Download Chapter 7, Part I

Chapter 9: Linear Momentum
THE COURSE THEME:
NEWTON’S LAWS OF MOTION!
• Chs. 4 & 5: Motion analysis with Forces.
• Ch. 6: Alternative analysis with Work & Energy.
– Work-Energy Theorem &
Conservation of Mechanical Energy: NOT new laws!
We’ve seen that they are Newton’s Laws re-formulated or
translated from Force Language to Energy Language.
• NOW (Ch. 7): Another alternative analysis using the concept of
(Linear) Momentum.
• Conservation of (Linear) Momentum: NOT a new law!
– We’ll see that this is just Newton’s Laws re-formulated or reexpressed (translated) from Force & Acceleration Language
to Force & (Linear) Momentum Language.
• In Chs. 4 & 5, we expressed Newton’s Laws of Motion using
the position, displacement, velocity, acceleration, & force concepts.
• Newton’s Laws with Forces & Accelerations: Very general. In principle,
could be used to solve any dynamics problem, But, often, they are very
difficult to apply, especially to very complicated systems. So, alternate
formulations have been developed which are often easier to apply.
• In Ch. 6, we expressed Newton’s Laws in
Work & Energy Language.
• Newton’s Laws with Work & Energy: Very general. In principle, could
be used to solve any dynamics problem, But, often (especially in collision
problems) it’s more convenient to use still another formulation.
• The Ch. 7 formulation uses Momentum & Force as the basic physical quantities.
• Newton’s Laws in
Momentum & Force Language
• Before we discuss these, we need to learn the vocabulary of this language.
Sect. 7-1: Momentum & It’s Relation to Force
• Momentum: The momentum of an object is
DEFINED as:
(a vector || v)
SI Units:
kgm/s = Ns
• In 3 dimensions, momentum has 3 components:
px = mvx
py = mvy
pz = mvz
• Newton called mv “quantity of motion”.
• Question: How is the momentum of an object changed?
• Answer: By the application of a force F!
• Momentum:
• The most general statement of Newton’s 2nd Law is:
(p/t)
(1)
– The total force acting on an object = time rate of change of
momentum. (1) is more general than ∑F = ma because
it allows for the mass m to change with time also!
• Example, rocket motion!
– Note: if m is constant, (1) becomes:
∑F = (p/t) = [(mv)/t] = m(v/t) = ma
• Newton’s 2nd Law (General Form!)
∑F = p/t
(1)
m = constant. Initial p0 = mv0. Final
momentum p = mv. (1) becomes:
∑F = p/t = m(v-v0)/t
= m(v/t) = ma
(as before)
Example 7-1: Force of a tennis serve
For a top player, a tennis ball may
leave the racket on the serve with a
speed v2 = 55 m/s (about 120 mi/h).
The ball has mass m = 0.06 kg & is
in contact with the racket for a time
of about t = 4 ms (4 x 10-3 s),
Estimate the average force Favg on
the ball. Would this force be large
enough to lift a 60-kg person?
Example 7-2: Washing a Car, Momentum Change & Force
Water leaves a hose at a rate of 1.5 kg/s with a
speed of 20 m/s & is aimed at the side of a
car, which stops it. (That is, we ignore any splashing
back.) Calculate the force exerted by the water
on the car. Newton’s 2nd Law:
∑F = p/t
Initial p = mv, Final p = 0, m = 1.5 kg each t = 1 s
(water instantaneously stops before splashing back)
p = 0 – mv
F = (p/t) = [(0 -mv)/t] = - 30 N
This is the force the car exerts on the water.
By N’s 3rd Law, the water exerts an equal &
opposite force on the car!
Conceptual Exercise B (p 169)
Water splashes back!
Section 7-2: Conservation of Momentum
(Collisions!!)
• An Experimental fact (provable from Newton’s Laws):
For 2 colliding objects, (zero external force) the total
momentum is conserved (constant)
throughout the collision.
That is,
The total (vector) momentum before the collision
= the total (vector) momentum after the collision.
 Law of Conservation of Momentum
Law of Conservation of Momentum
The total (vector) momentum before the collision
= the total (vector) momentum after the collision.
 ptotal = pA + pB = (pA)+ (pB) = constant
or:
ptotal = pA + pB = 0
pA = mAvA, pB = mBvB, Initial momenta
(pA) = mA(vA), (pB) = mB(vB), Final momenta
or:
 mAvA + mBvB = mA(vA) + mB (vB)
Momentum Conservation
can be derived from
Newton’s Laws.
We are mainly interested in analyzing
collisions between 2 masses, say mA &
mB. We assume that a collision takes a
short enough time that external forces
can usually be ignored, so that all that
matters is the internal forces between
the 2 masses during the collision. Since, by
Newton’s 3rd Law
the internal forces are equal & opposite
The Total Momentum Will Be
Constant.
Momentum Conservation in Collisions
A Proof, using Newton’s Laws of Motion.
If masses mA & mB collide, N’s 2nd Law (in terms of momentum) holds for each:
∑FA = (pA/t) & ∑FB = (pB/t). pA & pB, = momenta of mA & mB
∑FA & ∑FB = total forces on mA & mB, including both internal + external forces.
Define the total momentum: P = pA + pB and add the N’s 2nd Law equations:
(P/t) = (pA/t) + (pB/t) = ∑FA + ∑FB.
By N’s 3rd Law, internal forces cancel & the right side = ∑Fext = total external force. So
(P/t) = (pA/t) + (pB/t) = ∑Fext
So, in the special case when the total external force is zero (Fext = 0), this is:
(P/t) = 0 = (pA/t) + (pB/t) or
P = pA + pB = 0
So, when the total external force is zero, (ΔP/Δt) = 0, or the
total momentum of the 2 masses remains constant during the collision!
P = pA + pB = constant = (pA) + (pB)
mAvA + mBvB = mA(vA) + mB (vB)
a or
Another Proof, using Newton’s 2nd & 3rd Laws
Two masses, mA & mB in collision:
Internal forces: FAB = - FBA by
Newton’s 3rd Law
Newton’s 2nd Law:
Force on A, due to B, small t:
FAB = pA/t = mA[(vA) - vA]/t
Force on B, due to A, small t:
FBA = pB/t = mB[(vB) - vB]/ t
Newton’s 3rd Law:
FAB = - FBA = F

