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Physics 30 Unit 1 – Conservation Laws
Momentum and Impulse
Momentum (p) is the product of mass and velocity.
p = mv
p = momentum (kg.m/s)
m = mass (kg)
v = velocity (m/s)
Ex: A car’s mass is 1200 kg traveling east at 30 m/s.
Find its momentum.
p = mv
= (1200)(30)
= 36,000 kg.m/s
= 3.6 x 104 kg.m/s, East
*Significant Digits are 2 due to 30 m/s, lowest number of
digits*
*Direction is given*
Change in momentum (p) is the product of mass and
change in velocity.
p= mv

p = Change in momentum (kg.m/s)
m = mass (kg)
v = change in velocity (m/s)
v = vf - vi
Ex: An object has a mass of 3.0 kg. Its velocity changed
from 5.0 m/s to 15 m/s. Find the change in momentum.
p = mv
= (3.0)(10)
= 30 kg.m/s
Impulse is the product of force and time
Impulse = Ft
Impulse (N.s)
F = force (N)
t = time (s)
Impulse = Change in momentum
Ft = mv
Ex: A force of 20 N acts on a 10 kg object for 2.0 s.
(a) Find the impulse exerted.
(b) Find the change in momentum
(c) Find the change in velocity
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2
(a) Impulse = Ft
= 20 x 2.0
= 40 N.s
(b) Impulse=Change in momentum
= 40 N.s or kg.m/s
(c)
p = mv
40 =10(Δv)
Δv = 4.0 m/s
Impulse = Change in momentum
Impulse = p = Ft = mv
Ex: A force of 50 N acts on a 10 kg object for 2.0 s. Find:
a)
the impulse
b)
the change in momentum
c)
the change in velocity
a) Impulse = Ft
= 50 x 2.0
= 1.0 x 102 N.s
b) p = Impulse
1.0 x 102 N.s
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3
c) Impulse = p
Ft = mv
100 = 10 (v)
v = 10 m/s
Example
A person throws a 0.20 kg ball against a wall at a speed of 20
m/s. It rebounds with a speed of 15 m/s. Find the impulse
exerted by the wall on the ball.
Impulse=p=mv=0.20(-15-20)=
=0.20(-35)= -7.0 kg.m/s
Summary of formulas:
p = mv = Ft
p = mv
Impulse = mv = Ft
Impulse and momentum are vector quantities.
A direction is therefore needed.
Weight to mass - divide by 9.81
Mass to weight - multiply by 9.81
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4
A 4 kg mass is dropped from a height of 10 m. How far did the
object drop when its momentum is 8 kg m/s?
m = 4 kg
vi = 0
vf = 2 m/s
a = 9.81 m/s2
d=?
p = mv
vf = 8/4 = 2 m/s
d = vf2 – vi2
2a
d = 0.2 m
http://www.youtube.com/watch?v=OrLcZNG0N0I
p. 449 – Concept check
p. 451 #1, 2 p. 452 #1, 2 p. 453 #9 -11, 14
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5
Law of Conservation of Momentum
Linear Collisions
In an isolated system, the sum of momentum before a
collision is equal to the sum of momentum after the
collision.
*Isolated – nothing is coming in and nothing is leaving*
Friction must be negligible.
A system is a collection of two or more objects. An isolated
system is a system that is free from the influence of a net
external force that alters the momentum of the system. There
are two criteria for the presence of a net external force; it must
be...
a force that originates from a source other than the
two objects of the system
e.g. friction


a force that is not balanced by other forces.
A system in which the only forces that contribute to the
momentum change of an individual object are the forces acting
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between the objects themselves can be considered an isolated
system.
Consider the collision of two balls on the billiards table. The
collision occurs in an isolated system as long as friction is small
enough that its influence upon the momentum of the billiard
balls can be neglected. If so, then the only unbalanced forces
acting upon the two balls are the contact forces that they apply
to one another. These two forces are considered internal forces
since they result from a source within the system - that source
being the contact of the two balls. For such a collision, total
system momentum is
conserved.
