Rotational Motion
  • General Motion – combination of translational and rotation motion
calculating center of mass ex 1
Calculating Center of Mass: Ex 1.

xCM = m1x1 + m2x2

m1 + m2

Three people each massing about 60 kg sit on a banana boat as shown below. Calculate the center of mass.

slide53
xCM = m1x1 + m2x2 + m3x3

m1 + m2 + m3

xCM = (60 kg)(1m) + (60 kg)(5 m) + (60 kg)(6m)

180 kg

xCM = 4.0 m

slide54
Where is the center of mass for the Earth-Moon system. Assume the center of the Earth is the origin. Some values you need are:

mEarth = 5.97 X 1024 kg

mMoon = 7.35 X 1022 kg

Earth-Moon distance = 3.84 X 108 m

(try it with the moon as the origin.)

slide55
Calculate the center of mass for the Earth-Sun system. Assume the center of the Sun is the origin. Some values you need are:

mEarth = 5.97 X 1024 kg

mSun = 2.00× 1030 kg

Earth-Sun distance = 1.50 X 1011 m

slide56

Find the center of mass of the CO molecule if carbon has a mass of 12.0 u and oxygen has a mass of 16.0 u. The bond length is 1.13 Angstroms.

(0.645 A from the carbon)

slide57

A 150 kg man and a 450 kg owlbear sit on opposite sides of a 4 m plank. Calculate where the plank should be pivoted to produce a seesaw.

(3 m from the man)

slide58

Repeat the calculation, but now assume the plank has a mass of 10.0 kg and all the mass of the plank acts at the center, 2.00 m.

(2.98 m from the man)

slide59

Calculate the center of mass of the water molecule. Hydrogen’s mass is 1.00 g/mole and oxygen’s is 16.0 g/mole. The distance between the two nuclei is 0.942 Angstroms. You may do all your calculations in these units.

(0.745, 0.512)

slide60

Now calculate the center of mass of ozone:

(0.447, 0 (left O as origin))

slide61

Calculate the center of mass of SO2, whose bond angle is about 120o and whose bond length is 1.43 Angstroms.

momentum and force
Momentum and Force

Second Law of Motion

“rate of change of momentum of a body is equal to the net force applied on it.”

SF = Dp (this is the more general form)

Dt

derivation
Derivation

SF = Dp = mv-mvo

Dt Dt

SF = m(v-vo)

Dt

SF = mDv

Dt

SF = ma

impulse
Impulse
  • Impulse = Dp

Dp = FDt

  • Usually occurs over a very short timeframe
slide67
A 50-g golf ball is struck with a force of 2400 N. The ball flew off with a velocity of 44 m/s.
  • Calculate the impulse (2.2 kg-m/s)
  • How long was the club in contact with the ball? (9.1 X 10-4 s)

Dp = pf – pi

Dp = mvf – 0 (ball was still)

Dp = (0.050 kg)(44 m/s) = 2.2 kg-m/s

SF = Dp/Dt

Dt = Dp/SF

Dt = 2.2 kg-m/s/(2400 kg-m/s2) = 9.1 X 10-4 s

slide68
In a crash test, a 1500 kg car moving at –15.0 m/s collides with a wall for 0.150 s. The bumpers on the car cause it to rebound at +2.60 m/s.
  • Calculate the impulse (26,400 kg-m/s)
  • Calculate the average force exerted on the car. (176,000 N)
slide69

A 0.144 kg baseball is moving toward homeplate at -43.0 m/s. It is bunted with a force of 6500 N for 1.30 ms.

  • Calculate the initial momentum of the ball. (-6.19 kg m/s)
  • Calculate the impulse (the change in momentum) (-8.45 kg m/s)
  • Calculate the final momentum (-2.26 kg m/s)
  • Calculate the final velocity of the ball. (-0.144 m/s)
slide70

A 50.0 kg physics student jumps for joy over her grade. She jumps up with a speed of 2.1 m/s for 0.36 s.

  • Calculate her impulse (105 kg m/s)
  • Calculate the force she exerts on the floor (292 N)
  • Calculate the maximum height of her jump (use conservation of energy or kinematics). (22.5 cm)
impulse and area
Impulse and Area
  • Impulse = Area under Force vs. Time graph
connection to calculus
Connection to Calculus
  • You integrate to get area

Dp = FDt

dp = Fdt

p = F dt

tf

to

slide77

A 150.0 g baseball is thrown with a speed of -20.0 m/s. The interaction force vs. time is shown in the following graph.

  • Calculate the impulse. (4.50 kg m/s)
  • Calculate the return speed of the baseball. (10.0 m/s)
slide78

A 100.0 g golf ball is dropped from a height of 2.00 meters from the floor.

  • Calculate the speed of the ball just as it hits the ground. (6.26 m/s)
  • Calculate the impulse. (1.2 kg m/s)
  • Calculate the return speed of the ball just as it bounces back.(5.74 m/s)
  • Calculate the return height of the ball. (1.68 m)
slide79

2. -0.720 m/s opposite the direction of package

4. 4.8 m/s

6. 1.6 X 104 kg

8. 510 m/s

10. 4200 m/s

12. a) 4700 m/s and 6900 m/s b) 5.90 X 108 J

14. 130 N, not enough

16. a) 2.0 kg m/s b) 400 N

  • a) 460 kg m/s(E) b) -460 kg m/s(W)

c) 460 kg m/s (E) d) 610 N (E)

slide80

22. v1’ = -1.00 m/s (rebound) v2’ = 2.00 m/s

24. v1’ = 0.70 m/s v2’ = -2.20 m/s

  • a) v1’ = 3.62 m/s v2’ = 4.42 m/s

b) -396 kg m/s, +396 kg m/s

30. 0.61 m

32. 3000 J and 4500 J

38. 60o and 6.7 m/s

40. a) 30o b) v1’ = v/(3 ½ ) c) 2/3

44. A = 0o, v1’ = 3.7 m/s, v2’ = 2.0 m/s

46. 6.5 X 10-11 m 56. 4.66 X 106 m