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1206 - Concepts in
Physics
Monday, October 5th
Notes
• Assignment #3 has been posted
• Today’s tutorial will have two parts
first 15 min: problem solving strategies and
than assignment #2 will be discussed in
detail
• Assignment #2 will be available Wednesday
for pick-up (after class)
From last time ...
Definition of Impulse:
The impulse J of a force is the product of the average force Favg and the time interval
∆t during which the force acts:
J = Favg * ∆t
impulse J = area under the force curve.
Definition of Momentum:
The linear momentum p of an object is the product of the object’s mass m and
velocity
p=m*v
momentum p is a vector quantity and the unit is kg m/s
Impuls = change in Momentum
Favg ∆t = pf - pi
Note: this was how Newton formulated his second law
Conservation of linear momentum
Principle of conservation of linear momentum
The total linear momentum of an isolated system remains
constant (is conserved). An isolated system is one for which the
vector sum of the average external forces acting on the system
is zero.
This principle applies to a system containing any number of objects, regardless of the internal forces,
provided the system is isolated. Whether the system is isolated depends on whether the vector sum
of the external forces is zero. Judging whether a force is internal or external depends on which
objects are included in the system.
Internal forces - Forces that the objects within the system exert on each other.
External forces - Forces exerted on the objects by agents external to the system.
Earlier we learned about energy conservation. Conservation of Impuls is equally important and
used in many area’s of physics.
Deriving conservation of linear momentum
example: collision of two masses m1 and m2:
m1
m2
F12
v01
v02
F21
Before: t = t0
During: t = tc
vf2
After: t = tf
vf1
During the collision (t = tc) the forces F12 (force exerted on object 1 by object 2) and F21 (force
exerted by object 2 on object 1) are action - reaction forces and therefore equal in magnitude and
opposite in direction (Newton’s third law). They are also internal forces, since they are forces that the
two object exert on each other inside the system. Since the objects have wights (W1 and W2) due to
the force of gravity, we also have external forces acting. Other external force would be air resistance
and friction - we will ignore them for simplicity reasons in this example.
We apply the impuls - momentum theorem on every object and obtain:
Object 1:
Object 2:
(W1 + F12) Δt = m1 vf1 - m1 v01
(W2 + F21) Δt = m2 vf2 - m2 v02
We add these equation to produce a single result for the system as a whole and obtain:
Internal
forces
{
{
External
forces
{
{
(W1 + W2 + F12 + F21) Δt = (m1 vf1 + m2 vf2) - (m1 v01 + m2 v02)
total final
momentum pf
total initial
momentum p0
Internal
forces
{
{
External
forces
{
{
(W1 + W2 + F12 + F21) Δt = (m1 vf1 + m2 vf2) - (m1 v01 + m2 v02)
total final
momentum pf
total initial
momentum p0
We can write that in words:
(Sum of average external forces + Sum of average internal forces) * elapsed time =
Difference between final momentum and initial momentum
We know that the sum of internal forces is zero, so we can leave them out and write:
(Sum of average external forces) Δt = pf - pi
Suppose the sum of external forces is zero, which is true for every isolated system, then:
0 = pf - pi
or pf = pi
We obtained the principle of conservation of linear momentum!!
Example:
Imagine to balls colliding on a billiard table that is friction free.
Use the momentum conservation principle in answering the
following questions.
(a) Is the total momentum of the two-ball system the same before
and after the collision?
(b) Does this change if the system contains only one of the two
balls?
(a) Let’s collect all external forces involved into this two-ball system:
Weights W1 and W2 of the balls
Normal forces (upward) on each ball FN1 and FN2
Since the balls do not accelerate in the vertical direction, the normal forces must balance the weights,
so that the vector sum of the four external forces is zero. We defined the table as friction free.
Therefore we don’t have a net external force to change the total momentum of the two-ball system.
The total momentum of this two-ball system is conserved.
(b) YOUR TURN
Example:
Starting from rest, two skaters “push off” against each other on a smooth level ice, where friction is
negligible. One is a woman (m1 = 54 kg), and the other one is a man (m2 = 88 kg). The woman moves
away with a velocity of vf1 = +2.5 m/s. What is the final velocity of the man? (Also called recoil
velocity)
For a system consisting of the two skaters on level ice, the sum of external
forces is zero. Thus, we can use the principle of conservation of linear
momentum to determine the man’s recoil velocity.
The man has a larger mass, therefore (according to Newton’s second law) the
acceleration he experience will be smaller and thus a smaller recoil velocity.
The total momentum after the “push off” must be equal to before. When the
skaters start, they are at rest, therefore their total momentum is zero.
m1 vf1 + m2 vf2 = 0
m1 vf1 = - m2 vf2
vf2 = -(m1/m2) vf1 = -(54 kg/88 kg) 2.5 m/s = -1.5 m/s
As expected the recoil velocity of the man is smaller.
Note! The total linear momentum is conserved, even when the kinetic energy of the individual parts
of a system changes. Kinetic energy changes because work is done by the internal forces.