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p before collision = p after collision
or
m1v1 + m2v2 + … = m1v1' + m2v2' + …
' means, “prime”
m1, & m2 = mass of the first and second object (kg)
v1, & v2 = velocity of the first and second object (m/s)
before collision
v1' & v2' = velocity of the first and second object after
collision (m/s)
*is the sum of…*
Inelastic Collisions – Locking on Contact
If the objects lock together after the collision the above
equation can be written as:
m1v1 + m2v2 = mcvc
vc = combined velocity
mc = combined mass
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This type of a collision is called inelastic collision.
Ex-1: A 200 g object traveling at 30 cm/s collides with a 400
g object traveling in the same direction at 10 cm/s. After the
collision the 200 g object is traveling at 25 cm/s. Find the
velocity of the 400 g object.
m1 = 200g
m2 = 400g
v1 = 30 cm/s
v2 = 10 cm/s
v1' = 25 cm/s
v2' = ?
m1v1 + m2v2 = m1v1' + m2v2'
200 (30 ) + 400 (10) = 200(25) + 400 v2'
6000 + 4000 = 5000 + 400v2'
5000 = 400v2
v2' = 12.5 cm/s
The velocity of the 400 g object is 13 cm/s in the original
direction after collision.
Ex-2: A 0.25 kg steel ball is traveling east at a velocity of 4.5
m/s when it collides head on with a 0.30 kg steel ball,
traveling west at a velocity of 5.0 m/s. After the collision the
0.25 kg ball is traveling west at a velocity of 2.0 m/s. What is
the velocity of the 0.30 kg ball after the collision?
m1 = 0.25 kg
m2 = 0.30 kg
m1v1 + m2v2 = m1v1' + m2v2'
0.25 (4.5) + 0.30 (-5.0 ) = 0.25(-2.0) + 0.30v2'
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v1 = 4.5 m/s
v2 = -5.0 m/s
v1' = -2.0 m/s
v2' = ?
1.125 + (-1.5 ) = -0.5 + 0.30v2'
0.125 = 0.30v2'
v2' = 0.42 m/s
The velocity of the 0.30 kg object is 0.42 m/s east after
collision.
Ex-3: A 1.1x103 kg car traveling east at a velocity of 25
km/h collides head on with a 1.3 x 103 kg car-traveling west
at velocity of 15 km/h. During the collision the 2 cars lock
together. What is the velocity of the locked cars as they
move together immediately after the collision?
m1 = 1.1x103 kg
m2 = 1.3x103 kg
v1 = 25 km/h
v2 = -15 km/h
vc = ?
m1v1 + m2v2 = mcvc
1.1x103(25) + 1.3x103 (-15) =(2.4x103)vc
vc = 3.3 km/h
**positive answer means East**
The velocity of both cars after the collision was 3.3 km/h East.
Ex-4: A 0.10 kg bullet is shot into a stationary 5.0 kg block.
The bullet and the block travel at 8.0 m/s after impact.
Find the velocity of the bullet.
m1 = 0.10 kg
m1v1 + m2v2 = mcvc
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m2 = 5.0 kg
vc = 8.0 m/s
mc = 5.1 kg
v1 = ?
0.10v1 = (5.1)(8.0)
v1 = 408 m/s
Law of Conservation of Momentum – Special Case
Objects initially at rest go off in opposite directions in a
straight line.
mcvc = m1v1 + m2v2
Formula can be reduced to:
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Where
vc = 0
An internal force causes the objects to move in opposite
directions. This follows from Newton’s Third Law of
Motion: For every action there is an equal and opposite
reaction.
spring
Ex:
m1 = 4.0 kg
m2 = 2.5 kg
vc = 0
v2 = 2.0 m/s
v1 = ?
4.0 kg
2.5 kg
v1 =?
