2/24/2014
Linear Momentum of a Single
Particle
r
p
• Linear momentum:
r
r
p = mυ
Chapter 7
• It is a measure of the particle’s motion
• It is a vector, similar to the velocity
p x = mυ x p y = mυ y p z = mυ z
Linear Momentum
• It also depends on the mass of the object,
similar to the kinetic energy.
A few pointers:
Newton’s 2nd Law for a Single Body.
• The linear momentum of an object changes
(and therefore there is a net force acting
on it) if
• The time rate of the change of the
momentum of a particle is equal to the net
force acting on the particle.
r
Fnet
r
• The velocity changes in magnitude
r
r
r
∆υ ∆ (mυ ) ∆p
= ma = m
=
=
∆t
∆t
∆t
• The velocity changes in direction
• The momentum formulation of Newton’s law
is more inclusive as it is applicable to
bodies and systems with variable mass.
• The mass of the body changes
r
r
Fnet ∆t = ∆p
y
Example- a ball hits a wall
x
)
)
)
r r
r
r
∆p = p f − pi = −mυ i − mυ i = −2mυ i = −2 p
r r
r
∆p = p f − pi
∆px = −mυ sin θ − mυ sin θ = −2mυ sin θ
∆py = −mυ cosθ + mυ cosθ = 0
I
θ
)
)
r r
r
r
∆p = p f − pi = 0 − mυ i = −mυ i = − p
If the ball sticks to the wall the force is smaller than
if it bounces.
II
Collisions
• Before and after the collision the momentum of a
body changes.
• There is a net force acting on the body.
r
r
Fnet ∆t = ∆p
• The collision takes time (although quite small).
• The force varies during the time of the collision.
• The total change of the momentum of the body
during the collision is:
III
r r
r
∆p = p f − pi = Fnet ∆t
Impulse
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The point mass
Impulse for a single collision
• The point mass is a physical approximation
of a real object that has dimensions.
• How good is this this approximation???
• Can we apply Newton’s laws and all we
learned so far to a big, extended object, or
to a system of objects that are not even
attached to each other?
• The impulse of a force is a vector.
r
F∆t = area under the curve
The area below the two curves is
the same!
r
same impulse F∆t
r
r r
r
F ∆t = ∆p = p f − p i
The change of the linear momentum is
equal to the impulse of the acting force.
System of particles
The Center of Mass
• A body can have a complicated shape
• We can represent the body as a sum of
smaller parts
• Apply Newton’s law to each part.
r
r
Fi = mi ai
n
r
r
r
F = ∑ Fi = ∑ mi ai
i =1
r
r
F = MaCM
⇒
• The center of mass (CM) of a system of particles
is the point that moves as though
• the system’s mass is concentrated in this point and
• all external forces are applied there.
r
1
rCM =
M
r
1
rCM =
M
i i
n
M = ∑ mi
r
∑m r
i i
M = ∑ mi
r
mi , ri
r r
Fi , ai
r
∑m r
i =1
• Important note:
• We can substitute the entire system with a single point!
• CM is a geometrical point (it might be outside of the
body).
We can replace the object with
point mass at the center of mass!
⇒
CM
Example: 2 masses
• Two particles have masses m1 and m2 and
positions x1 and x2 respectively. Find the center
of mass of the system.
r
r
r
m r + m2 r2
rCM = 1 1
m1 + m2
xCM =
y
xcom
x1
x2
x
m1 x1 + m2 x2
m1 + m2
1)m1 = m2 ⇒ xCM = 0.5( x1 + x2 )
The motion of the bat is described by the motion of the CM.
2)m2 >> m1 ⇒ xCM ≈ x2
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A few pointers.
Center of Mass of a System
• If a body has a
symmetry and it has a
uniform density then
the CM is on the line
of symmetry.
