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Chapter 7: Linear Momentum
• Linear momentum is:
– the product of mass and velocity
– Represented by the variable p
– Equal to mv, where m is the mass of the object
and v is its speed. (p = mv)
– Conserved
– A vector
More on the momentum vector
• Because velocity is a vector, momentum is a
vector.
• The direction of the momentum depends on
the direction of the velocity.
• The magnitude of the vector is p = mv.
The units of momentum
• Because p = mv, we would expect the units of
momentum to be units of mass x units of
velocity.
• Indeed, the units of momentum in SI units is
kg*m/s, this unit has no special name.
• However, this tells you that you need
kilograms and meters per second.
Example
• A 100,000kg truck is traveling east at a speed
of 20m/s. Find the magnitude and direction of
the momentum vector.
Solution
• Remember the magnitude of momentum is
p = mv, so p = (100,000kg)(20m/s) =
2,000,000kgm/s.
• The direction of the momentum vector is the
same direction of the velocity vector, so east
in this example.
Changing momentum
• The only way to change the momentum of an
object is to either change its mass or change
its velocity.
• Remember that a change in velocity is called
acceleration, which requires a force.
• So, changing momentum requires a force.
Momentum and Newton’s 2nd
• Let’s start with ΣF = ma
– a = Δv/Δt
• So ΣF = m Δv/Δt
– Δv = v – v0
mv  mv0
• So ΣF = m(v – v0)/ Δt =
t
– p = mv
• So ΣF = Δp/Δt
Example of momentum change
• Water leaves a hose at a rate of 1.5kg/s at a
speed of 20m/s and is aimed at the side of a
car. The water stops when it hits the car. What
is the force exerted on the water by the car?
Solution
• Every second 1.5kg of water moves 20m/s.
• This means the water has a momentum of
p = mv = (1.5kg)(20m/s) = 30kgm/s, which
goes to 0 when it hits the car (because v = 0).
• F = Δp/Δt = (pfinal – pinitial)/Δt =
(0 – 30kgm/s )/1s = -30N
Question
• What happens if a car hits a semi head on?
Conservation of momentum
• Earlier I told you that momentum is
conserved.
• What that means is “the total momentum
before a collision equals the total momentum
after”
• Momentum before = momentum after
•
m1v1 + m2v2 = m1v1’ + m2v2’
• The ‘ is read as “prime” and means after.
The law of conservation of momentum
• The law states this “The total momentum of
an isolated system of bodies remains
constant.”
• System = a set of objects that interact with
each other
• Isolated system = the only forces present are
those between objects in the system.
Example
• A 10,000kg railroad car traveling at a speed of
24.0 m/s strikes an identical car at rest. If the
cars lock together as a result of the collision,
what is their speed afterward?
Solution
• We start with pinitial = pfinal
• pinitial = m1v1 + m2v2 = (10,000kg)(24m/s) +
(10,000kg)(0m/s) = 240,000kgm/s
• pfinal = (m1 + m2)v ’ (it’s m1 + m2 because the
cars linked up and became one object in the
eyes of physics)
• (m1 + m2)v ’ = 240,000
• v‘ = 240,000 / 20,000 = 12m/s
Example
• Calculate the recoil velocity of a 5.0kg rifle
that shoots a 0.050kg bullet at a speed of
120m/s.
Solution
• Momentum is conserved, so start with pi = pf
• What is pinitial?
• Ask yourself the following:
– What is the starting speed of the bullet before
firing?
– What is the starting speed of the rifle before
firing?
• pinitial = mBvB + mRvR = 0
Solution Cont.
• Now we need to set up pfinal and set it equal to
0.
• pfinal = mBvB’ + mRvR’ =
(0.050kg)(120m/s) + (5.0kg)(v ’R) = 0
• Solve for v ’R
• v ’R = 0.050kg 120m / s 

5.0kg
 1.2m / s
Interpretation of data
• Does our answer make sense?
• The rifle is moving 100 times slower than the
bullet. How does that work?
Impulse
• During a collision the force on an object
usually jumps from 0 to very high in a very
short amount of time and then abruptly
returns to 0.
• Let’s start with ΣF = Δp/Δt
• And solve it for Δp,
• Δp = FΔt = impulse
When do we care about impulse?
• Impulse is very helpful when we are working
with large forces that occur in a very short
amount of time.