2.0 m/s
mcvc = m1v1 + m2v2
0 = 4.0v1 + 2.5 (2.0)
v1 = -1.25 m/s
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Ex: A 100 kg mass explodes and in the process it breaks into
two pieces. A 40 kg piece travels west at 25 m/s. Find the
velocity of the other mass produced in the explosion.
m1 = 40 kg
m2 = 60 kg
vc = 0.0 m/s
mcvc = m1v1 + m2v2
0 = 40 (-25) + 60v2
1000 = 60v2
v2 = 16.6 m/s = 17 m/s
v1 = - 25 m/s
v2 = ?
The velocity of the 60 kg mass is 17 m/s east.
****Ex: A 50 g bullet is fired from a 5.0 kg gun. If the
velocity of the bullet is 275 m/s, what is the recoil velocity of
the gun?
m1 = 0.050 kg
m2 = 5.0 kg
vc = 0
v1 = 275 m/s
mcvc = m1v1 + m2v2
0 = (0.050)(275) + 5.0v2
v2 = - 2.75 m/s
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v2 = ?
Assignment:
p.451 #1,2 p.452#1,2 p.453 #14
p.458 #1 p.478 #1,2 p.482 #1,2
p.484 #1,2
Perpendicular collisions – inelastic collisions
Occur when two masses collide at 900 to each other.
Ex-1: A 0.25 kg steel ball is traveling east at a velocity of 4.5
m/s when it collides head on with a 0.30 kg steel ball
traveling south at a velocity of 5.0 m/s. After the collision
they stick together and travel with a common velocity. Find
the common velocity.
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Step 1 – sketch the collision.
p1
p2
normally drawn as
θ
pr
p2
p1
pr
Error!
Step 2 - find the momentum of each before the collision.
Reminder: only momentum is conserved.
m1 = 0.25 kg
v1 = 4.5 m/s
p1 = 1.125 kg.m/s
m2 = 0.30 kg
v2 = 5.0 m/s
p2 = 1.5 kg.m/s
*The red line is the resultant, to find it, use the Pythagoras
Theorem*
*The angle between the resultant and the first direction
must also be found (use Inverse Tan)*
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The Resultant
Angle at point of collision
c2 = a2 + b2
tan θ = Opp/Adj
tan θ = 1.5 / 1.125
θ = 53°
p c2 = (1.125)2 + (1.5)2
pc = 1.9 kg m/s
pc = 1.9 kg .m/s 53° S of E
Now the velocity…
pc = mcvc
1.9 = 0.55vc
vc = 3.5 m/s
The resultant velocity is 3.4 m/s 53° S of E
Ex-2: A 0.30 kg ball is traveling west at a velocity of 5.0 m/s
when it collides head on with a 0.40 kg ball traveling north
at a velocity of 3.0 m/s. After collision they stick together
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and travel with a common velocity. Find the common
velocity.
pc
p2
p1
m1 = 0.30 kg
v1 = 5.0 m/s
m2 = 0.40 kg
v2 = 3.0 m/s
p1 = m1v1
= (0.30)(5.0)
= 1.5 kg m/s
p2 = m2v2
= (0.40)(3.0)
= 1.2 kg m/s
pc2 = p12 + p22
pc = 1.92 kg m/s
vc = pc/mc = 1.92/0.70
vc =2.7 m/s
tan Ө = 1.2/1.5
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Ө = 39o
The resultant velocity is 2.7 m/s at 390 N of W
Assignment: p. 492 #1, 2
p. 503 #1
p. 504 #41 - 43
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Ex: Two cars collide at an intersection. The first car has a
mass of 775 kg and was traveling west. The second car has
a mass of 1125 kg and was traveling north. Immediately
after impact, the first car has a velocity of 65.0 km/h
330 W of N, while the second car had a velocity of 42.0 km/h
460 W of N. What were the velocities of the cars
immediately before collision?
