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Linear Momentum. p = mass X velocity p = m v Unit = kg-m/s Vector Quantity. 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. If a freshman running at 2 m/s has a momentum of 80 kg m/s, what is his mass?. Momentum and Force.

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linear momentum
Linear Momentum
  • p = mass X velocity
  • p = mv
  • Unit = kg-m/s
  • Vector Quantity
slide2
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 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
slide7

Force

area = impulse

time

impulse example 1
Impulse: Example 1

A 50-g golf ball is struck with a force of 2400 N. The ball flew off with a velocity of 44 m/s. How long was the club in contact with the ball?

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

impulse example 2
Impulse: Example 2

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, and the average force exerted on the car.

Impulse = Dp

pi = mvi = (1500 kg)(-15.0 m/s) = -22,500 kg-m/s

pf = mvf = (1500 kg)(+2.60 m/s) = +3900 kg-m/s

slide10
Dp = pf – pi

Dp = 3900 kg-m/s – (-22,500 kg-m/s)

Dp = 26,400 kg-m/s

To calculate the force:

SF = Dp

Dt

SF = 26,400 kg-m/s = 176,000 N

0.150 s

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 = J

tf

to

slide17

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)
slide18

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)
slide19
The Law of Conservation of Momentum

“The total momentum of an isolated system of bodies remains constant

m1v1 + m2v2 = m1v1’ + m2v2’

momentumbefore = momentumafter

slide20
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)?
slide21
A 10g and a 30 g ball are placed inside a tube with a compressed spring between them. When the spring is released, the 10 g ball exits the tube at 6.0 m/s.
  • Calculate the speed of the 30 g ball.
  • Does it have to be in the opposite direction?
slide22
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?
slide23

Mr. West (75 kg) sees a stationary cart (25.0 kg) 8.00 m ahead of him in the hallway. He accelerates at 1.00 m/s2 and jumps on the cart.

  • Calculate his speed just as he jumps on the cart
  • Calculate his speed as he and the cart zoom down the hall.
rocket ship
Rocket Ship
  • Can use the Third Law (“equal and opposite forces”)
  • Can also use Law of Conservation of momentum
  • Dp = 0 (since neither the gas or rocket were moving initially)
  • Which moves more, the rocket or the gas?
slide25
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? (-10.4 m/s)
  • How fast must you swing the bat to return the ball at 40.2 m/s? (-13.9 m/s)
types of collisions
Types of Collisions
  • Elastic
    • KE and momentum conserved
  • Inelastic
    • Momentum conserved
    • KE not conserved
    • Also, if energy is added (like in a chemical reaction)
  • Perfectly inelastic
    • Two objects stick together after the collision
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 Ion 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

slide28
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

slide30

A lab experiment, a 200 g air track glider and a 400 g glider collide and stick with velcro. The 200 g glider was moving to the right at 3.00 m/s. Afterwards, the system is moving to the left at 0.40 m/s. Calculate the initial speed of the 400 g glider. (-2.1 m/s)

slide31

A uranium atom (238 amu) decays into a small fragment and a “daughter” nucleus. The small fragment is ejected at 1.50 X 107 m/s and the daughter nucleus at 2.56 X 105 m/s. Calculate the mass of both products. (234 amu, 4 amu)

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:
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.

slide34
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

slide35
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)vcosq

vcosq = 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)vsinq

vsinq = 12.5 m/s

slide36
Divide the two equations

vsinq = 12.5 m/s

vcosq = 9.375 m/s

sinq = 1.33 tanq = 1.33 q = 53o

cosq

Now solve for v

vsinq = 12.5 m/s

v = (12.5 m/s)/sin (53o)

v = 16 m/s

slide37

A 5000 kg van travelling south at 10.0 m/s collides with a 2500 kg SUV travelling west. After the collision, the vehicles stuck together and made a 45o angle with the horizontal in a south-west direction.

a. Calculate the initial speed of the 2500 kg car.

b. Calculate the speed just after the collision.

(Remember, two equations, two unknowns).

2 d collisions example 3
2-D Collisions: Example 3

A pool ball moving at +3.0 m/s strikes another pool ball (same mass) that is initially at rest. After the collision, the two balls move off at + 45o. Calculate the speeds of the two balls. (2.1 m/s)

vred

q = +450

q = -450

vgreen

slide39
Let’s first work with the initial components

Spix = (m)(3 m/s)

Spiy = 0

After the collision (x-direction):

vredx = vredcos(45o) vgreenx = vgreencos(-45o)

Spix = Spfx

Spix = predfx + pgreenfx

(m)(3 m/s) = mvredcos(45o) + mvgreencos(-45o)

3 m/s = vredcos(45o) + vgreencos(-45o)

slide40
After the collision (y-direction):

vredy = vredsin(45o) vgreeny = vgreensin(-45o)

Spiy = Spfy

Spiy = predfy + pgreenfy

0 = mvredsin(45o) + mvgreensin(-45o)

mvredsin(45o) = -mvgreensin(-45o)

vredsin(45o) = -vgreensin(-45o)

slide41
Here are the two equations we will solve:

3 m/s = vredcos(45o) + vgreencos(-45o)

vredsin(45o) = -vgreensin(-45o)

vred = -vgreensin(-45o)

sin(45o)

vred = vgreen

3 m/s = vredcos(45o) + vredcos(-45o)

3 m/s = vred(cos(45o) + cos(-45o))

3 m/s = 1.414vred

vred = 2.1 m/s = vgreen

slide42

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.

1.63 m/s, 3.25 m/s

slide43

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)