CHAPTER 9 LINEAR MOMENTUM, IMPULSE AND COLLISIONS 103 Phy 1 9.1 Lnear Momentum The prncple o energy conervaton can be ued to olve problem that are harder to olve jut ung Newton law. It ued to decrbe moton o an object or a ytem o object. A new concept o lnear momentum can alo be ued to olve phycal problem, epecally the problem nvolvng collon o object. Lnear momentum o an object whoe ma m and movng at a velocty o v dened a p mv 103 Phy 2 1
Conder two partcle nteract wth each other By Newton 3rd law: F21 F12 F21 F12 0 ma 1 1ma 2 20 dv 1 dv 2 m1 m2 0 dt dt d( m1v1) d( m2v 2) 0 dt dt d( m1v1 m2v 2) 0 dt Lnear momentum p mv d p p dt 1 2 0.e., Total momentum reman contant p p p p tot 1 2 103 Phy 3 What can you tell rom th denton about momentum? Momentum a vector quantty. The heaver the object the hgher the momentum The hgher the velocty the hgher the momentum It unt kg.m/ What ele can ue ee rom the denton? Do you ee orce? 103 Phy 4 2
p mv VECTOR (kg m/ec.) p mv, p mv, p mv x x y y z z Relatonhp to orce: dp d m v m dv v dm dt dt dt dt ma F Ma uually doen t change wth tme Improved orm o 2 nd dp Law: F dt Probably wa Newton orgnal orm o 2 nd Law. Work even when ma doe change, e.g., rocket. 103 Phy 5 Momentum conerved n the abence o a net external orce. p p p tot 1 2 p p p p : Contant Conervaton o Total Momentum 1 2 1 2 103 Phy 6 3
Momentum Conervaton F EXT P t P 0 t F EXT 0 The concept o momentum conervaton one o the mot undamental prncple n phyc. Th a component (vector) equaton. - We can apply t to any drecton n whch there no external orce appled. You wll ee that we oten have momentum conervaton even when energy not conerved. 103 Phy 7 Example 9.1 An archer tandng at ret on rctonle ce. He re an arrow horzontally. m 1 =60kg,m 2 =0.5kg,v 2 =50m/ What the velocty o the archer ater rng the arrow? mv 1 1 mv 2 2 0 m 2 v1 v 2 0.42 m/ m1 I the arrow were hot at an angle wth theh horzontal, what the recol velocty mv 1 1 mv 2 2 coθ 0 m v v coθ 2 1 2 m1 READ Example 9.2 103 Phy 8 4
Example: Fgure below how a 2.0 kg toy race car beore and ater takng a turn on a track. It peed 0.50 m/ beore the turn and 0.40 m/ ater the turn. What the change P n the lnear momentum o the car due to the turn? (0.50 / ) ˆ v m j and v (0.40 m / ) ˆ P M v (2.0 kg )( 0.50 m / ) jˆ ( 1 kg m / ) jˆ P M v (2.0 kg )(0.40 m / ) ˆ (0.80 kg m / ) ˆ P P P P (0.80 kg m / ) ˆ( 1.0 kg m / ) jˆ (0.8 ˆ1.0 jˆ) kg m / 103 Phy 9 Example 9.2 A type o partcle, neutral kaon (K 0 ) decay (break up) nto a par o partcle called pon ( + and - ) that are oppotely charged but equal ma. Aumng K 0 ntally produced at ret, prove that the two pon mut have momenta that are equal n magntude and oppote n drecton. Th reacton can be wrtten a 0 K Snce th ytem cont o a K 0 n the ntal tate whch reult n two pon n the nal tate, the momentum mut be conerved. So we can wrte Snce K 0 produced at ret t momentum 0. 0 p K p 0 K p p 0 p Thereore, the two pon rom th kaon decay have the momenta wth ame magntude but n oppote drecton. p p K 0 p p p 103 Phy 10 5
9.2 Impule and Momentum The change o momentum n a gven tme nterval v v F k = 0 t v -v v -v t t F t m v m v p p p a, F m a m ( ) Impule = orce tme tme=change n momentum I Ft p ur r r r r r mv v 0 p mv mv 0 v r ur m ma F t t t t 103 Phy 11 Impule and Momentum A clac photograph taken by Arthur Edgerton at MIT n 1935 howng the moment o mpact between bat and otball. The huge orce exerted by the bat on the ball caue evere dtorton o the ball a t ht. 103 Phy 12 6
