TEST 1
Tuesday May 19
9:30 am
CNH-104
Kinematics, Dynamics
& Momentum
Physics 1D03 - Lecture 25
Momentum
• Newton’s original “quantity of motion”
• a conserved quantity
• a vector
-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 (or more)
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
10 min rest
Physics 1D03 - Lecture 25
Collisions
• Conservation of Momentum
• Elastic and inelastic collisions
Physics 1D03 - Lecture 25
Collisions
A collision is a brief interaction between two
(or more) objects. We use the word “collision”
when the interaction time Δt is short relative to
the rest of the motion.
During a collision, the objects exert equal and
opposite forces on each other. We assume
these “internal” forces are much larger than
any external forces on the system.
We can ignore external forces if we compare
velocities just before and just after the
collision, and if the interaction force is much
larger than any external force.
v1,i
m1
v2,i
m2
F1
F2 = -F1
v1,f
v2,f
Physics 1D03 - Lecture 25
Elastic and Inelastic Collisions
Momentum is conserved in collisions. Kinetic energy is sometimes
conserved; it depends on the nature of the interaction force.
A collision is called elastic if the total kinetic energy is the same
before and after the collision. If the interaction force is
conservative, a collision between particles will be elastic (eg:
billiard balls).
If kinetic energy is lost (converted to other forms of energy), the
collision is called inelastic (eg: tennis ball and a wall).
A completely inelastic collision is one in which the two colliding
objects stick together after the collision (eg: alien slime and a
spaceship). Kinetic energy is lost in this collision.
Physics 1D03 - Lecture 25
If there are no external forces, then the total momentum is
conserved:
v1,i
m1
v2,i
m2
p1,i + p2,i = p1,f + p2,f
v1,f
v2,f
This is a vector equation. It applies to each component of p
separately.
Physics 1D03 - Lecture 25
Elastic Collisions
In one dimension (all motion along the x-axis):
1) Momentum is conserved:
m1v1i  m2 v2i  m1 v1f  m2v2f
In one dimension, the velocities are represented by positive
or negative numbers to indicate direction.
2) Kinetic Energy is conserved:
1
2
m1v12i  12 m2 v22i  12 m1v12f  12 m2 v22f
We can solve for two variables if the other four are known.
Physics 1D03 - Lecture 25
One useful result: for elastic collisions, the magnitude of the relative
velocity is the same before and after the collision:
|v1,i – v2,i | = |v1,f – v2,f |
(This is true for elastic collisions in 2 and 3 dimensions as well).
An important case is a particle directed at a stationary target (v2,i = 0):
• Equal masses: If m1 = m2, then v1,f will be zero (1-D).
• If m1 < m2, then the incident particle recoils in the opposite direction.
• If m1 > m2, then both particles will move “forward” after the collision.
before
after
Physics 1D03 - Lecture 25
Elastic collisions, stationary target (v2,i = 0):
Two limiting cases:
1) If m1 << m2 , the incident particle
rebounds with nearly its original
speed.
v1
-v1
2) If m1 >> m2 , the target particle moves
away with (nearly) twice the original
speed of the incident particle.
v1
v1
2v1
Physics 1D03 - Lecture 25
Concept Quiz
A tennis ball is placed on top of a basketball and both are dropped.
The basketball hits the ground at speed v0. What is the maximum
speed at which the tennis ball can bounce upward from the
basketball? (For “maximum” speed, assume the basketball is
much more massive than the tennis ball, and both are elastic).
a) v0
b) 2v0
c) 3v0
?
v0
v0
Physics 1D03 - Lecture 25
Example
An angry 60.0 kg physicist standing on a frozen lake
throws a 0.5 kg stone to the east with a speed of
24.0 m/s.
Find the recoil velocity of the angry physicist.
Physics 1D03 - Lecture 25
Example – inelastic collision:
A neutron, with mass m = 1 amu (atomic mass unit),
travelling at speed v0, strikes a stationary deuterium
nucleus (mass 2 amu), and sticks to it, forming a
nucleus of tritium. What is the final speed of the
tritium nucleus?
Physics 1D03 - Lecture 25
An elastic collision:
Two carts moving toward each other collide and
bounce back. If cart 1 bounces back with v=2m/s,
what is the final speed of cart 2 ?
6 m/s
2 kg
5 m/s
4 kg
Physics 1D03 - Lecture 25
Example
A bullet of mass m is shot into a block of mass M that is
at rest on the edge of a table. If the bullet embeds in
the block, determine the velocity of the bullet if they
land a distance of x from the base of the table, which
has a height of h.
Physics 1D03 - Lecture 25
10 min rest
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:
J = FΔt = Δp
(change in p) = (total impulse from external forces)
(Newton’s Second Law again)
(Extra) In general (force not constant), we integrate:

J 
 Fdt
where 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.
The impulse – momentum theorem, J=FΔt=Δp, tells us that
we do not need to know the details of the force/interaction,
we only need to know the area under the curve=integral
of the force.
Compare with work: W = F Dx ; so the work-energy theorem
(derived from Newton #2) is DK = F Dx.
Physics 1D03 - Lecture 25
Quiz
A 100g rubber and a 100g clay ball are thrown at a wall
with equal speed. The rubber ball bounces back
while the clay ball sticks to the wall.
Which ball exerts a larger impulse on the wall ?
a)
b)
c)
d)
the clay bass b/c it sticks
the rubber ball b/c it bounces
they are equal because the have equal momenta
neither exerts an impulse b/c the wall doesn’t move
Physics 1D03 - Lecture 25
Example 1
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
Example 2
Use the impulse-momentum theorem to find how long a
falling object takes to increase its speed from 5.5m/s
to 10.4m/s.
Physics 1D03 - Lecture 25
Example 3
• A 150 g baseball is thrown with a speed of 20m/s. It
is hit straight back toward the pitcher at a speed of
40m/s. The interaction force is shown by the graph:
F
Fmax
6ms
t
What is the maximum force Fmax that the bat exerts
on the ball ?
Physics 1D03 - Lecture 25