Physics 1A
Lecture 7A
"Burns’ Hog-Weighing Method:
(1) Get a perfectly symmetrical plank and balance it
across a sawhorse.
(2) Put the hog on one end of the plank.
(3) Pile rocks on the other end until the plank is again
perfectly balanced.
(4) Carefully guess the weight of the rocks.”
--Robert Burns
Building Intuition for 2D
Collisions
Let’s go through a view simple examples, and
see if this helps you understand the general
case.
Two identical balls (I)
Two balls with same mass and speed bounce off
each other in an elastic collision, without any
change in angles.
(a) both have same speed after collision
(b) both do not have same speed after collision
Two identical balls (II)
Two balls with same mass and speed bounce off
each other in a perfectly inelastic collision.
What’s the angle after the collision?
Small ball hitting big ball
The small ball can bounce off the big ball at any
angle, depending on where exactly it hits on the
surface.
vi
The momenta of the two after the collision depend
on the angles they fly off.
2D Collisions
Example
A 10.0g golf ball initially traveling at 5.00m/s hits
a 0.500kg stationary basketball off center. The
two collide such that they finally travel at 50.0o
with respect to one another. Find the final
velocities of both balls.
vi
30o
20o
Answer
First, you must define a coordinate system.
Let’s say that original motion the golf ball is in
the positive x-direction.
Nomenclature
vi = initial speed of golf ball
vg = final speed of golf ball
vB = final speed of basketball
We are given:
vi
the masses of the two balls
the initial and final directions
Momentum
Answer
Let’s also define the two balls as a single system.
Every force between them is now internal.
Initially, the momentum of the system in the xdirection is:
pix = pgx + pBx
(
)
pix = 5 m s (0.01kg) + 0 = 0.05kg m s
Initially, the momentum of the system in
the y-direction is:
p = p + p =0
iy
For the final momentum of the
system we need to break the
final velocities into components.
gy
By
vg
θ
vgcosθ
vgsinθ
Momentum
Answer
Finally, the momentum of the system in the
x-direction is:
p fx = pgx, f + pBx, f
p fx = mg v gx + mB v Bx
p fx = mg v g cos 30° + mB v B cos20°
Apply momentum conservation:
pix = p fx
0.05kg m s = mg v g cos 30° + mB v B cos20°
Finally, the momentum of the system in the
y-direction is:
p fy = pgy, f + pBy, f
p fy = mg v gy + mB v By
p fy = mg v g sin 30° − mB v B sin20°
Answer
Momentum
Apply momentum conservation:
piy = p fy
0 = mg v g sin 30° − mB v B sin20°
mg v g sin 30° = mB v B sin20°
Let’s solve for vg:
mB sin20°
vg = vB
mg sin 30°
0.5kg) sin20°
(
vg = vB
= 34.2v B
(0.01kg) sin 30°
Momentum
Answer
Substitute this back into the x-direction:
0.05kg m s = (0.01kg)( 34.2v B ) cos 30° + (0.5kg)v B cos20°
0.05kg m s = (0.296kg)v B + (0.470kg)v B
0.05kg m s = (0.766kg)v B
0.05kg m s
vB =
= 0.0653 m s
(0.766kg)
Putting this back into the y-direction results:
v g = 34.2v B
(
v g = 34.2 0.0653 m
)s = 2.23m s
Note:
We used momentum conservation in X and Y.
We did not use energy conservation at all in
order to solve the problem.
We could thus ask whether or not this was
an elastic collision!
In class questions
The next two questions are very confusing, I
know.
Think carefully about momentum conservation
in X and Y, and you will arrive at the correct
answer.
In class question
A toy car of mass, M, is moving at an initial speed, v1i,
in the positive x-direction. A payload of mass, m, is
dropped directly down onto the car. The speed of the
toy car after the payload has been dropped on it is:
A) Same as its initial speed.
B) Faster than its initial speed.
C) Slower than its initial speed.
In class question
A toy car of mass, M, is moving at an initial speed, v1i,
in the positive x-direction as it carries a payload of
mass, m. The payload is then dropped into a hole in the
floor. The speed of the toy car after the payload has
been dropped is:
A) Same as its initial speed.
B) Faster than its initial speed.
C) Slower than its initial speed.
Chapter 7
Rotational Motion and gravity
Rotational Variables
Up until now, we have only dealt with bodies in
translation, which is motion along a line.
Now, we will deal with the consequences of
rotational motion.
We need to define useful
variables for rotational motion
as we did with translational
motion.
Angular position is measured
by θ [in radians].
Rotational Variables
Angular displacement is the change in angular
position:
Δθ = θf - θi
For ease, when dealing with angular displacement
we say that we are dealing with a rigid body.
This means that each point on the object
undergoes the same angular displacement.
The convention for angular displacement is that a
clockwise displacement is negative and a
counterclockwise displacement is positive.
Also, be careful, Δθ can be greater than 2πrad.
Rotational Variables
Average angular speed is given by:
It is angular displacement
over a time interval.
The units of angular speed
are [rad/s].
Angular speed will be positive if θ is increasing (ccw).
Angular speed will be negative if θ is decreasing
(cw).
Rotational Variables
Average angular acceleration is given by:
It is the change in angular
speed over a time interval.
The units of angular
acceleration are [rad/s2].
These variables correspond to translational variables.
θ --> x
ω --> v
α --> a
Rotational Variables
We can use the constant acceleration equations
with rotational variables:
Remember (just like linear motion) these equations
rely on the fact that angular acceleration remains
constant over the time period in question.
Rotational Variables
Example
A wheel has a constant angular acceleration of
3.0rad/s2. During a certain 4.0 second interval,
you measured that the wheel had an angular
displacement of 120 radians. If the wheel
started at rest, how long had it been in motion
at the start of the 4 second interval?
Answer
First, you must define a coordinate system.
Let’s say that the wheel starts moving
counterclockwise in the positive direction.
Rotational Variables
Answer
Let’s list the quantities we know:
ωo = 0 <-- it starts from rest
α = +3.0rad/s2
ω
<-- don’t know
Δθ
<-- don’t know
t
<-- finding
But we also know that in a 4sec. span it went
through an angle of 120rad.
We can use the third equation to find the angular
velocity at the beginning of that time span.
Rotational
Variables
Answer
Use:
Thus, it started out the 4 sec. interval with an
angular velocity of 24rad/s.
We have a new known: ω = 24rad/s.
Rotational
Variables
Answer
Since Δθ is missing we can use:
The only difference between translational kinematic
equations and rotational kinematic equations is that as
the object rotates it will gain angular displacement
even as it keeps repeating the same position.
Angular displacement is cumulative.
For Next Time (FNT)
Finish working on the Homework for
Chapter 6.
Keep reading Chapter 7.