Momentum

Essential idea: Conservation of momentum is an example of a law that is never violated.

Nature of science: The concept of momentum and the principle of momentum conservation can be used to analyse and predict the outcome of a wide range of physical interactions, from macroscopic motion to microscopic collisions.

Topic 2: Mechanics2.4 – Momentum and impulse

Understandings: • Newton’s second law expressed in terms of rate of

change of momentum • Impulse and force – time graphs • Conservation of linear momentum • Elastic collisions, inelastic collisions and explosions

Applications and skills: • Applying conservation of momentum in simple

isolated systems including (but not limited to) collisions, explosions, or water jets

• Using Newton’s second law quantitatively and qualitatively in cases where mass is not constant

• Sketching and interpreting force – time graphs • Determining impulse in various contexts including

(but not limited to) car safety and sports • Qualitatively and quantitatively comparing situations

involving elastic collisions, inelastic collisions and explosions

Topic 2: Mechanics2.4 – Momentum and impulse

Guidance: • Students should be aware that F = ma is the

equivalent of F = p / t only when mass is constant

• Solving simultaneous equations involving conservation of momentum and energy in collisions will not be required

• Calculations relating to collisions and explosions will be restricted to one-dimensional situations

• A comparison between energy involved in inelastic collisions (in which kinetic energy is not conserved) and the conservation of (total) energy should be made

Topic 2: Mechanics2.4 – Momentum and impulse

Data booklet reference: • p = mv• F = p / t• EK = p 2 / (2m)

• Impulse = F t = p

Topic 2: Mechanics2.4 – Momentum and impulse

International-mindedness: • Automobile passive safety standards have been

adopted across the globe based on research conducted in many countries

Theory of knowledge: • Do conservation laws restrict or enable further

development in physics? Utilization: • Jet engines and rockets• Martial arts• Particle theory and collisions (see Physics sub-

topic 3.1)

Topic 2: Mechanics2.4 – Momentum and impulse

Aims: • Aim 3: conservation laws in science

disciplines have played a major role in outlining the limits within which scientific theories are developed

• Aim 6: experiments could include (but are not limited to): analysis of collisions with respect to energy transfer; impulse investigations to determine velocity, force, time, or mass; determination of amount of transformed energy in inelastic collisions

• Aim 7: technology has allowed for more accurate and precise measurements of force and momentum, including video analysis of real-life collisions and modelling/simulations of molecular collisions

Topic 2: Mechanics2.4 – Momentum and impulse

Momentum

The linear momentum of an object of mass m moving with a velocity v is defined as the product of the mass and the velocity

The symbol for momentum is the letter pp = m v SI units for momentum are kg m/s or N

s Momentum is a vector quantity The direction of the momentum is the

same as the direction of the velocity

Momentum

So let’s say you have a 10 kg model car, and it is going 5 m/s.

What is it’s momentum? p = mv = (10 kg )(5 m/s) = 50 kg

m/sec

Impulse In order to change the momentum of an

object, a force must be applied The time rate of change of momentum of an

object is equal to the net force acting on it Gives an alternative statement of Newton’s

second law

Impulse cont. When a single, constant force acts on

the object Impulse = Δp = mΔv = FΔt FΔt is defined as the impulse (units N

sec) Δp then is also defined as the impulse The impulse is the average force x the

time interval over which it acts Impulse is a vector quantity, the direction

is the same as the direction of the force

This side mv the same

Impulse = Δp = mΔv = FΔt

Impulse-Momentum Theorem

The theorem states that the impulse acting on the object is equal to the change in momentum of the object FΔt = Δp

If the force is not constant, use the average force applied

You might also see it rearranged and used like this:

Average Force in Impulse

The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval

Average Force cont.

The impulse imparted by a force during the time interval Δt is equal to the area under the force-time graph from the beginning to the end of the time interval

Or, the average force multiplied by the time interval

ImpulseexampleThe force shown in the force-time diagram to the right acts on a 1.5 kg object. Finda) The impulse of the

force.Area under curve = (2)(3) + ½(2)(2) = 8 N-s

b) The final velocity of the object if it is initially at rest.

FΔt = mΔv8 N-s = 1.5 (vfinal – vinitial) = 1.5 vfinal

Vfinal = 8 N-s / 1.5 kg = 5.3 m/s

Impulse Applied to Auto Collisions

The most important factor is the collision time or the time it takes the person to come to a rest Lengthening the time of impact will

reduce the chance of death in a car crash

Ways to increase the time Seat belts, air bags

so for a given change of momentum, the force can be reduced if the time is extended.

