Transcript 1D Collisions - Berkner's Base for Physics and Math

Unit A Momentum
Objectives
 You will be able to state and apply the Law of
Conservation of Momentum to linear collisions.
 You will be able to define an isolated system, and be
able to determine if a system is or is not isolated.
Collisions
 There are two types of collisions we study:
1. Elastic Collisions – objects collide and bounce back
without deforming, there is no energy lost (pool balls)
Energy and momentum are conserved.
http://www.youtube.com/watch?v=pZqkaJDaz2A
Inelastic Collisions – objects collide but deform in some
way, losing energy (volleyball) Only momentum is
conserved
http://videosift.com/video/Golf-ball-deformation-at-70-000fps
Collisions and systems
 A collision is an interaction between two objects
where a force acts on each object for a period of time.
 A group of two or more objects that interact is known
as a system.
 A system where the system’s mass is constant, and no
external net force acts on the system is an isolated
system.
Isolated System
 Matter, energy and information do not enter or leave
the system
 - the system’s total energy and mass stay the same
 There is no such thing as an isolated system in real life,
it is completely theoretical (it is the model we use in
Physics 30)
Mometum is conserved in an
isolated system
 The sum of the momentum before an event is the
same as the sum of the momentum after
Step 1: Determine if the system is isolated, if it is
you can use the conservation of momentum
Step 2: Write the conservation statement.
Step 3: Note the initial momentum is zero because
there is no movement
Step 4: Sub in the values for the momentum and
watch your negatives.
Law of Conservation of Momentum.


total p before collision  total p after collision
1)





mAv A  mBvB  mAvA  mBvB
Use when objects do not stick together.
2)


mAv A  mBvB  mA  mB  v
Use when objects stick together.
3)
mAv A  mBvB
Use for explosions (momentum total is zero before).
Note: These formulas are not on your formula sheets and
you can use the first formula exclusively if you like.
*
Using “Conservation of Momentum”
Colliding Objects Stick Together
Example : A 1.50 g pellet is fired into a 12.3 g wood block. The
block and imbedded pellet fly off at 2.78 m/s E.
What was the pellet’s velocity before impact?


mAv A  mBvB  mA  mB  v

0.00150 kg  vA 
0  0.00150 kg  0.0123 kg  2.78 m
s
0.00150 kg  vA 
0  0.038364 kg  m
s

v p  25.6 m E

s
Answer is
“+”
*
Example : A 980 kg Toyota going at 52.8 km/h N collides with
and becomes entangled with a 738 kg Honda going
79.3 km/h S. What is the velocity of the wreckage
immediately after impact?
Note: It does not matter what units of mass or velocity are
used - just be consistent!


mAv A  mBvB  mA  mB  v
980 kg 
 52.8 kmh  738 kg  79.3 kmh   980 kg  738 kg  v'
 51744   58523

 1718 v'

v '  3.95 km
S
h
Answer was “–”
*
Example
 A 50 g bullet is fired into a 650 g block of wood, which
sits on a frictionless surface. After impact the bullet
and block move right at 30 m/s. Find the bullet’s
velocity just before impact.
Solution
 Before:
mB  0.050kg
 After
mBW  mB  mW  0.700kg
vBi  ?
vBW f  30m / s
mW  0.650kg
Because wood and
vWi  zero
bullet move as a single
 pWi  zero
object.
Solution
 0.42km / s [ R]
“Explosion” Problems
Example : When a 960 g plate is dropped, the first 370 g piece
flies off at 2.63 m/s S. What is the velocity of the
second piece?


mv  mv
370 g  2.63 ms   590 g  v

v  1.65
m
s
N
Note: In order for the total final momentum to be “0”, if
one piece flies north, the other must fly south.
*
Example:
 Fred sits in a stationary, frictionless wheelchair and
throws a 3.0 kg ball outward at 8.0 m/s. If the mass of
Fred and his chair is 75 kg, what will be the velocity of
Fred and the chair immediately after the ball is
thrown?
Solution
 Before:
 After
mB  3.0kg
mB  3.0kg
vB  zero
vB f  8.0m / s
mF  75kg
mF  75kg
vF  zero
psys  zero
Because both velocities are zero.
vFf  ??
Solution
 0.32m / s [ R]
http://www.coolschool.ca/content/junk/showcase/content/physics/crash.swf
Example
 A 0.25-kg volleyball is flying west at 2.0 m/s when it
strikes a stationary 0.58-kg basketball dead centre. The
volleyball rebounds east at 0.79 m/s. What will be the
velocity of the basketball immediately after impact?
(Textbook p. 478 #1)
Is the system isolated?
Write the conservation statement. Plug and chug.
(watch your negatives)
Example 3
 A 125-kg bighorn ram butts heads with a younger 122-
kg ram during mating season. The older ram is rushing
north at 8.50 m/s immediately before collision, and
bounces back at 0.11 m/s [S]. If the younger ram moves
at 0.22 m/s [N] immediately after collision, what was
its velocity just before impact? (Textbook p. 479 #2)
Assignment:
 Workbook p. 51 #2 – 5, 8
 Reading from the textbook: p. 468-485
Slo Mo for fun
 http://www.youtube.com/watch?NR=1&feature=endsc
reen&v=rU7iYYpSrlo