• Examples:
– A bat hitting a ball
– Two particles colliding
– Brief body contact
Example
• Calculate the impulse experienced when a
70kg person lands on firm ground after
jumping from a height of 3.0m. Then estimate
the average force exerted on the person’s feet
by the ground, if the landing is stiff-legged and
again with bent legs. With stiff legs the body
moves 1cm during impact. With bent legs the
body moves 50cm.
Solution
• We don’t know F so we can’t solve for impulse
directly. But we know that impulse = Δp, and
Δp = mv2 – mv1.
• This means we need to find v1 (we know that
v2 will be 0)
• We can find v using conservation of energy:
ΔKE = - ΔPE
1/2mv2 – 0 = -mgΔy
Solution
• Algebra gives us
v  2 g  y  y0   2(9.8m / s )(3.0m)  7.7 m / s
2
• Δp = 0 – (70kg)(7.7m/s) = -540Ns
Straight Legged
• In coming to a rest the body goes from 7.7m/s
to 0 in a distance of 0.01m.
• The average speed during this period is
(7.7m/s + 0)/2 = 3.8m/s = v
• Δt = d/v = 0.01m / 3.8m/s = 2.6E-3 s
• Impulse = FΔt = -540Ns so,
• F = -540Ns / 2.6E-3s = 2.1E5N
Bent Legs
• This is done just like the straight leg except
d = 0.50m so Δt = 0.50m / 3.8m/s = 0.13s so,
• F = 540Ns/0.13s = 4.2E3N
Elastic Collisions
• An elastic collision is a collision in which
kinetic energy is conserved.
• This means both kinetic energy and
momentum are conserved.
• This is handy, because it gives us 2 equations
we can solve simultaneously to find the two
unknowns (the speed of each object after the
collision)
The Math
• The two equations we need to solve are:
• v1 – v2 = v ’2 – v ’1 (derived from conservation of
kinetic energy) and
• m1v1 + m2v2 = m1v ’1 + m2v ’2 (the conservation
of momentum equation
• The strategy is to solve the first equation for
either v’2 or v ’1 plug that into the second
equation.
Example
• A billiard ball of mass m moving with speed v,
collides head-on with a second ball of equal
mass at rest (v2 = 0). What are the speeds of
the two balls after the collision, assuming it is
elastic?
Solution
• Conservation of momentum gives us:
• mv = mv’1 + mv ’2, which we can divide by m to
get:
• v = v ’1 + v ’2 (call this *)
• Now we use the first equation,
v1 – v2 = v ’2 – v ’1,
• v = v ’2 – v ’1 (call this #)
• * - # gives us 0 = 2v ’1, so v ’1 = 0
Solution
• We can now substitute v’1 = 0 into
v = v ’2 – v ’1 and solve for v ’2
• v‘2 = v + v ’1 = v + 0 = v
• To summarize,
• Before collision: v1 = v and v2 = 0
• After collision: v’1 = 0 and v ’2 = v
Inelastic Collisions
• An inelastic collision is a collision in which
kinetic energy is not conserved.
• If it is not conserved, then either KEf < KEi or
KEf > KEi
• In the former case, the energy of the objects is
wasted as heat energy, sound energy,
potential energy, or crushing the objects.
• In the later case, chemical or nuclear potential
energy is released. (Think explosives)
Completely Inelastic Collisions
• When two objects completely stick together
as a result of the collision, the collision is said
to be completely inelastic.
• When this happens, the conservation of
momentum becomes
m1v1 + m2v2 = (m1 + m2)vf
Ballistic Pendulum
• The ballistic pendulum is a device used to
measure the speed of a projectile, such as a
bullet. The projectile, of mass m, is fired into a
block of mass M, which is suspended like a
pendulum (M > m). As a result of the collision,
the pendulum-projectile combination swings
up to a maximum height h.
Solving the Ballistic Pendulum
• Let us determine the relationship between the
initial speed of the projectile, v, and the
height h.
• mv = (m + M)v’
(i)
• KE1 + PE1 = KE2 + PE2 or
• ½(m + M)v ’2 + 0 = 0 + (m + M)gh (ii) so
• v‘ = 2 gh
Solution Continued
• Combining (i) and (ii) gives us
mM '
mM
v
v 
m
m
2 gh