Air Bags

The air bag increases the time of the collision

It will also absorb some of the energy from the body

It will spread out the area of contact decreases the

pressure helps prevent

penetration wounds

Example Prob. 1 2200 kg vehicle with vi = 26 m/s Can be stopped in 21 sec by gently applying brakes. Can be stopped in 3.8 sec if slam on brakes. Can be stopped in 0.22 sec if hits brick wall. What average force is exerted on the vehicle each

time? FΔt = Δp = pf – pi =(2200)(0) – (2200)(26) =

-5.7 x104 kg m/sec So F = -5.7 x 104 / Δt Case 1 21 sec F = -2.7 x 103 N braking force Case 2 3.8 sec F = - 1.5 x 104 N Case 3 0.22 sec F = -2.6 x 105 N

Example Problem 2

725 kg car, going 115 km/hour east Convert Find momentum: p = mv = 725(31.9) =

23159.7 kg m/sec If a 2nd car with mass 2175 kg has the

same p, what is its velocity? p = mv 23159.7 = 2175 v Solve for v = 23159.7/2175 = 10.6 m/s

Example Prob. 2 (continued)

The driver of the 725 kg car suddenly applies the brakes hard for 2 sec. As a result, an average force of 5000 N is exerted on the car to slow it down.

What is the change in momentum? (in other words, what is the direction and magnitude of the impulse?)

FΔt = Δp so FΔt = (5000 N) (2 sec) = 10,000 N s to the west (he was driving east).

Example Prob. 2 (continued)

Make before and after sketches and find final momentum and velocity.

FΔt = Δp = -10,000 N s brakingVi = 31.9 m/s vf = 13160/725 = 18 m/spi = 23160 N s pf = 13160 N s

Example prob. 3 A driver of a 240 kg snowmobile accelerates, which

results in a force being exerted that speeds up the snowmobile from 6 to 28 m/s over a time interval of 60 seconds.

Give the initial and final situations Initial p = mvi = (240)(6) = 1440 kg m / sec Final p = mvf = (240)(28) = 6720 kg m / sec What is the change in momentum? Δp = 6720-1440 = 5280 kg m /sec (this is the

impulse) What is the magnitude of the average force that is

exerted on the snowmobile by the engine? F = Δp/Δt = 5280 / 60 sec = 88 N

Conservation of Momentum

Momentum in an isolated system in which a collision occurs is conserved A collision may be the result of

physical contact between two objects “Contact” may also arise from the

electrostatic interactions of two charged particles (more on that in January).

An isolated system will have no external forces

Conservation of Momentum The principle of conservation of

momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

The total momentum of an isolated system of objects is conserved regardless of the nature of the forces between the objects

Conservation of Momentum, cont.

Mathematically if m1 and m2 collide:

m1v1i + m2v2i = m1v1f + m2v2f Momentum is conserved for the

system of objects, not individual objects

The system includes all the objects interacting with each other

Assumes only internal forces are acting during the collision

Can be generalized to any number of objects

Types of Collisions Momentum is conserved in any

collision Inelastic collisions

Perfectly inelastic collisions occur when the objects stick together after they collide.

Elastic collisions Example: perfectly elastic balls that

bounce off each other without deforming.

Actual collisions Most collisions fall between elastic

and perfectly inelastic collisions

More About Perfectly Inelastic Collisions

When two objects stick together after the collision, they have undergone a perfectly inelastic collision

Conservation of momentum becomesm1v1i + m2v2i = (m1+ m2)vf

total

Great video on momentum from space

Richard Garriott Space Video Blog

Some General Notes About Collisions

Momentum is a vector quantity Direction is important Be sure to have the correct signs

In order for conservation of momentum to occur it needs to be a closed system that does not gain or lose mass.

Net external force on the system must be zero for cons. of momentum to hold.

Elastic Collisions

Momentum is conserved

m1v1i + m2v2i = m1v1f + m2v2f

Problem Solving for One -Dimensional Collisions

Set up a coordinate axis and define the velocities with respect to this axis It is convenient to make your axis

coincide with one of the initial velocities

In your sketch, draw all the velocity vectors with labels including all the given information

Sketches for Elastic Collision Problems

Draw “before” and “after” sketches

Label each object include the

direction of velocity

keep track of subscripts

Sketches for Perfectly Inelastic Collisions

The objects stick together

Include all the velocity directions

Don’t forget: The “after” collision combines the masses!!

m1v1i + m2v2i = (m1+ m2)vf total

Types of collisions

Problem Solving for One-Dimensional Collisions, cont.

Write the expressions for the momentum of each object before and after the collision Remember to include the appropriate

signs Write an expression for the total

momentum before and after the collision Remember the momentum of the

system is what is conserved (not the velocity…)

Perfectly Inelastic Collison Example on next page

Note that once the cars collide, they stick together so the total mass afterwards is the combined mass of the two cars.

