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
Mar 19, 2016
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'