Momentum

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Momentum

• Mr. Rodichok
• Regents Physics

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Momentum Ch 9
The product of mass velocity

• Momentum
• Formula
• Units
• Change in momentum
• Formula
• Units

Inertia in motion
p m v
kg m / s
Momentum is a vector. The change is added taking
into account direction.
?p m?v m ( vf vi)
kg m / s
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Momentum (contd)

• Ex 9-1 A 1000 kg car accelerates from rest to 30
  m/s. Find the following
• a. The cars initial momentum
• b. The cars final momentum
• c. The cars change in momentum

p m v
p 1000 kg ( 0 m/s)
p 0 kg m/s
m 1000 kg vi 0 m/s
p m v
p 1000 kg ( 30 m/s)
p 30,000 kg m/s
m 1000 kg vf 30 m/s
pi 0 kg m/s pf 30,000 kg m/s
?p pf - pi
?p 30,000 kg m/s 0 kg m/s
? p 30,000 kg m/s
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Impulse
A force acting during a specific interval of time

• Impulse
• Formula
• Units
• Ex 9-2 A ball is stopped by a 100 N force over a
  time period of 2 sec.
• Determine the impulse acting on the ball.

Impulse equals the change in momentum.
J F t ? p
N s kg m/s
F 100 N t 2 s
J F t
J (100 N) (2 s)
J 200 N s
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Impulse - Momentum Theorem
Impulse Momentum Theorem Ex 9-3 A ball is
stopped by a 100 N force over a time period of 2
sec. a) What was the balls change in momentum
in Example 9-2 above? b) What was the balls
velocity if its mass is 4 kg?
The impulse acting on an object is equal to the
objects change in momentum.
J F t ? p
200 N s ? p
J 200 N s
J 200 N s m 4 kg vi 0 m/s
J ? p m ? v
200 N s (4 kg) (vf 0 m/s)
vf 50 m/s
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Vector Addition

• Independence of Vectors (extremely important)
• Example 6-3 A boat is traveling 10 m/s East. It
  is on a river with a current of 5 m/s South.
  What is the boats resultant velocity (magnitude
  and direction) if a) drives upstream, b) drives
  downstream, c) crosses the river.

Vectors are separated into x y when solving
equations.
a) 5 m/s North
a) 15 m/s South
a) 15 m/s South
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Newton the apple

• Newtons 2nd Law in relation to gravity
•  
•  
• Direction
•  
•  Example 5-4

Using Newtons 2nd Law with acceleration due to
gravity weight can be calculated.
Since the acceleration is down (negative) the
force will also be down (negative).
Find the weight of a 35 kg mass.
m 35 kg a g -9.81 m/s2 Fg ?
Fg -343 N
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Newton and the apple (continued)

• Example 5-5
•   

Find the weight of a 35 kg mass is 100 N. What
is the acceleration rate due to gravity in that
location? Is it on the earth?
m 35 kg a g ? Fg 100 N
g 2.86 m/s2
No, this is not on the Earth.
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Changing Speed (continued)

• Example Problems for the 2005 Corvette Z51
• 1.  The corvette can start from rest and reach a
  speed of 27 m/s (60 mph) in 4.3 sec.   
• (a) What was the average acceleration of the car?
  
•  
•  
•  
•  (b) What is the average velocity of the car?

13.5 m/s
14 m/s
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Combined Kinematic Equations

• Uniformly Accelerated Motion
• Combined Kinematic Equations
• Depending on the given information different
  combinations of the following formulas will have
  to be used.
• Old Standbys
• If problem has the following four variables d,
  vi, a, and t as the givens and unknown use the
  following formula

d vi t ½ a t2
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Combined Kinematic Equations

• Example The corvette traveling 40 m/s (100 mph)
  applies its brakes and decelerates at a rate of 6
  m/s2 for 3 seconds after seeing a speed trap. How
  far did it travel while the brakes were applied?

vi 40 m/s d ? a - 6 m/s2 t 3 s
d vi t ½ a t2
d (40 m/s) (3 s) ½ (- 6 m/s2) (3 s)2
d 93 m
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Combined Kinematic Equations

• If problem has the following four variables d,
  vi, vf, and a as the givens and unknown use the
  following formula
• Example The corvette traveling at 30 m/s (70
  mph) has to quickly stop for a squirrel. The
  driver stops the car with an acceleration of -9
  m/s2. Find the distance covered while stopping
  the car.

vf2 vi2 2 a d
vi 30 m/s d ? a - 9 m/s2 vf 0 m/s
vf2 vi2 2 a d
(0 m/s)2 (30 m/s)2 2 (-9 m/s2) d
d 50 m