m2 =
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 p1 = mv1
= 775 kg (65.0 km/h)
= 50375 kg.km/h, 33.0° West of North
 p2 = mv2
= 1125 kg (42 km/h)
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20
= 47250 kg.km/h, 46° West of North
Px
Px
Py
Py
*With the resultant known, momentum of each car going each
way needs to be found, finding the x and y component*
Momentum
X-component Y-component
Sin 33° = Px / Cos 33° = Py /
P1 = 50375
50375
50375
kg.km/h
= 27436.19
= 42248.03
kg.km/h
kg.km/h
Sin 46° = Px / Cos 46° = Py /
P2 = 47250
47250
47250
kg.km/h
= 33988.8
= 32822.61
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Total
kg.km/h
61424.99 kg
.km/h
kg.km/h
75070.64 kg
.km/h
*All the horizontal momentum comes from the first car*
*All the vertical momentum comes from the second car*
First Car
Second Car
p1 = m1v1
p2= m2v2
61424.00 kg km/h =775 kg v
75070.64 = 1125 kg v2
=79.3 km/h, West
= 66.7 km/h, North
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Glancing Collisions
Ex: A 5.0 kg object is moving east at a velocity
3.0 m/s when it collides with a 3.0 kg stationary object.
After collision, the 3.0 kg is traveling 30° South of East and
the 5.0 kg travels 60° North of East. What is the velocity of
each object after collision?
5.0
kg
5.0
kg
3.0
kg
60°
30°
0.0 m/s
3.0
 3.0 m/s
kg






PBefore Collision = m1v1 + m2v2
= (5.0) (3.0 )
= 15 kg m/s = PAfter Collision = 15 kg m/s
*By definition of Conservation of Momentum*
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*Momentum of each side (Right Triangle so cos can be used)*
cos 60° = PA/15
PA = mv1
7.5 = 5.0v
13 kg m/s = 3.0v
v = 1.5 m/s, 60° N of E
v = 4.3 m/s, 30° South of East
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Example
A 2.0 kg object traveling east at 6.0 m/s collides with
another 2.0 kg stationary object. After collision, the first
object travels at 40o north of east and the second object
travels 50 o S of E. Find the speed of each object after
collision.
v1’=4.6 m/s v2’=3.9 m/s
Ex: A 4.0 kg object is moving east at an unknown velocity
when it collides with a 6.1 kg stationary object. After the
collision, the 4.0 kg object is traveling at a velocity of 2.8 m/s
32 North of East and the 6.1 kg object is traveling at a
velocity of 1.5 m/s 41 South of East. What was the velocity
of the 4.0 kg object before collision?
4.0
kg
? m/s
6.1
kg
0.0 m/s
- -
4.0
kg
2.8 m/s
6.1
kg
1.5 m/s
32°
41°
25
PA = mv1’
= 4.0 kg (2.8 m/s)
= 11.2 kg m/s
PB = mv2’
= 6.1 kg (1.5 m/s)
= 9.15 kg m/s
*32 + 41 – 180 = 107*
c2 = a2 + b2 – 2abCos C
(cosine law)
c2 = (11.2)2 + (9.15)2 – 2(11.2)(9.15Cos (107))
c2 = 269.09 kg m/s
c = 16.4 kg m/s
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The law of cosines is used because the triangle is not a right
triangle.
P = mv
16.4 kg m/s = 4.0v
v = 4.1 m/s
The velocity of the 4.0 kg object before collision was 4.1 m/s,
East.
Ex: A 15 kg object is moving east at a velocity of 7.0 m/s
when it collides with a 10.0 kg stationary object. After the
collision the 15.0 kg object is traveling at a velocity of 4.2
m/s 20.0 South of East. Find the velocity of the 10 kg
object after collision.
15.0
kg
10.0
kg

7.0 m/s
0.0 m/s



PBefore Collision = m1v1 + m2v2
= 15.0(7.0) + 10.0(0.0)
- -
10.0
kg
? m/s
15.0
kg
4.2 m/s
?
