Linear Momentum

• p = mass X velocity• p = mv• 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

Second Law of Motion

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

F = p (this is the more general form)

t

Derivation

F = p = mv-mvo

t t

F = m(v-vo)

t

F = m v

t

F = ma

Impulse

• Impulse = p

p = Ft

• Usually occurs over a very short timeframe

Force

time

area = impulse

Impulse: Example 1A 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?

p = pf – pi

p = mvf – 0 (ball was still)p = (0.050 kg)(44 m/s) = 2.2 kg-m/sF = p/tt = p/Ft = 2.2 kg-m/s/(2400 kg-m/s2) = 9.1 X 10-4 s

Impulse: Example 2In 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 = p

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

p = pf – pi

p = 3900 kg-m/s – (-22,500 kg-m/s)

p = 26,400 kg-m/s

To calculate the force:

F = p

t

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

0.150 s

Impulse and Area

• Impulse = Area under Force vs. Time graph

• Can use the average Force to approximate an answer

Average Force

Area = Favet

Connection to Calculus• You integrate to get

area

p = Ft

dp = Fdt

p = F dt = Jtf

to

Calculate the impulse (p) given the following graph.

Force vs. Time

0

2

4

6

8

10

12

0 2 4 6 8 10 12

Time (s)

Fo

rce(

N)

Calculate the impulse (p) given the following graph.

Calculate the impulse (p) given the following graph.

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.

a. Calculate the impulse. (4.50 kg m/s)

b.Calculate the return speed of the baseball. (10.0 m/s)

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

a. Calculate the speed of the ball just as it hits the ground. (6.26 m/s)

b. Calculate the impulse. (1.2 kg m/s)

c. Calculate the return speed of the ball just as it bounces back.(5.74 m/s)

d. Calculate the return height of the ball. (1.68 m)

The Law of Conservation of Momentum

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

m1v1 + m2v2 = m1v1’ + m2v2’

momentumbefore = momentumafter

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)?

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.

a. Calculate the speed of the 30 g ball.

b.Does it have to be in the opposite direction?

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?

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.

a. Calculate his speed just as he jumps on the cart

b.Calculate his speed as he and the cart zoom down the hall.

Rocket Ship• Can use the Third Law (“equal and opposite

forces”)

• Can also use Law of Conservation of momentum

p = 0 (since neither the gas or rocket were moving initially)

• Which moves more, the rocket or the gas?

A 0.145 kg ball is thrown at +40.2 m/s. The bat has a mass of 0.840 kg.

a. How fast must you swing the bat to return the ball at 20.1 m/s? (-10.4 m/s)

b. How fast must you swing the bat to return the ball at 40.2 m/s? (-13.9 m/s)

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 1An 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

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

KE = 60,000 J – 180,000 J = -120,000J

Perfectly Inelastic Collision: Ex 2Two 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

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)

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)

• A moving marble collides with a stationary marble.

• Which of the following situations is/are not possible after:

2-D Collisions: Example 1At 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.

Let’s first work with the initial components

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

pix = 37,500 kg-m/s

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

piy = 50,000 kg-m/s

After the collision:v

vx = v cos

vy = v sin

After the collision:

pix = pfx

37,500 kg-m/s = mvx

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

37,500 kg-m/s = (4000 kg)vcosvcos = 9.375 m/s

piy = pfy

50,000 kg-m/s = mvy

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

50,000 kg-m/s = (4000 kg)vsinvsin = 12.5 m/s

Divide the two equationsvsin = 12.5 m/s vcos = 9.375 m/s

sin = 1.33 tan = 1.33 q = 53o

cos

Now solve for vvsin = 12.5 m/s v = (12.5 m/s)/sin (53o)v = 16 m/s

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 3A 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)

= +450

= -450

vred

vgreen

Let’s first work with the initial components

pix = (m)(3 m/s)

piy = 0

After the collision (x-direction):

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

pix = pfx

pix = predfx + pgreenfx

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

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

After the collision (y-direction):

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

piy = pfy

piy = predfy + pgreenfy

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

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

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

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

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

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.

a) Calculate the x and y components of the green puck’s velocity after the collision. (-0.843 m/s, -0.707 m/s)

b) 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)

c) Find the speed and direction of the red puck. (1.08 m/s, 60.7o north of east)

d) Can you express your answer to (c) in “i and j” notation? ((0.527 i + 0.940j) m/s)