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Momentum

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Tuesday – November 9, 2010
• Title: Collision: Sticky and non-sticky
• Purpose: Develop an understanding of collisions
and the impact of both mass and velocity on
momentum
– Why: Because collisions are vital to understanding everything
from car crashes to chemical reactions.
– GLEs: Science Systems, PS1 A-F

• To Do & Question:
– Pick up a collision worksheet.
– What is momentum?
Collisions
• Define each of the following:
– Sticky Collisions
• Occurs when two (or more) objects run into each
other and stay together as one mass
– Elastic Collisions
• Occurs when two (or more) objects run into each
other and separate after the collision
Momentum Equations
• Momentum is the product of mass and velocity
– Equation: r =

• When a collision occurs it can be:
– Non-sticky
Before      After
m1v1 + m2v2 = m1v1 + m2v2
– Sticky
Before      After
m1v1 + m2v2 = (m1 + m2)v2
Problem
• A basketball with a mass of 2.00kg, at
rest, is hit by a soccer ball with a mass of
2.50kg traveling with a velocity of 5.30
m/s. What is the velocity of the two balls if
the bounce off each other with equal
velocities?
Problem #2
• During an ice skating routine, the male
figure skater with a mass of 81.8 kg is
skating with a velocity of 4.3m/s. The
female skater, with a mass of 50.0 kg has
a velocity of 1.3 m/s in the opposite
direction. What is there combined velocity
after the collision?
Conclusion
• Reflection:
– What is the relationship between mass and
momentum?
– What is the relationship between velocity and
momentum?

• Homework:
– Complete the momentum worksheet
– ERP Research Outline is due FRIDAY
– Complete and submit all missing work
To Crash or Not Crash

The collision of KE, friction, and
normal force (opposite of weight)
Friction
• Friction is the force that opposes forward
motion.
– It also allows forward motion to occur.

• The coefficient of friction (m) describes
how much friction exists between two
objects.
To crash or not to crash
• A car with a weight of 3200 lb is traveling at a velocity of 34.0 m/s. If
the brakes have locked-up and the coefficient of friction between the
tires and road is 0.43, will the car be able to stop before reaching the

w = mg                            FfK = m FN
w = (1451kg)(-9.8 m/s)            FfK = (0.43)(14219.8 N)
w = -14219.8 N                    FfK = 6114.514 N

FN = -w
FN = - (-14219.8 N)
FN = 14219.8 N
A car with a weight of 3200 lb is traveling at a velocity of 34.0 m/s. If the
brakes have locked-up and the coefficient of friction between the tires
and road is 0.43, will the car be able to stop before reaching the
w = -14219.8 N                     KE = ½ mv2
FN = 14219.8 N                     KE = ½ (1451 kg)(34.0 m/s)2
FfK = 6114.514 N                   KE = ½ (1677356 kg m2/v2)
KE = 838678 J
W = KE
W = 838678 J

W = Fd
838678 J = (6114.514 N)d
137.161841481 m = d
140m = d
To crash or not to crash
• A car with a weight of 3200 lb is traveling at a velocity of 34.0 m/s. If
the brakes have locked-up and the coefficient of friction between the
tires and road is 0.43, will the car be able to stop before reaching the
• A truck with a mass of 44000 kg is
traveling at a velocity of 96.0 km/hr. If
the brakes have locked-up and the
coefficient of friction between the tires
and road is 0.57, will the ursidae 298
• A truck with a mass of 52000 kg is
traveling on wet roadway with a
coefficient of friction with the tires of
0.12. What is the minimum velocity of
the truck if there is a blind corner 230m
before a large red stain on the road?
There were Alcer alcer antlers found
near the stain.
Solve…
• At what maximum speed can a car safely
negotiate a horizontal unbanked turn (r =
51 m) in dry weather (m = 0.95) and in icy
weather (m = 0.10)?

• Hint:
FN = mg
Fc = (mv2)/r
Ff = mFN

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