20°
27
= 105 kg. m/s = P After Collision
c2 = a2 + b2 – 2abCosC
a/Sin A = b/Sin B
c2 = 1052 + 632 – 2(105)(63)Cos(20)
50.6/Sin 20 = 63.0/Sin B
B = 25.2
c2 = 2561.87
= 50.61 kg m/s
P = mv
50.6 kg m/s = 10v
v = 5.06 m/s
The velocity of the 10.0 kg mass is 5.1 m/s, 25North of East
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Case 5: Objects initially at rest go off in different directions
after an explosion.
Ex: An object explodes into three equal masses. One mass
moves east at a velocity of 15.0 m/s. If a second mass moves
at a velocity of 10.0 m/s 45.0° South of East, what is the
velocity of the third mass?
*Assume the masses to be 1.0 kg*
Momentum of the first piece
= m1v1
= 1.00 kg (15.0 m/s)
= 15.0 kg m/s
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29
Momentum of the second piece
= m2v2
= 1.00 kg (10.0 m/s)
= 10.0 kg m/s
Sum of momentum of the first two pieces =
Momentum
Px (kg.m/s)
Py (kg.m/s)
(kg m/s)
15.0
15.0
0
Cos 45° =
Sin 45° =
10.0
Px/10
Py/10.0
= 7.07
=-7.07
Total
22.07
-7.07
22.07
7.07
c2 = a2 + b2
c2 = (22.07 kg m/s)2 + (7.07 kg m/s)2
c = 23.17 kg m/s
tan = 7.07 kg m/s / 22.07 kg m/s
= 17.7° South of East
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30
The resultant momentum of the two pieces is 23.2 kg.m/s
17.8° South of East. The momentum of the third piece has
to be exactly the opposite so that they add up to 0.
Momentum of the third piece is 23.2 kg m/s 17.8°North of
West
p = mv
23.2 kg m/s = 1.00v
v = 23.2 m/s at 17.8°, North of West
Assignment (Pearson)
p. 458 #1, 2
p. 467 #9, 10, 12
p. 476, 477, 478, 479, 484, 491 #1, 2
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31
Work, Energy and Power
In physics work is done when an object moves in
the direction of the force applied.
W = Fd
W = Work (N·m or J)
F = Force (N)
d = Displacement (m)
Ex: A 200 g object is lifted through a height of 2.0
m. Find the work done.
W = Fd
= (0.200) (9.81) (2.0)
= 3.9 J
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32
Work done at an angle
W = Fdcosθ
Fa
Fa = applied
Ex: An object is pulled by a force of 100 N at an angle of
60˚ a distance of 2.0 m. Find the work done.
W = Fdcosθ
= (100)(2) cos 60˚
= 1.0 x 102 J
Work done is the area
under the graph.
F (N)
A=W
d (m)
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Power - is the rate at which work is done.
P = W/t
W = Work (J)
t = Time (s)
P = Power (J/s or Watt-W)
Ex: A 200 kg mass is lifted by a machine through a
distance of 2.0 m in 50 s. Find the power.
P = W/t
= (200 kg x 9.81 m/s2 x 2.0 m) / 50 s
= 78 W
Energy – the capacity to do work.
Energy Consumed = Work Done
Forms of Energy:
1. Mechanical Energy
a)
Potential energy is the energy possessed
by a body by virtue of its position or
condition
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Gravitational Potential Energy –energy due to
gravity
-Energy is taken with respect to a reference
point
Ep = mgh
Ep = Potential Energy (J)
m = mass (kg)
g = 9.81 m/s2
h = height (m)
b. Kinetic Energy – the energy possessed by
a body due to its movement
Ek = 1/2mv2
Ek = Kinetic Energy (J)
m = Mass (kg)
v = Velocity (m/s)
Total Mechanical Energy = Ep + Ek
2. Elastic Potential Energy – a stretched spring
Hooke’s Law
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35
Hooke’s Law - The extension of a spring is
proportional to the force applied.