System of particles
r
1
rCM =
M
r
∑ mi ri
⇒
M = ∑ mi
xCM =
1
M
∑m x
yCM =
1
M
∑m y
zCM =
1
M
∑m z
i i
i
i
• The center of
symmetry coincides
with the CM.
i i
• The CM might be
outside the object
Newton’s Second Law for a System
of Particles
Example
• Find the CM of a 4x8
uniform sheet of
plywood with the upper
right quadrant
removed (a=4ft).
a a
m1 = 2 M ; ( x1 , y2 ) = ( , )
2 2
m2 = M ; ( x2 , y2 ) = (
xCM
y
a
x
a
a
3a a
, )
2 4
m1 y1 + m2 y2 2M ⋅ (a / 2) + M ⋅ (a / 4) 5a 5
=
=
= ft
m1 + m2
2M + M
12 3
r
r
Fnet = MaCM
2a
m x + m2 x2 2M ⋅ (a / 2) + M ⋅ (3a / 2) 5a 10
= 1 1
=
=
=
ft
m1 + m2
2M + M
6
3
yCM =
• The sum of all external forces Fnet acting on the
system is equal to the product of the total mass M
of the system and the acceleration of the center
of mass aCM.
• Fnet - sum of forces that are external to the system (?)
• M is the total mass
• It does not give information on the acceleration of the
individual objects of the system.
• If no external forces are present the center of mass will
stay at rest or move with constant velocity!
Example Problem
• A 10 m long boat stays at rest in a quite
lake. A turtle with a mass m, is originally
standing at the right end of the boat. The
turtle walks to the other end of the boat.
Does the boat move? If so how far?
Solution
• Yes the boat will move, but the center of
mass of the boat+turtle system will stay at
rest.
before
xCM =
0.5ML + mL
M +m
after
xCM =
M (0.5L + x ) + mx
M +m
M(0.5L + x) + mx = 0.5ML + mL
Mx + mx = mL
m
x=
L
M+m
m
M
0
L/2
L
L/2
L
x
0
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Example Problem
Solution
• A cannon shell is fired with an initial velocity of
50 m/s at an angle 45o to the horizontal. When
the shell is at the highest point of its
trajectory, it explodes into two pieces, m1 and
m2. m1 falls straight down from the highest
point after the explosion. If m1 = 2m2, where
does m2 land relative to the initial firing point?
• The CM will follow the trajectory of the
original cannon shell and final xcm can be
found with the range equation.
xCM
45o
1
2
Linear Momentum of a System of
particles
• The vector sum of the linear momentum of
all particles in the system!
r r r
r
r
P = p1 + p2 + p3 + ... + pn
r
r
r
r
r
r
P = m1υ1 + m2υ 2 + m3υ 3 + ... + mnυ n = ∑ miυi
Can show with some work
r
r
P = MυCM
υ02
m1 = 2m2
sin 2θ = 255m
g
0.5m1R + m2 x m2 R + m2 x R + x
=
=
=
m1 + m2
2m2 + m2
3
xCM = R =
R+x
=R
3
x = 3R − R = 2 R = 510m
Newton’s 2nd Law for a system of
particles
• The net force (the vector sum of all external
forces) acting on the system of particles is equal
to the rate of change of the the total linear
momentum of the system.
r
r
Fnet ∆t = ∆P
• The linear momentum of a system remains
constant if no net external force is acting on the
system.
CM velocity
Total mass
Conservation of Linear Momentum
Example
• A cannon fires a shell at 60o angle to
the horizontal. The shell initial velocity
is vs=100 m/s. If the ratio of the
masses of the cannon and the shell is
100 find the velocity of the cannon
after the shot was fired.
• If a system is closed and isolated the
momentum of the system is constant.
r
P = constant
r
∆P = 0
• If no net external force is acting on the
system then the momentum is conserved
y
r
r
r
r
r
r
p1 f + p2 f + ...+ pnf = p1i + p2i + ...+ pni
Pxi = Pxf
• If the net force on a closed system is zero
in a given direction (axis) the momentum in
this direction cannot change.
pc = mcυ c
⇒
0 = pc + psx
pc = − psx = −msυ sx = −msυ s cosθ
υc = −
ms
υ s cosθ
mc
υ c = −cos60 = −0.5 m /s
ps
pc
θ
x
External forces in y: Fg FN
Momentum is conserved in x
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Momentum and Kinetic Energy
conservation Laws
Elastic Collisions
• Elastic collisions: no deformations, no
heating of the bodies, no explosions ,no
crashing sound.
