Conserv. of Momentum (Applications)
Announcements:
•  Next midterm a week from
Thursday (3/15). Chapters 6–9
will be covered
•  LA information session at 6pm
today, UMC 235.
•  Will do some longer examples
today.
•  Material from Chapter 9.10-9.11
•  Wednesday – Chapter 9.1-9.5
•  Friday – Chapter 9.12
Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/
Collision summary
In all collisions,
momentum is conserved:
This is a vector equation.
Components are conserved.
In elastic collisions, kinetic
energy is conserved.
In completely inelastic collisions, final Components:
velocities are the same
Clicker question 1
Set frequency to BA
Ball A strikes a stationary ball B in a 2D collision. The initial
momentum of ball A, , and the final momentum of ball B,
shown on the graph. Which vector gives the final ball A
momentum
?
A
B
C
, are
D
Clicker question 1
Set frequency to BA
Ball A strikes a stationary ball B in a 2D collision. The initial
momentum of ball A, , and the final momentum of ball B,
shown on the graph. Which vector gives the final ball A
momentum
?
The initial momentum of the system
is .
By conservation of
momentum, the final total momentum
must be the same so
or
Can also do component addition:
A
B
C
, are
D
Clicker question 2
Set frequency to BA
Suppose the entire population of Earth gathers in one
location and, at a pre-arranged signal, everyone jumps up.
About a second later, 7 billion people land back on the
ground. After the people have landed, the Earth's momentum
is..
A: the same as it was before the people jumped.
B: different than it was before the people jumped.
C: impossible to know whether the Earth's momentum
changed/don't know.
Clicker question 2
Set frequency to BA
Suppose the entire population of Earth gathers in one
location and, at a pre-arranged signal, everyone jumps up.
About a second later, 7 billion people land back on the
ground. After the people have landed, the Earth's momentum
is..
A: the same as it was before the people jumped.
B: different than it was before the people jumped.
C: impossible to know whether the Earth's momentum
changed/don't know.
Same as before. Started zero, ended zero. Conservation of
momentum, there are no EXTERNAL forces!
Kick of a Rifle (Cons. of Momentum)
A 3.24 kg Winchester Super X rifle initially at rest, fires a
11.7 g bullet with a muzzle speed of 800 m/s. What is the
recoil velocity of the rifle? What is the ratio of the kinetic
energies of the bullet and the rifle ?
0 = −mrv r + mb v b
mb
11.7 ×10−3 kg
vr =
vb =
(800m /s) = 2.89m /s
mr
3.24kg
€
€
1 2 p2
p is conserved, K r m r = K b m b
K = mv =
2
2m
K r mb 11.7 ×10−3 kg
=
=
= 0.38%
K b mr
3.24kg
€
A collision example
A 10 g bullet hits a 2 kg block of ballistic gel
and is embedded. The gel comes to rest
after traveling 40 cm along the floor with a
coefficient of kinetic friction of 0.5. What is
the initial speed of the bullet?
.01 kg
Two sequential problems:
22 kg
kg
.4 m
1. Completely inelastic collision between bullet and gel
2. An energy problem (gel kinetic energy is lost to friction)
The collision part of the example
.01 kg
2 kg
The collision part:
2 kg
.4 m
The collision conserves momentum:
Collision is completely inelastic and the gel is initially at rest:
Need vgel to get the bullet velocity
The energy part of the example
.01 kg
The work-energy part:
2 kg
2 kg
.4 m
Conservation of energy:
No potential energy and no final kinetic energy:
Work due to friction:
Putting it together:
So
The collision part again
.01 kg
2 kg
2 kg
.4 m
Our analysis of the completely
inelastic collision gave
We obtained vgel from analyzing the work-energy
part of the problem
Clicker question 3
Set frequency to BA
Suppose the 0.01 kg bullet from the previous example hit the
same 2 kg gel block at 400 m/s but passed through the gel, exiting
the gel at 100 m/s. What speed would the gel have immediately
after the bullet exits?
A.  0.5 m/s
B.  1.0 m/s
.01 kg
2 kg
C.  1.5 m/s
D.  2.0 m/s
E.  300 m/s
Clicker question 3
Set frequency to BA
Suppose the 0.01 kg bullet from the previous example hit the
same 2 kg gel block at 400 m/s but passed through the gel, exiting
the gel at 100 m/s. What speed would the gel have immediately
after the bullet exits?
A.  0.5 m/s
B.  1.0 m/s
.01 kg
2 kg
C.  1.5 m/s
D.  2.0 m/s Momentum conservation:
E.  300 m/s
Initial gel velocity is 0:
An elastic example
A tennis ball is put on top of a basketball and both
are dropped from a height of 1 m. How high will the
tennis ball go? Assume all collisions are elastic.
Can use conservation of energy to get velocity
after falling 1 m.
becomes
so
Basketball hits first. What happens?
If no friction losses then by conservation of energy the upward
speed after bounce equals the downward speed before bounce
Or, elastic collisions conserve kinetic energy. Same answer.
Is momentum conserved? Only if you consider Earth
as part of the system
An elastic example
So we have a 1D elastic collision between basketball moving
at vB = 4.5 m/s and tennis ball moving at vT = –4.5 m/s
The collision conserves momentum:
Elastic collision conserves
kinetic energy:
Solve this problem in the reference
frame where the basketball is at rest
This gives
and
An elastic example
We have
and
Factoring:
Divide the equations to get
Substitute in to get
Rearrange:
becomes
Solve:
Substitute back in to get
An elastic example
So in the basketball rest frame we have
a post collision tennis ball velocity of
Masses and initial velocities:
This is with respect to the basketball which is moving
up at 4.5 m/s. Therefore the tennis ball speed with
respect to the ground is 6.3 m/s + 4.5 m/s = 10.8 m/s.
Basketball and tennis ball
Before collision, the tennis ball is moving down at 4.5 m/s
After collision the tennis ball is moving up at 10.8 m/s
What height will the tennis ball reach?
19 foot height from just a 3 foot drop?!
Elastic Collisions in 2D (Not Head-on)
v1
θ1
u1
v2
∑p :
∑p :
∑ KE :
x
y
A proton moving at
speed 5 km/s makes
an elastic collision with
another proton. If we
know one proton has
an angle of 37
degrees, what are the
velocities, v1, v2, θ1
m1u1 + 0 = m1v1 cosθ1 + m2v 2 cos(37)
0 = m1v1 sin θ1 − m2v 2 sin(37)
1
1
1
2
2
m1u1 = m1v1 + m2v 22
2
2
2
Elastic Collisions in 2D (Not Head-on)
v1 m1u1 + 0 = m1v1 cosθ1 + m 2v 2 cos(37)
θ1 0 = m1v1 sin θ1 − m 2v 2 sin(37)
1
1
1
2
2
m1u1 = m1v1 + m2v 22
2
2
2
u1 = 5 km/s
v2 Divide out mass as m1=m2
€
u1 = v1 cosθ1 + v 2 cos(37)
v1 cosθ1 = 5 − .8v 2
0 = v1 sinθ1 − v 2 sin(37) v1 sinθ1 = .6v 2 v12 = 25 − 8v 2 + v 22
u12 = v12 + v 22
25 = v12 + v 22
v1= 4 m/s ;
v2= 3 m/s ;
θ1= 53 degrees
€
€