Lecture 16 Chapter 11 Physcs I Energy Dsspaton Lnear Momentum Course webste: http://aculty.uml.edu/andry_danylov/teachng/physcsi Department o Physcs and Appled Physcs
IN IN THIS CHAPTER, you wll learn to use the concepts o mpulse and momentum. Today we are gong to dscuss: Chapter 10: Energy Prncple wth Losses: Secton 10.7-8 Chapter 11: Newton s 2 nd law (more general orm): Secton 11.1 Lnear Momentum: Secton 11.1 Impulse: Secton 11.1 Department o Physcs and Appled Physcs
Energy conservaton wth nonconservatve orces Consder an object experences conservatve and nonconservatve orces: F net F C F Snce the work-ke prncple works or ANY orces, the total work done on the object W net K Recall, rom the prevous class, a orce s conservatve, then K W C U U W C U WNC K W ( U U ) W NC K K NC K K U W NC NC y E F NC E F C ds W NC x Total Mech Energy at the nal pont As a nonconservatve orce, we are gong to have a rcton orce Total Mech Energy at the ntal pont W Department o Physcs and Appled Physcs r s k Mech. Energy converted to Thermal Energy (losses) (Work done by rcton s usually negatve)
Example Block sldng down Inclned Plane A block o mass m sldes down a plane o length l and angle α. Fnd the speed o the block at the bottom o the nclne plane assumng that t starts rom rest and the coecent o rcton μ s constant. k (Thermal energy s generated) α mg α Department o Physcs and Appled Physcs
Revew Knetc energy Gravtatonal potental energy U mgy K 1 mv 2 2 Potental energy o a sprng Total Mechancal Energy E K U U sprng 1 2 kx2 Conservaton o Mechancal Energy K U K U Wthout losses (no rcton) K U K U W NC Wth losses (wth rcton) Department o Physcs and Appled Physcs
Lnear Momentum Ch.11 Now we know how to use the energy approach to solve problems, so let s go back to our old good Newton s 2nd law, F=ma. F=ma worked great or pont objects, but t als to explan systems such as rockets, where mass changes wth moton (Rockets accelerate by ejectng mass backward). So, obvously, we need to mody (make t more general) our N. 2 nd law somehow. How? Department o Physcs and Appled Physcs
Let s rewrte N. 2 nd law n terms o lnear momentum m Let s pck an object and start descrbng t usng our old N. 2 nd law. v F ma m d v d dt (m v) dt d p dt We got a new structure mv, so let s gve t a nce name and symbol Lnear momentum s dened as the product o an object s mass and velocty: p mv Unts o momentum: kg m s Momentum s a VECTOR! So, nally, the Newton s 2 nd law wll look ths way: F Department o Physcs and Appled Physcs dp dt That s what actually Newton wrote. It s more general than our earler verson F=ma The rate o change o momentum s equal to the net orce So, a orce s requred to change momentum o an object.
ConcepTest The cart s change o lnear momentum Δp x s ntal nal x x Two Boxes/Momentum A) 20 kg m/s B) 10 kg m/s C) 0 kg m/s D) 30 kg m/s 10 1 102 Δp x = 10 kg m/s ( 20 kg m/s) = 30 kg m/s Fnal momentum Intal momentum Negatve ntal momentum because moton s to the let and v x < 0.
Impulse Ch.11 Now we have moded Newton s 2nd law, let s apply t to get somethng,.e. mpulse How? Department o Physcs and Appled Physcs
Consder Ball and Wall Collson (Beore) The gure shows a partcle wth ntal velocty n the x-drecton The partcle experences an mpulsve orce o short duraton Δt. (A large orce exerted or a small nterval o tme s called an mpulsve orce) The partcle leaves wth nal velocty n the +x-drecton. Instant o maxmum compresson Now, let s descrbe the collson usng our new moded N. 2 nd law (next slde) F d p dt Department o Physcs and Appled Physcs
F d p dt t p Fdt dp t p Impulse Durng the collson, objects are deormed because o the large orces nvolved. How to relate those orces wth a change n momentum? From Newton s 2 nd law: Integrate t: Fdt p p dp p p Impulse= change n momentum Dene Impulse as: t t Fdt Beore collson p Ater collson Force exerted on the ball p Recall: Integral Graphcal meanng s an area So, Impulse= area under F-vs-t curve Work-KE Prncple Smlarty Momentum Prncple p Impulse t t Fdt Department o Physcs and Appled Physcs End o Class
Example Tenns Ball/Racket Collson Force (kn) The orce exerted by a tenns racket on the ball (mass 56 g) durng a serve ( v 0 ) can be approxmated by the F vs tme plot below. What s the mpulse on the ball? What s the speed o the serve? 0 2 B A C 1 2 3 5 7 Tme (ms) 10 Momentum Prncple Department o Physcs and Appled Physcs Area under orce-tme curve s an mpulse: Area p 2 Area( ABC) v t t Fdt p mv v m p p 2kN 2ms 2( ) 2 4 N s 71m/s 0.056 kg 4N s
ConcepTest A 2.0 kg object movng to the rght wth speed 0.50 m/s experences the orce shown. What are the object s speed and drecton ater the orce ends? v =0.5m/s x Kckng Ball A) 0.5 m/s let B) At rest C) 0.5 m/s rght D) 1.0 m/s rght v -? x E) 2.0 m/s rght Δp x = x p x - p x = x p x = p x + x.. /
Impulse/Average orce The exact varaton o F wth tme s very oten not known (too complcated). How to get at least somethng? It s easer to nd an average orce. Let s keep the same mpulse And the same tme nterval Δt But nstead o a real orce we wll use an average orce t t Fdt F avg F avg t t p p Department o Physcs and Appled Physcs
Example How to avod broken legs or a cat? F avg t p Intal lnear momentum ( ) F, that s what can break cat s bones and the cat eels that and tres to reduce F as much as t can. Snce the cat alls rom a certan heght, Δp s gven and the cat cannot do anythng about that durng the all. By bendng legs and ncreasng an mpact tme, Δt. Havng a certan p, a cat by bendng ts lags tres to ncrease t (mpact tme), so that an mpact orce would be reduced. (ntutve knowledge o Physcs ) Department o Physcs and Appled Physcs Fnal lnear momentum
Example Average Force on a baseball The speed o a astball s about 40 m/s, and the speed o the ball comng o o player s bat or a home run s about 54 m/s. The ball (0.145kg) s n contact wth the bat or 1ms. What s the average Force exerted by the player? x F average F avg t p F average p t p p m( v t ( v )) t Pay attenton to drectons!!!!!!!! ( 0.145kg)[54 40] m / s F average 0. 001s F average 136, 300N n the drecton o v or x Department o Physcs and Appled Physcs
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ConcepTest Two Boxes/Momentum Two boxes, one heaver than the other, are ntally at rest on a horzontal rctonless surace. The same constant orce F acts on each one or exactly 1 second. Whch box has more lnear momentum ater the orce acts? A) the heaver one B) the lghter one C) both the same We know: p F av F p t av t p p t In ths case F and t are the same or both boxes! Both boxes wll have the same nal momentum. p 0 F lght F heavy
ConcepTest Two Boxes/velocty In the prevous queston, whch box has the larger velocty ater the orce acts? A) the heaver one B) the lghter one C) both the same Mv h p mv l Snce M m, then v l v h