Linear Momentum
Units = kg·m/s
• Single Particle:
The linear momentum of a particle is its mass times its velocity vector.
r
r
r
r
r
dv d (mv ) dp
Fnet = ma = m
=
=
dt
dt
dt
r
r
p = mv
r
r
dp
Fnet =
dt
The time rate of change of the momentum is equal to the net force acting on the
particle. If Fnet = 0 then p is constant in time (i.e. does not change).
• System of Particles:
The overall linear momentum of a system of particles is the vector sum of the
r
momentums of all the particles.
N
r
r
r
r
r
r
r
Ptot = ∑ pi = m1v1 + m2 v2 + m3v3 + L + mN v N = M tot vcm
i =1
r
r
dPtot
= M tot acm
Fnet =
dt (system of particles)
• Momentum Conservation:
If the net external force acting on a system of particles is zero (i.e. isolated system)
then Ptot is constant in time (i.e. does not change) and hence the total momentum at
some initial time ti is equal to the total momentum at some final time tf.
r tot r tot
Pi = Pf
R. Field
University of Florida
(conservation of linear momentum, isolated system)
PHY 2053
Page 1
Momentum Conservation: Examples
• Example 1:
Near the surface of the Earth, a bullet with mass m moving directly
upward at speed v = 1,000 m/s strikes and passes through the
center of mass of a block of mass M initially at rest as shown in the
figure. The bullet then emerges from the block moving directly
upward at speed v‘ = 500 m/s. If the mass of the block is 250 times
the mass of the bullet, to what maximum height does the block then
rise above its initial position? Answer: 20.4 cm
m
V=
(v − v ' )
M
V 2  m  (v − v ' ) 2
h=
= 
2g  M 
2g
2
• Example 2:
A man with mass M = 75 kg is riding in a cart with mass m = 25 kg at a speed of v =
2.5 m/s along a frictionless horizontal surface. He jumps off in such a way as to land
on the ground with no horizontal velocity. What is the new speed of the cart?
Answer: 10 m/s
v' =
R. Field
University of Florida
M +m
v
m
PHY 2053
Page 2
Momentum Conservation: Examples
• Completely Inelastic Collision:
Before
A projectile with mass M1 and speed v1 traveling to
the right along the x-axis collides with a target
particle at rest with mass M2. If the two particles
stick together to form a single particle of mass
M1+M2, what is its velocity? What is the velocity of
the center-of-mass of the two particles system
before the collision? What is the change in the
kinetic energy before and after the collision?
V = Vcm =
M1
v1
M1 + M 2
∆KE = KE f − KEi = −
v1
v2 =0
M1
M2
x-axis
After
V
M1+M2
x-axis
1 M 1M 2 2
v1
2 M1 + M 2
• Elastic Collision:
An elastic collision is one in which the kinetic energy is conserved (i.e. the initial
total kinetic energy is equal to the final total kinetic energy). If the two particles in
the previous problem collide elastically, what are each of the particles velocity after
the collision? What is the velocity of the center-of-mass of the two particle system
before and after the collision?
v '1 =
R. Field
University of Florida
M1 − M 2
2M 1
M1
v1 v '2 =
v1 Vcm =
v1
M1 + M 2
M1 + M 2
M1 + M 2
PHY 2053
Page 3
Conservation of Mass
y
Mass is conserved in classical mechanics:

 

 ∑ mn  =  ∑ mn 
 n =1 i  n =1  f
• Proof:
N
y'
y
Before Collision
y'
After Collision
V
N
mA
O
O'
O
x
z
V
mB
z'
MA
MB
x
x'
O'
x'
z
z'
The proof comes from momentum conservation and the Galilean transformation. We
follow the collision of two particles with masses mA and mB before the collision and
masses MA and MB after the collision in both the O-frame and the O'-frame, where O
and O' are inertial frames with relative velocity V.
Momentum Conservation O-frame
r
r
r
r
p A + pB = p A + pB
r
r
r
r
m A v A + mB v B = M A v A + M B v B
Momentum Conservation O'-frame
r
r
r
r
p' A + p'B = p ' A + p 'B
r
r
r
r
m A v ' A + mB v ' B = M A v ' A + M B v ' B
Galilean Transformation (i.e. velocity addition)
r r
r
r
r
r
Hence: r
v ' A = v A − V v ' B = vB − V
r
r
r
r
r
r
r
r r
r
r
r
r
m A ( v A − V ) + m B ( v B − V ) = M A (v A − V ) + M B ( v B − V )
v 'A = vA − V
v ' B = vB − V
r
r
r
r
r
m Av A + mB vB = M Av A + M B vB + (m A + mB − M A − M B )V
R. Field
University of Florida
PHY 2053
m A + mB = M A + M B
Page 4
Momentum Change: Impulse
The time rate of change of the momentum is equal to the net force acting on an
object.
r
r
dp
F (t ) =
dt
r r
dp = F (t )dt
r
r r
r
∆p = p f − pi = F∆t
• Impulse:
The change in the linear momentum is equal to the impulse.
The impulse J is defined as follows:
v r
J = F∆t
(constant force)
r
r r
r
∆p = p f − pi = J
(definition of impulse for constant force)
• Average Constant Force:
Often one is interested in the average constant force necessary to produce an
r
impulse J within the time interval ∆t = t2 – t1. r
r r
J = Fave ∆t
Fave =
J
∆t
Example: A ball of mass M drops vertically onto the floor,
hitting with speed v1. If the ball is in contact with the floor
for a time ∆t and if it rebounds with speed v2, what is the
magnitude of the average force on the floor from the ball?
R. Field
University of Florida
PHY 2053
Fave = M (v1 + v2 ) / ∆t
Page 5