YOUR TURN:
A freight train is being assembles in a switching yard. Let’s look at two boxcars. Car 1 has a mass of
m1 = 65 x 103 kg and moves at a velocity of v01 = 0.80 m/s. Car 2 has a mass of m2 = 92 x 103 kg and a
velocity of v02 = 1.30 m/s. It is faster, overtakes car 1 and couples to it.
Neglecting friction, find the common velocity vf of the car after they become coupled.
v02
v01
vf
YOUR TURN:
A freight train is being assembles in a switching yard. Let’s look at two boxcars. Car 1 has a mass of
m1 = 65 x 103 kg and moves at a velocity of v01 = 0.80 m/s. Car 2 has a mass of m2 = 92 x 103 kg and a
velocity of v02 = 1.30 m/s. It is faster, overtakes car 1 and couples to it.
Neglecting friction, find the common velocity vf of the car after they become coupled.
v02
v01
before
vf
after
The boxcars form a system. The sum of external forces acting on the system is zero.
So, we can use the linear momentum conservation principle.
(m1 + m2) vf = m1 v01 + m2 v02
vf = (m1 v01 + m2 v02)/(m1 + m2)
vf = {(65 x 103 kg)(0.80 m/s) + (92 x 103 kg)(1.30 m/s)}/(157 x 103 kg) = 1.1 m/s
Note! The value for the final velocity is between the two given initial velocities
YOUR TURN: from last time
The diagram below depicts the before- and after-collision speeds of a car
which undergoes a head-on-collision with a wall. In Case A, the car bounces
off the wall. In Case B, the car crumples up and sticks to the wall.
a. In which case (A or B) is the change in velocity the greatest? Explain.
b. In which case (A or B) is the change in momentum the greatest? Explain.
c. In which case (A or B) is the impulse the greatest? Explain.
d. In which case (A or B) is the force which acts upon the car the greatest
(assume contact times are the same in both cases)? Explain.
The diagram below depicts the before- and after-collision speeds of a car
which undergoes a head-on-collision with a wall. In Case A, the car bounces
off the wall. In Case B, the car crumples up and sticks to the wall.
a. In which case (A or B) is the change in velocity the greatest? Explain.
b. In which case (A or B) is the change in momentum the greatest? Explain.
c. In which case (A or B) is the impulse the greatest? Explain.
d. In which case (A or B) is the force which acts upon the car the greatest
(assume contact times are the same in both cases)? Explain.
Case A has the greatest velocity change. The velocity change is -9 m/s in
case A and only -5 m/s in case B.
Case A has the greatest momentum change. The momentum change is
dependent upon the velocity change; the object with the greatest velocity
change has the greatest momentum change.
The impulse is greatest for Car A. The impulse equals the momentum
change. If the momentum change is greatest for Car A, then so must be the
impulse.
The impulse is greatest for Car A. The force is related to the impulse
(I=F*t). The bigger impulse for Car A is attributed to the greater force upon
Car A. Recall that the rebound effect is characterized by larger forces; car A
is the car which rebounds.
Collisions in one dimension
We know, that the total linear momentum is conserved when two object collide, provided they
constitute an isolated system. When object are atoms or subatomic particles, the total kinetic
energy of the system is often conserved as well. So, the kinetic energy gained by one particle is lost
by the other (in case of a two object system).
In contrast, when macroscopic objects collide (like cars), the total kinetic energy after the collision
is generally less than that before the collision.
The kinetic energy is mainly lost in two ways:
(1) it can be converted into heat because of friction
(2) it is spent in creating permanent distortion or damage (deformation of the objects) very hard objects suffer less permanent distortion than softer objects. Therefore softer object
loose more kinetic energy.
Collisions are often classified according to wether the total kinetic energy changes during the
collision:
1.) Elastic collision - total kinetic energy of the system after the collision is equal to the total
kinetic energy before the collision
2.) Inelastic collision - total kinetic energy of the system after the collision is not the same
than before the collision; if object stick together after colliding, the collision is called completely
inelastic
The boxcars were an example for a completely inelastic collision.
Example
A ball of mass m1 = 0.250 kg and velocity v01 = 5.00 m/s collides head-on with a ball of mass
m2 = 0.800 kg that is initially at rest (v02 = 0 m/s). No external forces act on the balls. Assume the
collision is elastic, what are the velocities of the balls after the collision?
No external forces, therefore the total linear momentum of the two-ball system is conserved.
(This is true no matter if the collision is elastic or not)
m1 vf1 + m2 vf2 = m1 v01 + 0
Since the collision is elastic, we also know that the total kinetic energy before and after the collision is
the same.
1/2 (m1 vf12 + m2 vf22) = 1/2 m1 v012 + 0 --> m1 vf12 + m2 vf22 - m1 v012 = 0
We have two equations and two unknowns, which means we can solve the problem.