mA[(vA) - vA]/t = - mB[(vB) - vB]/t
or: mAvA + mBvB = mA(vA) + mB (vB)  Proven!
So, for Collisions: mAvA + mBvB = mA(vA) + mB (vB)
This is known as the
Law of Conservation of Linear Momentum
“When the total external force on a system of
masses is zero, the total momentum of
the system remains constant.”
Equivalently,
“The total momentum of an isolated
system remains constant.”
• Collision: mAvA+ mBvB = mA (vA) + mB (vB) (1)
• Another simple Proof, using Newton’s Laws of Motion:
Newton’s 2nd Law for each mass:
Force on A, due to B, small t:
FAB = (pA/t) = mA[(vA) - vA]/t
Force on B, due to A, small t:
FBA = pB/t = mB[(vB) - vB]/ t
Newton’s 3rd Law: FBA = - F1AB

mA[(vA) - vA]/t = mB[(vB) - vB]/t
or
mAvA+mBvB = mA(vA) + mB (vB)  Proven!
Example: 2 pool balls collide (zero external force)
The vector sum is constant!
Momentum before =
Momentum after!
Ex. 7-3: Railroad Cars Collide: Momentum Conserved
Simplest possible example!! Car A, mass mA = 10,000 kg, traveling at
speed vA = 24 m/s strikes car B (same mass, mB = 10,000 kg), initially at rest (vB = 0).
The cars lock together after the collision. Calculate their speed v immediately after the collision.
Conservation of Momentum in 1dimension
Initial Momentum = Final Momentum
vA = 0, (vA) = (vB) = v so, mAvA+mBvB = (mA + m2B)v

v = [(mAvA)/(mA + mB)] = 12 m/s
Example: An explosion as a “collision”!
Momentum Before = Momentum After
mAvA + mBvB = mA(vA) + mB (vB)
Initially:
A
A
mA explodes, breaking up into mB & mC. So:
0 = mBvB + mCvC
Given 2 masses & 1 velocity, can calulate the other velocity
B
C
B
C
Example: Rocket Propulsion
Momentum Before = Momentum After
Momentum conservation works for a rocket if we consider the rocket & its
fuel to be one system, & we account for the mass loss of the rocket (Δm/Δt).
Example 7-4: Rifle recoil
Calculate the recoil velocity of a rifle, mass mR = 5 kg, that shoots a
bullet, mass mB = 0.02 kg, at speed vB = 620 m/s.
Momentum Before = Momentum After
Momentum conservation works here if we consider rifle & bullet as one system
mB = 0.02 kg, mR = 5.0 kg
(vB) = 620 m/s
Conservation of Momentum
mAvA + mBvB = mR(vR) + mB (vB)
This gives:
0 = mB (vB)  + mR(vR)
 (vR) = - 2.5 m/s
(to the left, of course!)
Example: An Archer
• A man (mM = 60 kg, vM = 0) stands on frictionless ice.
He shoots an arrow (mA = 0.5 kg, vA = 0) horizontally
& it leaves the bow at (vA) = 50 m/s to the right. What
velocity (vM) does he have as a result?
mM = 60 kg, vM = 0, mA = 0.5 kg, vA = 0
(vA) = 50 m/s, (vM) = ?
• The total momentum before the arrow is shot is 0 &
momentum is conserved so
The total Momentum after the arrow is shot is also 0!
pA + pM = 0 = (pA) + (pM)
or: mAvA + mMvM = mA(vA) + mM(vM)
mA(0) + mM(0) = mA(vA) + mM(vM)
0 = mA(vA) + mM(vM)
or: (vM) = - (mA/mM)(vA)
(vM) = - 0.42 m/s
(The minus means that the man slides to the left!)