Transcript Potential Energy - McMaster University

Test – What to study
• The test will cover things we have done in class up to
Lecture 9
• Momentum will NOT be on the test
• Please look over old test (available online)
• Make sure you’re up on your reading !
• Review error calculations (done in lab)
Physics 1D03 - Lecture 25
Momentum
• Newton’s original “quantity of motion”
• a conserved quantity
• a vector
Today:
-Newton’s Second Law in another form
-Momentum and Momentum Conservation
Physics 1D03 - Lecture 25
Definition: The linear momentum p of a particle
is its mass times its velocity:
p  mv
Momentum is a vector, since velocity is a vector.
Units: kg m/s (no special name).
(We say “linear” momentum to distinguish it from angular momentum,
a different physical quantity.)
Physics 1D03 - Lecture 25
Example 1 & 2
• A car of mass 1500kg is moving with a velocity of
72km/h. What is its momentum ?
• Two cars, one of mass 1000kg is moving at 36km/h
and the other of mass 1200kg is moving at 72km/h in
the opposite direction. What is the momentum of the
system ?
Physics 1D03 - Lecture 25
Concept Quiz
Which object will have the smallest momentum ?
a)
b)
c)
d)
A 1.6x10-19kg particle moving at 1x103m/s
A 1000kg car moving at 2m/s
A 100kg person moving at 10m/s
A 5000kg truck at rest
Physics 1D03 - Lecture 25
Newton’s Second Law
If mass is constant, then the rate of change of (mv) is equal to m
times the rate of change of v. We can rewrite Newton’s Second
Law:



dv d(mv )

F  ma  m

dt
dt
or


dp
ΣF 
dt
Net external force = rate of change of momentum
This is how Newton wrote the Second Law. It remains true in cases
where the mass is not constant.
Physics 1D03 - Lecture 25
Example 3
Rain is falling vertically into an open
railroad car which moves along a
horizontal track at a constant speed.
The engine must exert an extra force
on the car as the water collects in it
(the water is initially stationary, and
must be brought up to the speed of
the train).
v
F
Calculate this extra force if:
v = 20 m/s
The water collects in the car at the rate of 6 kg per minute
Physics 1D03 - Lecture 25
Solution
Plan: The momentum of the car
increases as it gains mass (water).
Use Newton’s second law to find F.
F
dp d (mv ) dm
dv


vm
dt
dt
dt
dt
v
F
v is constant, and dm/dt is 6 kg/min or 0.1 kg/s (change to SI units!),
so
F = (0.1 kg/s) (20 m/s) = 2.0 N
F and p are vectors; we get the horizontal force from the rate of
increase of the horizontal component of momentum.
Physics 1D03 - Lecture 25
The total momentum of a system of particles is the
vector sum of the momenta of the individual particles:
ptotal = p1 + p2 + ... = m1v1 + m2v2 + ...
Since we are adding vectors, we can
break this up into components so that:
px,Tot = p1x + p2x + ….
Etc.
Physics 1D03 - Lecture 25
Example 4
• A particle of mass 2kg is moving with a velocity of
5m/s and an angle of 45o to the horizontal.
Determine the components of its momentum.
Physics 1D03 - Lecture 25
Newton’s 3rd Law and Momentum Conservation
Two particles interact:
Dp1= F12 Dt
Dp2= F21 Dt
F12
m1
F21 = -F12
Newton’s 3rd Law: F21 = -F12
The momentum changes are equal and
opposite; the total momentum:
m2
p = p1 + p2=0
doesn’t change.
The fine print: Only internal forces act. External forces would
transfer momentum into or out of the system.
eg: particles moving through a cloud of gas
Physics 1D03 - Lecture 25
Conservation of momentum simply says that the
initial and final momenta are equal:
pi=pf
Since momentum is a vector, we can also express it
in terms of the components. These are independently
conserved:
pix=pfx
piy=pfy
piz=pfz
Physics 1D03 - Lecture 25
Concept Quiz
A subatomic particle may decay into two different
particles. If the total momentum before is zero
before the decay before, what is the total after ?
a)
b)
c)
d)
0 kg m/s
depends on the masses
depends on the final velocities
depends on b) and c) together
Before
After
Physics 1D03 - Lecture 25