s Compare the inertia of a semi truck and a roller skate Will the truck always have more inertia than the roller skate Will it always have more momentum than the roller skate WarmUp Estimate the amount of work the engine performed on a 1200kg car as it accelerated at 12 ms ID: 620121
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Slide1
Chapter 6 – Linear Momentum and Collisions
Compare the inertia of a semi truck and a roller skate…
Will the truck always have more inertia than the roller skate?
Will it always have more momentum than the roller skate?Slide2
Warm-Up
Estimate the amount of work the engine performed on a 1200-kg car as it accelerated at 1.2 m/s
2
over a 150-meter distance.
Slide3
Warm-Up Solution
m = 1200 kg
a = 1.2 m/s
2d = 150 m
F = maW = ∆KE=Fcosq
d W = ma×dW = (1200 kg)(1.2 m⁄s^2 )(cos0o)(150 m)W = 2.2x10^5 JSlide4
A gallon of gasoline contains about 1.3 x 10
8
joules of energy. A
2000-kg car traveling at 20 m/s skids to a stop. Estimate how much gasoline it will take to bring the car back to the original speed.
To
complicate matters further, consider the fact that only about 15% of the energy extracted from gasoline actually propels the car. The rest gets exhausted as heat and
unburnt
fuel.Slide5
Linear Momentum:
Inertia
in Motion
Linear momentum
of an object is equal to the product of its mass and velocitySI Units of momentum are kgm/sA lowercase p is used for the momentum of an individual particleSlide6
Total Linear Momentum of a System
The total linear momentum
of a system is the
vector sum
of the
momenta of the individual particles
The total linear momentum of a system, expressed by an uppercase , is used for the total momentum of the individual particles of the systemSlide7
Sum of Linear MomentumSlide8
Example 1: Which has more momentum:
a) a 1500 kg car moving at 25 m/s or
b) a 40 000 kg truck moving at 1.00 m/s?
Given:
mcar = 1500 kg vcar
= 25 m/s mtruck = 40 000 kg vtruck = 1 m/spcar=mcarvcar= (1500 kg)(25 m/s) = 3.75 x 10
4 kgm/sb) ptruck
= mtruckvtruck
=(40000kg)(1m/s)
=4.00 x 10
4
kgm
/s
The much slower truck has more momentum than the faster car because the truck has much greater mass.Slide9
Example 2: Two identical 1500 kg cars are moving perpendicular to each other. One is moving with a speed of 25.0 m/s due north, and the other is moving at 15.0 m/s due east. What is the total linear momentum of the two car system?
Given:
m
1
= m2 = 1500 kg v1
= 25.0 m/s v2 = 15.0 m/s
p1 = m
1 v1 = (1500kg)(25.0m/s) = 3.75 x 104kgm/s
p
2
= m
2
v
2
= (1500kg)(15.0m/s) = 2.25 x 10
4
kgm/s
Continued on next slide…Slide10
From vector addition, the magnitude of the total momentum is
And the direction is given bySlide11
Check Your Understanding
1. Determine the momentum of a ...
60-kg halfback moving eastward at 9 m/s.
1000-kg car moving northward at 20 m/s.
40-kg freshman moving southward at 2 m/s.2. A car possesses 20 000 units of momentum. What would be the car's new momentum if ...
its velocity were doubled. its velocity were tripled.its mass were doubled (by adding more passengers and a greater load)both its velocity were doubled and its mass were doubled.Slide12
Solutions
1. a) p = m*v = 60 kg*9 m/s
p = 540
kg•m
/s, east b) p = m*v = 1000 kg*20 m/s p = 20 000 kg•m/s, north
c) p = m*v = 40 kg*2 m/s p = 80 kg•m/s, southSlide13
Solutions
2. a)
p = 40 000 units
(doubling the velocity will double the momentum)
b) p = 60 000 units (tripling the velocity will triple the momentum)
c) p = 40 000 units (doubling the mass will double the momentum) d) p = 80 000 units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is doubled twice -quadrupled)Slide14
Newton’s Laws and Momentum
Newton’s second law, F=ma, can also be expressed in terms of momentum.
If the mass is constant, then
If an object’s momentum changes, then a force
must have acted upon it. Slide15
Practice
Solve the problems on “Chapter 4” Momentum and Energy”
Momentum p. 27 Worksheet Slide16
A 100.-kg crate is sliding down an plane inclined at an angle of 30. degrees. The coefficient of friction between the crate and the incline
is 0.30.
Determine the net force and acceleration of the crate.Slide17
Impulse and Momentum
Impulse is the change in momentum.
When a moving object stops, its impulse depends only on its change in momentum. This can be accomplished by a large force acting for a short time, or a smaller force acting for a longer time.Slide18
Impulse and the Bouncing Ball Experiment
http://www.youtube.com/watch?v=ciageLSYXUE
Predict what will happen when two identical balls are dropped from the same height onto the same type of surfaceSlide19
Practice
Solve the problems on
“Chapter 4” Momentum and Energy”
Impulse-Momentum p.31
Use the Rally Coach Method
to work through the Examples 3, 4, 5, and 6.Slide20
Example 3
A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck.Slide21
Solution to Example 3: A hockey player applies an average force of 80.0 N to a 0.25 kg hockey puck for a time of 0.10 seconds. Determine the impulse experienced by the hockey puck.
Given:
F=80.0 N m=0.25 kg t=0.10 s I =
F
DtSlide22
Example 4:
When bunting, a baseball player uses the bat to change both the speed and direction of the baseball.
(a) How will the magnitude of the momentum of the baseball before and after the bunt change? (Describe in words, not numbers)
(b) The baseball has a mass of 0.16 kg; its speeds before and after the bunt are 15m/s and 10 m/s respectively; the bunt lasts 0.025 s. What is the change in momentum of the baseball?
(c) What is the average force on the ball by the bat?Slide23
Solution to Example 4
(b) Choose the direction of motion before the bunt as positive.
v
=
10 m/s, vo = 15 m/s.
p = mv – mvo = (0.16 kg)(10 m/s) – (0.16 kg)(15 m/s) = 4.0 kgm/s in direction opposite v0(c) Slide24
Example 5
A boy catches—with bare hands and his arms rigidly extended—a 0.16-kg baseball coming directly toward him at a speed of 25 m/s. He emits an audible “Ouch!” because the ball stings his hands. He learns quickly to move his hands with the ball as he catches it.
If the contact time of the collision is increased from 3.5 ms to 8.5 ms in this way, how do the magnitudes of the average impulse forces compare?Slide25
Solution to Example 5
Given: m=0.16 kg, v
0
=25 m/s, v=0m/s, t1=3.5ms, t
2=8.5msFavg t =
mv mvo = mvo, So the magnitude isSlide26
Example 6
A karate student tries
not
to follow through in order to break a board. How can the abrupt stop of the hand (with no follow-through) generate so much force?
Assume that the hand has a mass of 0.35 kg and that the speeds of the hand just before and just after hitting the board are and 0, respectively. What is the average force exerted by the fist on the board if (a) the fist follows through, so the contact time is 3.0 ms, and (b) the fist stops abruptly, so the contact time is only 0.30 ms? Slide27
Solution to Example 6
By stopping, the contact time is short. From the impulse momentum theorem (
F
avg
t = p =
mv mvo), a shorter contact time will result in a greater force if all other factors (m, vo, v) remain the same.