Rear-EndExample A 1875 kg car with a 23 m/s velocity rear-ends a

1025 kg car going 17 m/s on ice (FF =0), so momentum is conserved. The cars stick together and go off in the same direction afterwards. Find their final velocity.

m1v1i + m2v2i = (m1+ m2)vf total

Head-onExample

A 1875 kg car with a 23 m/s velocity has a head-on collision with a 1025 kg car going -17 m/s on ice (FF =0), so momentum is conserved. The cars stick together and go off in the same direction afterwards. Find their final velocity.

Recoil Example

• Initially two skaters are at rest, and then they push off each other and go in opposite directions on the frictionless ice.

p1i + p2i = p1f + p2f

0 + 0 = p1f + p2f

so - p1f = p2f

• The backward motion of the skater on the left is an example of recoil.

• Another example is how a gun kicks backwards when the bullet is shot out in the forward direction.

Recoil example – lab carts

Two lab carts on a track are pushed together with a spring mechanism compressed between them. Upon release, the 6 kg cart recoils one way with a velocity of 0.1 m/s while the 3 kg cart goes the opposite direction. What is the velocity of the 3 kg cart?

m1v1i + m2v2i = m1v1f + m2v2f

0 + 0 = (6)(-0.1) + (3)(v2f )

0.6 = 3v2f

v2f = 0.2 m/s

Propulsion in space gas

astronaut

An astronaut at rest in space fires a thruster pistol that expels 35g of hot gas at 875 m/s. The combined mass of astronaut and pistol is 84 kg. How fast and in what direction is the astronaut moving after firing the pistol?

pi = pcf + pdf

0 = mcf vcf + mdf vdf

vcf = - (mdf vdf ) / mc

Vcf = -(0.035)(-875)/84

vcf = 0.36 m/s

Uranium decay – 1D A single uranium atom has a mass of 3.97 x

10-25 kg. It decays into the nucleus of a thorium atom by emitting an alpha particle at a speed of 2.10 x 10-7 m/s. The mass of an alpha particle is 6.68 x 10-27 kg. What is the recoil speed of the thorium nucleus?

mUvU = mαvα + mThvTh and vUi = 0 m/s so and Substitute in the values from above

Rocket Propulsion

The operation of a rocket depends on the law of conservation of momentum as applied to a system, where the system is the rocket plus its ejected fuel This is different than propulsion on

the earth where two objects exert forces on each other road on car train on track

Rocket Propulsion, cont.

The rocket is accelerated as a result of the thrust of the exhaust gases

This represents the inverse of an inelastic collision Before: rocket + fuel After: rocket in one direction, fuel in the

other Momentum is conserved Kinetic Energy is increased (at the expense

of the stored energy of the rocket fuel)

Two Dimensional Momentum Problems The law of conservation of momentum holds for all

closed systems in 1-D, 2-D and 3-D. For a general collision of two objects in three-

dimensional space, the conservation of momentum principle implies that the total momentum of the system in each direction is conserved m1v1ix + m2v2ix = m1v1fx + m2v2fx

m1v1iy + m2v2iy = m1v1fy + m2v2fy

Use subscripts for identifying the object, initial and final, and x,y components.

Glancing Collisions

Now since ball C and D are not moving in a simple x or y direction following the collision, we need to find the x-component and y-component of their velocities.

Glancing Collisions

The “after” velocities have x and y components

Momentum is conserved both in the x direction and in the y direction Apply separately to each direction

Problem Solving for Two-Dimensional Collisions

Set up coordinate axes and define your velocities with respect to these axes It is convenient to choose the x axis to coincide

with one of the initial velocities In your sketch, draw and label all the velocities and

include all the given information Write expressions for the x and y components of the

momentum of each object before and after the collision

Write expressions for the total momentum before and after the collision in the x-direction, then repeat for the y-direction

To solve 2-D momentum problems, you may have up to three equations, conservation of momentum in x-direction, conservation of momentum in y-direction, and conservation of energy (if elastic).

Example

Car C collides with moving car D. The two cars stick together. In what direction and with what speed do they move after the collision?

