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

• So ΣF = m(v – v0)/ Δt =– p = mv

• So ΣF = Δp/Δt

tmvmv

0

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 = sm

kgsmkg /2.1

0.5/120050.0

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 = - ΔPE1/2mv2 – 0 = -mgΔy

Solution

• Algebra gives us

• Δp = 0 – (70kg)(7.7m/s) = -540Ns

smmsmyygv /7.7)0.3)(/8.9(22 20

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 exceptd = 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 orKEf > 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‘ = gh2

Solution Continued

• Combining (i) and (ii) gives us

ghmMmv

mMmv 2'