Use the first equation and re-arrange to find vf2
vf2 = (m1 v01 - m1 vf1)/m2 = (v01 - vf1)*(m1/m2)
Substitute this in second equation: m1 vf12 + m2 [(v01 - vf1)*(m1/m2)]2 - m1 v012
after simplification (your home work) we get:
vf1 = {(m1-m2)/(m1+m2)} v01
vf2 = {(2m1)/(m1+m2)} v01
And putting in numbers: vf1 = -2.62 m/s and vf2 = 2.38 m/s
Collisions in two dimensions
A “head-on” collision is one-dimensional, because the velocities of the objects all point along a single
line before and after contact. However in reality collisions occur in two of three dimensions. We will
take a look at a two-dimensional case. Assume a system of two balls on a billiard table - we don’t
have to deal with external forces (each weight is balance by a normal force, the sum is zero) and we
neglect friction. Momentum is a vector quantity and as for forces we can treat the x and y
components separately. This means the components (x and y) of the total momentum are conserved
separately. So we can write:
x Component: pfx = p0x
y Component: pfy = p0y
m1 = 0.150 kg
v01 = 0.900 m/s
y
YOUR TURN:
vf1 = ?
50°
Θ
35°
x
m2 = 0.260 kg
v02 = 0.540 m/s
vf2 = 0.700 m/s
Use momentum conservation to determine the
magnitude and direction of the final velocity of ball 1
after collision.
The magnitude and direction of the final velocity of ball 1 can be obtained one the
components vf1x and vf1y are known. The momentum conservation principle can be
used. (linear momentum for each component)
m1 = 0.150 kg
v01 = 0.900 m/s
m2 = 0.260 kg
v02 = 0.540 m/s
vf2 = 0.700 m/s
50 degrees to vertical incoming (ball 1) 35 degrees before and down from horizontal for ball 2 after
x Component: pfx = p0x
p = mv
y Component: pfy = p0y
put in numbers:
v1fx = 0.63 m/s and vf1y = 0.12 m/s
v1fy
Θ
v1fx
vf1 = sqrt(vf1x2 + vf1y2) = 0.64 m/s
direction given by Θ = tan-1 ((vfly/vf1x)) = 11°
Basic geometry
Let’s look at plane geometry - 2 dimensions
In this document lowercase letters a, b, c, ... denote the sides of a polygon and uppercase letters A, B, C
denote the vertices of a polygon. A polygon here is any geometric form with sides and vertices that can
be made out of linear functions (for example: triangle, square, rectangular, ...)
Lowercase greek letters α, β, γ, ... are uses for angles.
Sum of angles: α + β + γ = 180° = π
Perimeter: a + b + c
Area: 1/2*h*a (h height to vertex A,
note the right angle 90° between h
and a)
Right triangle 90° at C (or π/2
radians) between h and a)
Area: 1/2*a*b (b=h)
Perimeter: a + b + c
Pythagorean theorem: c2 = a2 + b2
Two triangles are called similar, when the corresponding angles are equal.
Rectangle with sides a and b and Square, where a = b
Perimeter: 2a + 2b = 2(a+b)
Area: a*b
Diagonal: d = √a2+b2
Perimeter: 2a + 2a = 4a
Area: a*a = a2
Diagonal: d = √a2+a2 = √2 a
Note the diagonal divides the rectangle into two triangles, for each one the area is given by
1/2 * a * b (where b = h). To get the area of the diagonal, we have to add the area of the two
triangles: 2*(1/2*a*b) = a*b.
We have just shown that the area of a right angle triangle and a rectangle are connected! As a
consequence knowing one of them also means knowing the other.
This is the case with many relations in physics as well - sometime it is not quite as obvious, but
it is always worth thinking about it and finding them. It will deepen your understanding and safe
you from learning multiple formulas for the “same thing”.
Sum of angles in any polygone with four sides is 360° or 2π radians.
Trapezoid with sides a, b, c and d and height h. Sides a and c are parallel
Perimeter: a + b + c + d
Area: 1/2(a+c)h
h
a1
a2
Let’s show how this formula can be derived from starting off with trianglars and a rectangle:
rectangle: c*h
triangle left: 1/2*h*a1
triangle right: 1/2*h*a2
Now add them all up: 1/2*h(2c+a1+a2)
At the end we don’t want to use a1 and a2, so express them through a and c:
a1 + a2 = a - c and substitute above:
Area: 1/2*h(2c+a-c) = 1/2(a+c)h
The sum of angles in a polygon with n sides is given by (n-2)π radians or (n-2)180 degrees,
so 180° (or π) in a triangle, 360° (or 2π) in square, rectangle, parallelogram, trapezoid,
540° (or 3π) in a pentagon, etc.
Now circles and ellipse:
knowing the radius of a circle
determines circumference and
area as well.
diameter: d = 2 r
circumference: 2πr
area: πr2
Shaded area between two
circles r1 < r2
area: πr22 - πr12 = π(r22 - r12)
Ellipse with semi-axis a and b
and centre C.
area: πab
Assume three lines, a and b are parallel and c intersects both of them
α = β and
α = α’ and
β = β’