Impule and Momentum Newton 2nd Law: F t ( p) p F t p p p F t Impule I Ft Impule-momentum theorem A clac photograph taken by Arthur Edgerton at MIT n 1935 howng the moment o mpact between bat and otball. The huge orce exerted by the bat on the ball caue evere dtorton o the ball a t ht. 103 Phy 13 Impule and Momentum I t t In general, Fdt F F t F, t 1 t F Fdt t t I Ft Impule = area under curve Smple cae: contant Force I Ft L:\103 Phy LECTURES SLIDES\103Phy_Slde_T1Y3839\CH9Flah 103 Phy 14 7
103 Phy 15 Example; (a) Calculate the mpule experenced when a 70 kg peron land on rm ground ater jumpng rom a heght o 3.0 m. Then etmate the average orce exerted on the peron eet by the ground, the landng (b) t-legged and (c) wth bent leg. In the ormer cae, aume the body move 1.0cm durng the mpact, and n the econd cae, when the leg are bent, about 50 cm. We don t know the orce. How do we do th? Obtan velocty o the peron beore trkng the ground. 1 2 KE PE mg y y mgy v 2 mv Solvng the above or velocty v, we obtan gy 29.83 7.7 m/ 2 Then a the peron trke the ground, the momentum become 0 quckly gvng the mpule I Ft p p p 0 mv 70kg 7.7 m / 540N 103 Phy 16 8
Example; cont d In comng to ret, the body decelerate rom 7.7m/ to 0m/ n a dtance d=1.0cm=0.01m. 0 v The average peed durng th perod v 7.7 3.8 m / 2 2 d The tme perod the collon lat t v 0.01m 3 2.610 3.8 m / Snce the magntude o mpule I Ft 540N The average orce on the eet durng F I th landng t 540 5 2.110 N 3 2.610 2 2 How large th average orce? Weght 70kg 9.8 m / 6.910 N 5 2 F 2.1 10 N 304 6.9 10 N 304Weght I landed n t legged, the eet mut utan 300 tme the body weght. The peron wll lkely break h leg. 103 Phy 17 Example ; cont d What the knee are bent n comng to ret? The body decelerate rom 7.7 m/ to 0 m/ n a dtance d=50 cm=0.5 m. 0 v The average peed durng th perod tll the ame v 7.7 3.8 m / 2 2 The tme perod the collon lat d t change to v 0.5m 1 1.310 3.8 m / Snce the magntude o mpule I Ft 540N The average orce on the eet durng F I th landng t 540 3 4.110 N 1 1.310 2 2 How large th average orce? Weght 70kg 9.8 m / 6.910 N 3 2 F 4.110 N 5.96.910 N 5.9Weght It only 6 tme the weght that the eet have to utan! So by bendng the knee you ncreae the tme o collon, reducng the average orce exerted on the knee, and wll avod njury! 103 Phy 18 9
Example 9.3 A gol ball o ma 50 g truck by a club. The orce exerted on the ball by the club vare rom 0, at the ntant beore contact, up to ome maxmum value at whch the ball deormed and then back to 0 when the ball leave the club. Aumng the ball travel 200m, etmate the magntude o the mpule caued by the collon. m B v B The range R o a projectle v 2 B n 2θ R g B 200m Let aume that launch angle =45 o. Then the peed become: v 200g 1960 44 m / B Note the deormaton o the ball due to the large orce rom the club 103 Phy 19 Conderng the tme nterval or the collon, t and t, ntal peed and the nal peed are v 0 (mmedately beore the collon) v 44 m / (mmedately ater the collon) Thereore the magntude o the mpule on the ball due to the orce o the club I p mv mv 00.0544 2.2 kg m / B B What the average orce on the ball durng the collon wth the club? I F t 200 N 103 Phy 20 10