Given: Unknown:mC = 1325 kg vfx

mD = 2165 kg vfy

vCiy = 27.0 m/s θvDix = 11.0 m/s

Example (continued)

pCi = mC vCiy = (1325)(27.0) = 3.58 x 104 kg m/sec (north)This is also equal to p(C+D)fy

pDi = mD vDix = (2165)(11.0)

= 2.38 x 104 kg m/sec (east)This is also equal to p(C+D)fx

Find p(C+D)f total using Pythagorean

4.30 x 104 kg m/secFinal speed = pf / mtotal = 4.30 x 104 / (1325+2165) = 12.3 m/sNow solve for θ θ = tan-1 (pfy / pfx ) = tan-1 (3.58 x 104 / 2.38 x 104 ) = 56.4o

Uranium decay 1 2D example 3

An unstable radioactive nucleus with a total mass of 15 x 10-27 kg initially at rest experiences a radioactive decay and splits into 3 particles.

m1 = 5 x 10-27 kg moves along y-axis at 6x106 m/s m2 = 7 x 10-27 kg moves along the –x-axis at 3x106

m/s Therefore m3 = 15x10-27 - 5x10-27 - 7x10-27 = 3x10-27

kgDetermine the angle and velocity of the third mass x-dir: 0 = 0 – m2v2f + m3v3fx y-dir: 0 = m1v1f + 0 + m3v3fy

So v3fx = m2v2f / m3 = (7E-27)(3E6)/(3E-27) = 7E6 m/s

v3fy = -m1v1f / m3 = (5E-27)(6E6)/(3E-27) = 10E6 or 1E7 m/s Pythagorean v3fx and v3fy to get v3f = 1.22E7 m/s

And θ = tan-1 ( 10E6/7E6) = 55o

2θ

Example –

car and truck collision A 985 kg car traveling south at 20 m/s hits

a ?? kg truck traveling west at 18 m/s. After the collision they stick together and travel with a final momentum of 4E4 kg m/sec at an angle of 45 degrees. What is the mass of the truck?

X- direction: mc vcix + mt vtix = (mc + mt )vf = pf

0 + 18mt = 4x104 cos 45 So mt = 4x104 cos 45 / 18 = 1571 kg or

about 1600 kg

45

car

truck

Example – 2 cars , 2D

How fast and in what direction are the cars going (stuck together) after the collision?

X direction: m2 v2ix = (m1+m2) vfx

vfx = m2 v2ix / (m1+m2) = (1732)(31.3)/(1732+1383)vfx = 17.4 m/s

Y direction: m1 v1iy = (m1+m2 ) vfy

vfy = m1 v1iy / (m1+m2) = (1383)(11.2)/(1732+1383) vfy = 4.97 m/s

= 18.1 m/s final velocity

θ

m1 =1383 kg, 11.2 m/sm2 =1732 kg, 31.3

m/s

Angular Momentum Momentum resulting

from an object moving in linear motion is called linear momentum.

Momentum resulting from the rotation (or spin) of an object is called angular momentum.

Angular momentum

Angular momentum is important because it obeys a conservation law, as does linear momentum.

The total angular momentum of a closed system stays the same.

Calculating Angular MomentumAngular momentum is calculated in a similar way to linear momentum, except the mass m and velocity v are replaced by the moment of inertia I and angular velocity ω

Angularvelocity

(rad/sec)

Angularmomentum(kg m/sec2)

L = I w

Moment of inertia(kg m2)

SI Unit of Angular Momentum: kg·m2/s

Requirement: The angular speed mustbe expressed in rad/s.

Moments of Inertia

~0

1. You are asked for angular momentum.

2. You are given mass, shape, and angular velocity. Hint: both rotate about y axis.

3. Use L= I, Ihoop = mr2, Ibar = 1/12 ml2 (where l is the length of the bar)

An artist is making a moving metal sculpture. She takes two identical 1 kg metal bars and bends one into a hoop with a radius of 0.16 m. The hoop spins like a wheel. The other bar is left straight with a length of 1 meter. The straight bar spins around its center. Both have an angular velocity of 2 rad/sec. Calculate the angular momentum of each and decide which would be harder to stop.

Angular Momentum example

3. For hoop: Ihoop= (1 kg) (0.16 m)2 = 0.026 kg m2

Lhoop= I ω = (0.026 kg m2) (2 rad/s) = 0.052 kg m2/s

4. Solve bar: Ibar = (1/12)(1 kg) (1 m)2 = 0.083 kg m2

Lbar = I ω = (0.083 kg m2) (2 rad/s) = 0.166 kg m2/s

5. The bar has more than 3x the angular momentum of the hoop, so it is harder to stop.

Example A Satellite in an Elliptical Orbit

An artificial satellite is placed in an elliptical orbit about the earth. Its pointof closest approach is 8.37x106mfrom the center of the earth, andits point of greatest distance is 25.1x106m from the center ofthe earth.

The speed of the satellite at the perigee is 8450 m/s. Find the speedat the apogee.

Angular momentum conserved

Substitute in

and

Angular Momentum Example (continued)

sm2820

m 1025.1

sm8450m 1037.86

6

A

PPA r

vrv