unit 4 - uplift education · unit 4 conservation of momentum. conservation of momentum m 1 v 1i + m...

UNIT 4

UNIT 4

MOMENTUM & IMPULSE

IMPULSE-MOMENTUM THEOREM

J = F(∆t) = ∆p = m∆v = (mvf – mvi)

The impulse, J, that acts on an object is equal to the

change in the object’s momentum, ∆p.

Remember, ∆

means

final – initial

∆p = pf – pi

∆v = vf – vi

EXAMPLE

A baseball player bunts (hits softly) a 0.144 kg baseball

thrown at 43.0 m/s. If the bat exerts an average force

of 6500 N on the ball for 0.00122 s, what is the final

speed of the ball?

Given: m = 0.144 kg

vi = -43.0 m/s

F = 6500 N

t = 0.00122 s

Unknown: vf = ?

Equation: F(∆t) = ∆p = (mvf – mvi)

Substitute: (6500 N)(0.00122s) = 0.144kg(v + 43.0 m/s)

Solve: vf = 12.1 m/s

YOU TRY

A soccer ball has a momentum of 2.5 kg*m/s. You

then apply and impulse of 10 N*s. What is the soccer

ball’s final momentum?

Given: pi = 2.5 kg*m/s

J = 10 N*s

Unknown: pf = ?

Equation: J = ∆p = (pf – pi)

Substitute: 10 N*s= pf – 2.5 kg*m/s

Solve: pf = 12.5 kg*m/s

UNIT 4

CONSERVATION OF

MOMENTUM

CONSERVATION OF MOMENTUM

m1v1i + m2v2i = m1v1f + m2v2f

Before collision After collision

ELASTIC COLLISIONSThe objects bounce off each other!



INELASTIC COLLISIONSThe objects stick together!


= (m1+m2) vf


EXAMPLE

A 0.014kg red bouncy ball is traveling at a velocity

of 1.2 m/s to the right. It collides elastically with a

initially stationary blue bouncy ball (mass of 0.011

kg). If the red ball has a final velocity of 0.5 m/s to

the left, what is the final velocity of the blue ball? G: m1 = 0.014kg

v1i = 1.2 m/s

v1f = -0.5 m/s

m2 = 0.011 kg

v2i = 0 m/s

U: v2f =?

E: m1v1i + m2v2i = m1v1f + m2v2f

S: v2f = 2.16 m/s

#5 FROM THE MOCK

EOC WITH YOUR

PARTNER

UNIT 4

WORK & POWER

WORK, W (Units of joules (J))

Work is ONLY done when a

force moves an object a

distance.

WORK, W (units of joules (J))

W = F (d) cosθ• W = work (N*m or J)

• F = Force (N)

• d = displacement (m)

• θ = angle between the force and displacement vectors (°)

d

F

d

F

d

F

θ = 0 θ = 90 θ = θ

θ

#32 FROM THE MOCK

EOC WITH YOUR

PARTNER

EXAMPLE

Your mom tells you to get to work cleaning your

room. Being the smart-alec you are and using your

knowledge of physics, you lift up one sock (mass =

0.04kg) straight up 1m. This is done at a constant

speed. How much work did you do?

G: Fg +Fa = m * ay

(0.04kg * -9.8) + Fa = 0

Fa = 0.392N

d = 1m

θ = 0

U: W =?

E: W = F*d*cosθ = 0.392*2*cos(0)

S: W = 0.392 J

YOU TRY

Your mom tells you to get to work cleaning your

room. Being the smart-alec you are and using your

knowledge of physics, you drag your laundry basket

with a force of 5N at an angle of 45° with the

horizontal for 5m. How much work do you do?

G: F = 5N

d = 5m

θ = 45 °

U: W = ? F = ?

E: W = F*d*cosθ = 5*5*cos(45)

S: W = 17.68 J

POWER, P (units of watts (W))

P = W/t • P = Power (J/s or W)

• W = work (N*m or J)

• t = time

UNIT 4

ENERGY

KINETIC ENERGY, KE

(Units of joules (J))

KE = ½mv2

Mass (in kg) Velocity (in m/s)

KE EXAMPLE

A constant 15,000 N force accelerates a 1700kg Maserati

from rest. What is the kinetic energy of the car after

traveling 50 m?

G: F = 15000N

d = 50m

m = 1700kg

vi = 0 m/s

U: KE = ? vf = ? a = ?

E: F = m*a a = 8.82 m/s2

a = (vf2 – vi2) / (2Δd) vf = 29.70 m/s

KE = ½ mv2

S: KE = 749776.5 J

WORK & KINETIC ENERGY


Wnet= ΔKE

Kinetic Energy (in J)

#22 FROM THE MOCK

EOC WITH YOUR

PARTNER

KE EXAMPLEA worker does 1000 J of work on a 100 kg box.

If the box loses 505 J of heat to the floor through the

friction between the box and the floor, what is the velocity

of the box after the work has been done on it?

G: Wnet= 1000-505J

m = 100kg

vi = 0 m/s

U : Wnet= ΔKE

E: Wnet = ½ mvf2 - ½ mvi

2

S: vf = 3.15 m/s

PE = mgh

Mass (in kg)+ 9.8 m/s2

Height off the

ground (in m)

GRAVITATIONAL POTENTIAL ENERGY, PEg


PEg EXAMPLE

A 10kg goat is in a tree, 15m above the ground. How much

potential energy does it have?

G: m = 10 kg

h = 15m

U: PE = ?

E: PE = mgh = 10*9.8*15

S: PE = 1470 J

KEi + PEi = KEf + PEf

CONSERVATION OF ENERGY

In a closed system, the total mechanical

energy before (initial) must be equal to the

total mechanical energy after (final).

#12, 29, 39 FROM

THE MOCK EOC WITH

YOUR PARTNER


Your friend is upstairs (10m above you) and calls for

you to toss them your 2kg physics textbook. How fast

do you have to throw it so it reaches them?


Your friend is upstairs (10m above you) and calls for

you to toss them your 2kg physics textbook. How fast

do you have to throw it so it reaches them?

G: m = 2kg

hi = 0m

hf = 10m

vf = 0m/s

U: vi = ?

E: KEi + PEi = KEf + PEf

½ mvi2 + mghi = ½ mvf

2 + mghf

½ (2)vi2 + (2)(9.8)(0) = ½ (2)(0)2 + (2)(9.8)(10)

S: vi = 14m/s


YOU TRY80kg Rapunzel rests on top of a 50m hill. What will

be the KE of Rapunzel be when she is 5m from the

bottom?


80kg Rapunzel rests on top of a 50m hill. What will

be the KE of the Rapunzel be when she is 5m from

the bottom?

G: m = 80kg

vi = 0m/s

hi = 50m

hf = 5m

U: KEf = ?


½ mvi2 + mghi = KEf + mghf

½ (80)(0)2 + (80)(9.8)(50) = KEf + (80)(9.8)(5)

S: KEf = 35,280 J


With what minimum speed must the cat leave the

top of the car in order to lift its mass up 0.75 m to

the roof with a speed of 3m/s?


With what minimum speed must the cat leave the

top of the car in order to lift its mass up 0.75 m to

the roof with a speed of 3m/s?

G: hi = 0m

hf = 0.75m

vf = 0.25m/s

U: vi = ?



2 + mghf


2 + mghf

½ vi2 + (9.8)(0) = ½ (3)2 + (9.8)(0.75)

S: vi = 4.87m/s

EXIT TICKET

5 question quiz

unit 4 - uplift education · unit 4 conservation of momentum. conservation of momentum m 1 v 1i + m...

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