Non-stretched spring
F
x
x = amount of
stretch in m
F = - kx
k = spring constant in N/m
Ep = 1/2kx2
Ep = Potential Energy (J)
k = Spring Constant (N/m)
x = Extension/Compression (m)
1. A spring (k = 200 N/m) is extended 5.0 cm. What
is its elastic potential energy?
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36
Ep = 1/2kx2
= 1/2(200)(5.0 x 10-2)2
= 0.25 J
2. A 60 kg person goes up an escalator 20 m long.
The escalator makes an angle of 20o with the
horizontal. Find the work done against gravity.
20 m
d
(vertical height)
d = sin 20 (20)
W= Fd = 60(9.81)(6.84)=4.0x103 J
Law of Conservation of Energy
Energy cannot be created or destroyed but can be
converted from one form into another.
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37
Energy Conversions
Mechanical energy is the sum of potential energy
and kinetic energy.
Mechanical Energy = Ek + Ep
Mechanical Energy = 1/2mv2 + mgh
Conversion between gravitational potential energy and
kinetic energy
100 m
60 m
0m
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38
Find the potential, kinetic and total mechanical energy
at 100 m, 60 m, and 0 m if the 5 kg is dropped from a
height of 100 m.
At 100 m:
PE = mgh = (5)(9.81)(100) = 4905 J
KE = 0
ET = 4905 J
At 60 m:
At 0 m:
PE = (5)(9.81)(60) = 2943 J
KE = 1962 J
ET = 4905 J
PE = 0
KE = 4905 J
ET = 4905 J
An object of mass 2.0 kg is dropped from a
height of 20 m. Find the speed of the object just as it
reaches the ground.
mgh = 1/2mv2
PE = KE
v= 2gh
v = 20 m/s
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39
A ball is thrown upward with
a speed of 30 m/s. How high will it rise?
KE = PE
mgh = 1/2mv2
h = 45.9 m
A 5.00 kg object is dropped from a height above the
ground. When the object is 4.00 m from the ground, it
has a speed of 9.00 m/s. The potential energy of the
object is chosen to be zero at ground level and the
effects of air resistance are ignored.
(a) What is the total mechanical energy of the
object?
(b) What was the initial height of the object?
(c) What is the speed of the object just as it
reaches the ground?
a) ET = PE + KE = mgh + 1/2mv2 = 400 J
b) PE = mgh
h = 400/(5x9.81) = 8.2 m
c) v2 = 2KE/m
v = 12.6 m/s
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40
Roller-coaster problems
A 1000 kg roller coaster, with its passengers, starts
from rest at point A on a frictionless track whose
profile is shown in the diagram.
(a) What is its maximum speed?
Max. speed will occur at the lowest point i.e. at D
It drops 9.0 m to reach D. Thus,
PE = KE
mgh = 1/2mv2
v = 13.3 m/s
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41
(b) With what speed does the roller coaster
arrive at point E?
At E it drops a distance of 4 m so the corresponding
PE will change to KE.
v= 8.86 m/s
(c)What constant braking force would have to be applied
to the roller coaster at point E, to bring it to rest in a
horizontal distance of 5.0m?
W = KE = Fd
F = W/d
F = - 7850 N
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42
A 250 kg roller-coaster car starts at position 1
and continues through the entire track. The brakes
are applied after the car passes position 4.Assume
that the effects of friction are negligible on the
roller coaster.
(a) What is the roller coaster’s speed at position 4?
v = 17.2 m/s
(b)After the car passes position 4, the brakes stop the
car in 3.03 s. What is the magnitude of the average
frictional force applied by the brakes
to stop the car?
Ft = mΔv
F = -1420 N
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43
Example
A 30 kg girl goes down a slide at an amusement
park, reaching the bottom with a velocity of 2.5
m/s. The slide is 10.0 m long and the top end is 4.0
m above the bottom end, measured vertically.
