Physics 1A
Lecture 6C
"And in knowing that you know nothing, that
makes you the smartest of all.”
--Socrates
Collisions
Elastic collisions are collisions in which kinetic energy
is conserved.
Inelastic collisions are collisions in which kinetic
energy is not conserved.
All real life collisions are inelastic, to a certain
degree. This means that some kinetic energy will be
lost in the collision.
A collision in which the two objects end up sticking
together is known as a perfectly inelastic collision.
If you have a collision you should turn to momentum
conservation.
Collisions
When confronted with a collision problem it makes
sense to draw “before” and “after” diagrams.
Label each object in the
diagram.
Include the direction of
velocity.
Keep track of your subscripts.
For perfectly inelastic
collisions, remember that they
will have the same final
velocity and have a combined
mass.
Momentum
Example
A 10g bullet is fired into a 1.0kg wood block,
where it lodges. Subsequently, the block slides
4.0m across a wood floor (μk = 0.20 for wood
on wood). What was the bullet’s speed?
Bullet
v
Block
Answer
First, you must define a coordinate system.
Let’s choose the direction of bullet as the positive
x-direction.
Momentum
Answer
Let’s also define the bullet and the block as a single
system. Every force between them is now internal.
We will examine momentum of the system just
before the collision and just after the collision.
Initially, the momentum of the system is:
Finally, the momentum of the system is:
Momentum
Answer
Since there are no external net forces, momentum
will be conserved. So:
We know the masses in the above equation, if we just
find vtot then we will have vbul.
We can turn to mechanical energy to solve for vtot
(Note we can also turn to kinematics to solve for vtot).
Momentum
Answer
Are there any non-conservative forces present?
Yes, friction. So we turn to:
Answer
Momentum
Now we can input this value into the previous
equation:
Recall that we used momentum conservation to compare
“before” and “after” the collision AND used mechanical
energy for the motion following the collision.
In class question
In an inelastic collision:
A) impulse is conserved.
B) momentum is conserved.
C) force is conserved.
D) energy is conserved.
E) elasticity is conserved.
Explosions
Just like collisions, explosions will also conserve
momentum (as long as we define our system
properly).
Let’s say that we have a bomb sitting on a table
and then it explodes into two pieces. What is the
final momentum of the system?
The final momentum of the system should be zero
since there are no external net forces affecting
the bomb.
The potential energy that was released by the
bomb will not affect the momentum of the pieces.
Explosions
Example
A 0.5kg cart and a 2.0kg cart are attached and
are rolling forward with a speed of 2.0m/s.
Suddenly a spring-loaded plunger pops out and
blows the two carts apart from each other. The
smaller cart shoots backward at 2.0m/s. What
are the speed and direction of the larger cart?
2.0m/s
0.5kg
2.0kg
Answer
First, you must define a coordinate system.
Let’s say that original motion of the two trains is
in the positive x-direction.
Answer
Momentum
Let’s also define the two trains and the spring as a
single system. Every force between them is now
internal.
Initially, the momentum of the system is:
pi = mAivAi + mBivBi
pi = (mA + mB)vi
pi = (0.5kg + 2.0kg)2.0m/s = 5.0kg(m/s)
Finally, the momentum of the system is:
pf = mAfvAf + mBfvBf
pf = (0.5kg)(–2.0m/s) + (2.0kg)vBf
pf = -1.0kg(m/s) + (2.0kg)vBf
Momentum
Answer
Are there any net external forces on the system?
No, the spring is part of our system (note: FN and Fg
cancel out).
This means that momentum will be conserved such
that:
pi = pf
5.0kg(m/s) = -1.0kg(m/s) + (2.0kg)vBf
(2.0kg)vBf = 6.0kg(m/s)
vBf = [6.0kg(m/s)]/(2.0kg) = +3.0m/s
The speed of the larger cart is 3.0m/s and the
positive sign indicates that it will be moving in the
positive x-direction.
Collisions
The harder cases are collisions (or explosions) in
two dimensions (or three).
Just remember that perpendicular directions (x
and y) are independent.
If for external forces, ΣFx = 0 and ΣFy = 0, then
px and py are conserved.
Apply
conservation of
momentum
separately to
each direction.
Collisions
So, in the case of the picture below, the initial
momentum in the y-direction is zero.
Thus, the final momentum in the y-direction is also
zero since there are no external forces in the ydirection.
This means that: m1v1f sinθ = m2v2f sinϕ.
Also, in the x-direction:
pi = m1v1i
pf = m1v1f cosθ + m2v2f cosϕ
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.
For Next Time (FNT)
Keep working on the Homework for
Chapter 6.
Start reading Chapter 7.