Momentum & Impulse

Momentum (p) “inertia of motion” p = mv Units for momentum  Kg*m/s Vector Quantity One way of looking at it…How much an object in motion… wants to stay in motion Lot of momentum  hard to stop

How can you change an object’s momentum?? Newton’s 2 nd Law states a net force causes an acc. An acc. Changes the velocity Changing the velocity, changes the momentum

Impulse Momentum Theorem Applying a force over a time interval changes the momentum ▫F changes v, therefore (mv) changes Never looked at a relationship between ‘F’ and ‘t’ F x t = Impulse ▫Since an impulse changes ‘v’, this changes momentum Ft = Δ(mv)  Impulse-Momentum Theorem

Newton’s 2 nd Law reworked… F = ma and a = (Δv/t) F= m(Δv/t) then multiply both sides by ‘t’ Ft = mΔv which is the same thing as Ft = Δ mv Impulse- Momentum Theorem is just Newton’s 2 nd Law written a different way

Examples Boxing gloves vs. MMA gloves deo/future-cars-nido/ deo/future-cars-nido/ Features on a car?? Pillow punch vs. brick punch Bungee jump w/ elastic cord vs. rigid cord Egg toss competition “rolling with a punch”

Bouncing? Greater Δ(mv) than just stopping an object?? Why??  ….greater Δv ▫ Going from -5 m/s to 5 m/s is a greater velocity change than going from -5 m/s to 0 m/s, therefore greater Δmv and impulse

Pelton Wheel Example Paddles are cups instead of just flat planks Allows water to change directions Greater Δmv of water which means more impulse and wheel is turned much more effectively

Example Problem

Other Examples Karate chop eature=relatedhttp:// eature=related whttp:// w Mr. Schober gets assaulted by strangers… Story

Conservation of Momentum If no outside force is applied, then the total amount of momentum in a closed system will remain constant. ▫Only external forces can change momentum. Σp i = Σp f m 1 v 1i +m 2 v 2i …= m 1 v 1f + m 2 v 2f …

Conservation of Momentum p ai = m(v) p bi = m(0) p af = m(0) p bf = m(v)

Conservation of Momentum Momentum is conserved for all objects in the interaction, even if one doesn't stop p ai + p bi = p af + p bf

Is momentum conserved here? Yes, due to the vector nature of momentum.

Is momentum conserved? Initial velocities of both objects is 0. p ai = m a (0) p bi = m b (0) Σp i = 0 AB

Is momentum conserved? p af = m a (-v a ) p bf = m b (v b )  p f = 0 Σp i = Σp f, so momentum is conserved!! A B  p f = m a (-v a ) + m b (v b )

Why do internal forces result in momentum being conserved? When Girl A pushes on Girl B, according to Newton’s 3rd Law, Girl B pushes on Girl A ▫How much? These forces are equal in magnitude and opposite in direction The time over which these forces act is exactly the same ▫Only while the girls are in contact, in this case

How does a gun work?

How does the gun work? Only forces are internal (no net external forces are adding impulse to the system) The momentum of both will add up to zero (bullet is +, gun is -)

Why do internal forces result in momentum being conserved? Impulse is equal in magnitude but opposite in direction ▫I = ( ΣF)(Δt) ▫ Forces are equal and opposite, times are equal Δp is equal in magnitude, opposite in direction, resulting in Σp = 0!!

Collisions Inelastic ▫Any collision in which momentum is conserved but kinetic energy is not ▫Most ‘real’ collisions are of this kind ▫KE is not conserved because some is lost to the deformation ▫ m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f Perfectly Inelastic ▫Objects collide and stick together ▫KE not conserved ▫ m 1 v 1i + m 2 v 2i = (m 1 + m 2 ) v f Elastic ▫Both momentum and KE are conserved ▫“perfectly “elastic collisions only occur in real life at the subatomic level, but will treat any collision labeled as “elastic” as being ‘perfectly’ elastic ▫Collisions between billiard balls or between air molecules and the surface of a container are both highly elastic ▫No Energy lost to deformation ▫ m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f And ▫ ½m 1 v 1i 2 + ½m 2 v 2i 2 = ½m 1 v 1f 2 + ½m 2 v 2f 2 ▫When combining these two and reducing we get…. V 1i – v 2i =-(v 1f – v 2f ) ball-slow-motion.htmhttp:// ball-slow-motion.htm - Golf Ball during a surprising inelastic collision h?v=pQ9NiazPYI8 --- baseball h?v=pQ9NiazPYI8

Example problem

Problem Solving #1 A 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is at rest. Find the velocity of the fish immediately after “lunch”. System is both fish, and collision is perfectly inelastic so ….. Σp i = Σp f (m 1 v 1i ) + (m 2 v 2i ) = (m 1 + m 2 )v f 6(1) + (2)(0) = (6+2) v f V f =6/8 =.75 m/s

Problem Solving #2 Now the 6 kg fish swimming at 1 m/sec swallows a 2 kg fish that is swimming towards it at 2 m/sec. Find the velocity of the fish immediately after “lunch”. System is both fish, so…. Σp i = Σp f (Σ(mv)) i = (Σ(mv)) f (m 1 v 1i ) + (m 2 v 2i ) = (m 1 + m 2 )v f (6 kg)(-1 m/s) + (2 kg)(2 m/s) = (6 kg + 2 kg)(v f ) -6 kg. m/s + 4 kg. m/s = (8 kg)(v f ) v f = -2 kg. m/s / 8 kg v f = -.25 m/s

Collisions in 2-D (more to be posted later) Σp xi = Σp xf Σp yi = Σp yf Momentum is a vector, so momentum must be conserved in the x-direction, and in the y- direction

Inelastic in 2-D 1 kg.5 kg 2.2 m/s 33° 1.5 m/s ??

Perfectly Inelastic in 2-D 1 kg.5 kg 2.5 m/s 1.3 m/s 1.5 kg