(a)What is her gravitational potential energy at the
top of the slide, relative to the bottom?
PE=1177 J
(b)What is her kinetic energy when she reaches the
bottom?
KE = 1/2mv2 = 94 J
(c) How much energy is lost due to friction?
Ef = 1177 – 94 = 1083 J
(d) Calculate the average frictional force acting on
her as she goes down the slide?
W = Fd
F = 1083/10 = - 108 N
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Law of Conservation of Energy
Archery Problems
Work=Energy
Ex: An archer puts a 0.25 kg arrow to the
bowstring. An average force of 200 N is used to
draw the string back 1.2 m.
a)
Assuming no frictional loss, at what speed
does the arrow leave the bow?
W = Ek
Fd = 1/2mv2
200 (1.2) = 1/2(0.25 )v2
= 44 m/s
b)
If the arrow is shot straight up, how high
does it rise?
W = Ep
Fd = mgh
200 (1.2) = 0.25 (9.81)h
= 98 m
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Hooke’s Law
Hooke’s Law - The extension of a spring is
proportional to the force applied.
Non-stretched spring
F
x
x = amount of
stretch in m
F = - kx
k = spring constant in N/m
Ep = 1/2kx2
Ep = Potential Energy (J)
k = Spring Constant (N/m)
x = Extension/Compression (m)
http://webphysics.davidson.edu/applets/animator4/demo_hook.html
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1. A spring (k = 200 N/m) is extended 5.0 cm. What
is its elastic potential energy?
Ep = 1/2kx2
= 1/2(200)(5.0 x 10-2)2
= 0.25 J
2. A linear spring can be compressed 10.0 cm by an
applied force of 5.0 N. A 4.5 kg crate of apples,
moving at 2.0 m/s, collides with the spring. What
will be the maximum compression of the spring?
Ek  EP(e)
1/2mv2 = 1/2kx2
k = F/x = 5.0 /(10.0 x 10-2 ) = 50 N/m
(4.5)(2.0 )2 = (50)x2
x = 0.60 m
3. A ball bearing of mass 50 g is sitting on a vertical
spring (k=120 N/m). By how much must the spring
be compressed so that, when released, the ball rises
to a maximum height of 3.1 m above its release
position?
Ep(e)  Ep(g)
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1/2kx2 = mgh
1/2(120)x2 = (5.0 x 10-2)(9.81)(3.1)
x = 0.16 m
4. A 3.0 kg ball is dropped from a height of
0.80 m onto a vertical spring of force constant
1200 N/m. What is the maximum compression of
the spring?
Ep(g)  Ep(e)
mgh = 1/2kx2
3.0 (9.81)(0.80) = 1/2(1200)x2
= 0.20 m
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Law of Conservation of Energy
Archery Problems
Work=Energy
Ex: An archer puts a 0.25 kg arrow to the
bowstring. An average force of 200 N is used to
draw the string back 1.2 m.
b)
Assuming no frictional loss, at what speed
does the arrow leave the bow?
W = Ek
Fd = 1/2mv2
200 (1.2) = 1/2(0.25 )v2
v = 44 m/s
b)
If the arrow is shot straight up, how high
does it rise?
W = Ep
Fd = mgh
200 (1.2) = 0.25 (9.81)h
h = 98 m
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Pendulum Type Problems
Ex: A pendulum is dropped from a height of 0.25 m
above the equilibrium position. What is the speed
of the pendulum as it passes through the
equilibrium position (maximum speed)?
