Transcript Potential Energy - McMaster University

Momentum and Impulse
• Newton’s original “quantity of motion”
• a conserved quantity
• a vector
Today:
-Newton’s Second Law in another form
-momentum and impulse
Serway & Jewett 9.1 – 9.3
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
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
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
Relation between kinetic energy
K and momentum p:
2
p
K
2m
(practice problem)
Physics 1D03 - Lecture 25
Example
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
Impulse
dp
 F , or dp = F dt
Newton #2:
dt
For a constant force, Dp = F Dt . The vector quantity F Dt is
called the Impulse:
I = FΔt = Δp
(change in p) = (total impulse from external forces)
(Newton’s Second Law again)
(Extra) In general (force not constant), we integrate:
I
 Fdt
Recall, the integral gives the area under a curve…
Physics 1D03 - Lecture 25

F

F
ti
tf
t

Area = F (Dt )
ti
tf
t
Impulse is the area under the curve. The average force
is the constant force which would give the same impulse.
Compare with work: W = F Dx ; so the work-energy theorem
(derived from Newton #2) is DK = F Dx.
Physics 1D03 - Lecture 25
Example
A golf ball is launched with a velocity of 44 m/s. The
ball has a mass of 50g. Determine the average force
on the ball during the collision with the club, if the
collision lasted 0.01 s.
Physics 1D03 - Lecture 25
Quiz
A rubber ball and a snowball, each 100 grams, are thrown
at a school bus window at identical speeds. The
snowball sticks, the rubber ball bounces off. Which one
transfers the larger impulse to the window?
A) The rubber ball
B) The snowball
C) They both transfer the same amount
Physics 1D03 - Lecture 25
Example
Figure below shows an approximate plot of force magnitude F versus
time t during the collision of a 58 g ball with a wall. The initial velocity of
the ball is 34 m/s perpendicular to the wall. The ball rebounds directly
back with approximately the same speed, also perpendicular to the
wall. What is Fmax, the maximum magnitude of the force on the ball from
the wall during the collision?
Physics 1D03 - Lecture 25
Newton #3 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:
p = p1 + p2=0
doesn’t change.
m2
The fine print: Only internal forces act. External forces would
transfer momentum into or out of the system.
Physics 1D03 - Lecture 25
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
Application
• A subatomic particle may decay into two different
particles. If the momentum before is zero, it must
also be zero after, so:
pi=0
pf=0
Given the mass and velocity of one of the particles,
if we measure the velocity of the other one we can figure
out it’s momentum and hence the mass.
Physics 1D03 - Lecture 25
Example: the fire extinguisher
How can we calculate the thrust on a fire extinguisher when gas
of mass Dm is ejected out of the nozzle in time Dt ?
Plan: Calculate the change in
momentum of the gas; this is
equal to the impulse the gas
gets from the extinguisher
nozzle.
gas
thrust
From Newton’s Third Law, the nozzle gets an equal but opposite
impulse from the gas. Since impulse is thrust times Dt, divide by
Dt to get the average thrust (force).
Result: thrust = (speed of gas relative to nozzle) times
(mass of gas ejected per unit time)
Physics 1D03 - Lecture 25
Summary
• Definitions: momentum, impulse
• Newton’s Second Law, in terms of momentum and
impulse
• conservation of momentum
Physics 1D03 - Lecture 25