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UNIT 4
UNIT 4
MOMENTUM & IMPULSE
IMPULSE-MOMENTUM THEOREM
Remember, ∆
means
final – initial
∆p = pf – pi
∆v = vf – vi
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.
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 COLLISIONS
The objects bounce off each other!
m1v1i + m2v2i
Before collision
=
m1v1f + m2v2f
After collision
INELASTIC COLLISIONS
The objects stick together!
m1v1i + m2v2i
Before collision
=
m1v1f + m2v2f
=
(m1+m2) vf
After collision
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 (°)
θ=0
F
d
θ = 90
θ=θ
F
F
d
θ
d
#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 =
2
½mv
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
(Units of joules (J))
Wnet= ΔKE
Kinetic Energy (in J)
#22 FROM THE MOCK
EOC WITH YOUR
PARTNER
KE EXAMPLE
A 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 mvi2
S: vf = 3.15 m/s
GRAVITATIONAL POTENTIAL ENERGY, PEg
(Units of joules (J))
PE = mgh
Mass (in kg)
+ 9.8 m/s2
Height off the
ground (in m)
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
CONSERVATION OF ENERGY
KEi + PEi = KEf + PEf
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
CONSERVATION OF ENERGY
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?
CONSERVATION OF ENERGY
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 mvf2 + mghf
½ (2)vi2 + (2)(9.8)(0) = ½ (2)(0)2 + (2)(9.8)(10)
S: vi = 14m/s
CONSERVATION OF ENERGY
YOU TRY
80kg Rapunzel rests on top of a 50m hill. What will
be the KE of Rapunzel be when she is 5m from the
bottom?
CONSERVATION OF ENERGY
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 = ?
E: KEi + PEi = KEf + PEf
½ mvi2 + mghi = KEf + mghf
½ (80)(0)2 + (80)(9.8)(50) = KEf + (80)(9.8)(5)
S: KEf = 35,280 J
CONSERVATION OF ENERGY
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?
CONSERVATION OF ENERGY
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 = ?
E: KEi + PEi = KEf + PEf
½ mvi2 + mghi = ½ mvf2 + mghf
½ mvi2 + mghi = ½ mvf2 + mghf
½ vi2 + (9.8)(0) = ½ (3)2 + (9.8)(0.75)
S: vi = 4.87m/s
EXIT TICKET
5 question quiz
EXIT TICKET
5 question quiz