• Mechanical energy conservation
• Conditions:
• isolated system, (no external forces)
• closed system, (do not lose bodies)
• conservative forces. (no friction, drag…)
• Examples: two balls collide and separate after
the collision (pool games).
• Conservation laws:
• linear momentum is conserved
• kinetic energy is conserved.
• Linear momentum conservation
• Conditions:
• isolated system,
• closed system.
Inelastic collisions
Inelastic Collisions in 1D
• Inelastic collisions: there is deformation,
or heating, or explosion, or sound …
• What happens before and after the collision?
• Examples: a bullet hits a target and goes
through the target;
• Completely inelastic collisions: two balls collide
and stick together; a bullet hits a target and
remains inside; a ball is thrown in a cart and
stays there…
• Conservation laws:
• Linear momentum is conserved
• Kinetic energy is NOT CONSERVED
v1i
v2i
before
vf
• What happens before and after the collision?
• Linear momentum of the system is conserved
• Kinetic energy of the system is not conserved
• After the collision the two bodies move together with
a common velocity V.
m1υ1i + m 2υ 2i = (m1 + m2 )V
m υ + m 2υ 2i
V = 1 1i
m1 + m 2
VCM = V
p1i + p2i = p1 f + p2 f
m1υ1i + m2υ 2i = m1υ1 f + m 2υ 2 f
• We have only 1 equation.
• We have 6 parameters. We need to know 5 of them to
find the 6th.
• If we know all initial conditions (masses of the bodies,
velocities before the collision) we still cannot find
both velocities after the collision!
after
Completely inelastic collision 1D
p1i + p2i = p1 f + p2 f
• Linear momentum of the system is conserved
• Kinetic energy of the system is not conserved
v1i
v2i
vf
Example Problem:
ballistic pendulum
• If you know the height
and the masses, find the
initial velocity of the
bullet.
During the collision:
Momentum conservation
h
After the collision:
Momentum conservation
mυ 0 = (m + M)V
m+ M
υ0 =
V
m
1
(m + M)V 2 = (m + M)gh
2
V = 2gh
υ0 =
m+ M
2gh
m
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Example: stationary target
playing pool
Elastic Collisions in 1D
• What happens before and after the collision?
• Linear momentum of the system is conserved
• Kinetic energy of the system is conserved
p1i + p2i = p1 f + p2 f
m1υ1i = m1υ1 f + m 2υ 2 f
m1υ1i + m 2υ 2i = m1υ1 f + m2υ 2 f
m1υ1i = m1υ1 f + m2υ 2 f
2
K1i + K 2i = K1 f + K 2 f
1
1
1
1
2
2
2
2
m1υ1i + m 2υ 2i = m1υ1 f + m 2υ 2 f
2
2
2
2
• If we know the initial velocities and the masses then
we can find both final velocities.
v1f
v2f
v1i
v2i
before
Special cases
υ1 f = 0
2
v1i
⇓
m − m2
υ1 f = 1
υ1i
m1 + m2
2m1
υ2f =
υ1i
m1 + m 2
v1f
v2f
Example: Moving target
• Both bodies are moving before the collision.
• This is the most general case and you can
derive the rest of the cases from these
formulas
v1i
υ 2 f = υ1i
• m1<<m2 massive target (like wall)
υ2 f
2
after
• m1=m2 the two masses are the same
υ1 f ≈ −υ1i
• The target does not move v2i=0
v1i
• m1>>m2 massive projectile (cannon ball and ping-pong ball)
υ1 f ≈ υ1i
m1 − m 2
2m1
υ1i +
υ 2i
m1 + m 2
m1 + m2
2m1
m − m2
=
υ1i + 1
υ 2i
m1 + m2
m1 + m 2
υ1 f =
2m
≈ 1 υ1i
m2
υ2f
v1i
υ 2 f ≈ 2υ1i
Elastic Collisions in 2D
r
r
r
r
p1i + p2i = p1 f + p2 f
v1i
K1i + K 2i = K1 f + K 2 f
• 2 momentum equations
(for x and y)
• 1 energy equation.
• We have total of 8
parameters.
• We need to know 5 of
the parameters to
determine the rest of
them.
v2i
v1f
β
α
v2f
6