(Ep + Ek)top= (Ep + Ek)bottom
mgh + 0.0 = 0.0 + 1/2mv2
(9.81) (0.25) = 1/2v2
v = 2.2 m/s
Ex: A pendulum is 1.20 m long. What is the speed
of the pendulum bob as it passes through the
equilibrium position if it is pulled aside until it
makes an angle of 25 .0° to the vertical position?
cos 25° = x/1.20
x = 1.0875 m
Height = 1.20 – 1.0875
h = 0.1125 m
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(Ep + Ek)top = (Ep + Ek)bottom
mgh + 0.0 = 0.0 + 1/2mv2
9.81(0.1125) = 1/2v2
v = 1.49 m/s
Elastic and Inelastic Collisions
An elastic collision is one in which both momentum
and kinetic energy is conserved.
Elastic collisions occur at the atomic level. In
ordinary collisions, kinetic energy is converted into
sound, heat, and light.
Inelastic collisions are collisions in which only
momentum is conserved.
Ex: A 5.0 kg metal ball at rest is hit by a metal ball
(1.0 kg) moving at 4.0 m/s. The 5.0 kg ball moves
forward at 1.0 m/s and the 1.0 kg ball bounces back
at 2.0 m/s. Is this an elastic collision? Prove your
answer.
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In an elastic collision, kinetic energy is conserved.
Ek1 + Ek2 = Ek1' + Ek2' (true if collision is elastic)
before
after
So, if KEbefore is equal to KEafter then the collision is
elastic.
KEbefore = 1/2m1v12 + 1/2m2v22
1/2(5.0)(0)2 + 1/2(1.0)(4.0)2 = 8.0 J
KEafter = 1/2(5.0 )(1.0)2 + 1/2(1.0 )(2.0)2 = 4.5 J
Kinetic energy before collision (8.0 J) is not equal
to kinetic energy after collision (4.5 J).
Thus, collision is inelastic due to loss of energy in
the form of heat (friction), light, and sound.
Example
A 6.0 kg ball of putty moving at 10.0 m/s runs
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head on into another 6.0 kg ball of putty at rest
They stick together and move ahead at 5.0 m/s. Is
this an elastic collision? Show calculations
KEbefore = 300 J
KEafter = 150 J
Thus, the collision is inelastic.
Ballistic pendulum problems
The ballistic pendulum is a device used to measure the
speed of a projectile, such as a bullet. The projectile is
fired into a large block (of wood or other material),
which is suspended like a pendulum. As a result of the
collision, the pendulum and projectile together swing
up to a maximum height. By using the conservation
laws, the speed of the projectile can be found.
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mgh = 1/2mv2c
(using law of conser. of energy)
h
m1v1 + m2v2 = mcvc
v2 = 0
(law of conser. of momentum)
A bullet’s speed may be determined by firing
it into a sandbag pendulum and measuring
the vertical height to which it rises. The bullet stays
in the sandbag.
(a)What is the change in the gravitational potential
energy of the sandbag and bullet during the
swing?
(b)What is the velocity of the sandbag-bullet
combination at the start of the swing?
(c) What was the original velocity of the bullet?
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(d) Is the collision between the bullet and the
sandbag elastic or inelastic? Justify your answer.
Assignment
1.A chronograph is a sophisticated device
commonly
used by technicians to find the speed of a bullet
fired from a gun. Before this technology was
developed, the speed of a bullet was determined
using different methods. A simplification of one
method of determining the speed of a bullet
requires
shooting the bullet into the end of a stationary
wooden block suspended by two long strings. This
method is illustrated below.
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In this case, a bullet(m=10.89 g) is shot from a
handgun and becomes embedded in a 2.20 kg
block of wood. The block of wood swings upward
a vertical distance of 69.1 cm. Using the
information provided, the Law of Conservation of
Energy, and the Law of Conservation of
Momentum, calculate the speed of the bullet just
before the impact with the block.
2. A 56.7-g steel ball is shot from a spring
gun into a 203-g pendulum of a ballistic
pendulum. The pendulum swings aside
and is kept at its maximum height by
mechanical means. The change in height
is measured to be 13.1 cm. What is the
velocity of the steel ball when fired from
the spring gun?
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Review
Momentum(p)=mv
Change in momentum
Impulse
Law of conservation of momentum
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