Question

Mechanics of Melhines -lon object Robles through 360° it travels a distance of a r chie circunference, do circle. 7 So sub into o 55 =) o. 2 a=251 Relains SET2 Mechanics of machines Assessment1 Attempt both questions Packages in an automobile parts supply house are transported to the louding dock by pushing them along on a roller track with very little friction. At the instant shown packages and are at rest and package A has a velocity of 2 ms. Knowing the coefficient of restitution between the packages is 0.3, determine a) the velocity of package Cafter A hits B and B hits C. ) the velocity of A after it hits B for the second time eithese loss of energy in the system (18 marks) (9 marks) (6 marks) 2 m/s Looke load is pulled up 4 plane inclined at 3r to the horizontal by a horizontal force P.

Answer 1

Solution Known: 3 masses A, B and c of 8kg, 48 and 8 kg respectively, Assumption: friction is negligible. given: Cofficient og Re sitution (e) = 0.3; VA= 2 ms. Analysis: A (given VB=0 Veo A B [initially] maaskg my=4kg Me=8kg. go, By momentum conservation Blw Amd B. Ma VA tmy VB = Mava tmavé . . VAANA & Velocity of A and B after . first collisim. 8x2 + 4 X0 = 8V + 4V6 + 16 = 8 VA! +4VB? D és VB'-VA VA-VB 0.3 = Vp!-VA. → We'VA= 0.6 2-0 . By Ey? (1) and (2) VA? = 1:13 mo and VR' = 1.73 mis and I Vel =VC) No collisim), VA'= .3mjs VR!= 1.73 ms V=0 14 B o Cafter a maa 8kg 1. mg 24kg me arug first collisim Now, Collision Blw Band C will occurs; B/w A and B so, By momentum conservation between 6 cmnd c 4x1.73 + 8X0 = 48% V" + 8xve" here vg" and voll is velocity of Band e respect CS$canned with CamScanner ively after and collisim. ME

4V2'+8V! = 6.92 -☺ e Vie! - VAT 23% V" -ve" 3 V6 17–Vg" = 0.BDO 7.730 1 By Eq? (3) and (4), we get VR' = 0), 2306 mis md Vell = 0.7496 mjs and TVA EVA] No Collision. Now, VA> Ve! So, A will callide to B' thus again By momentum conservation befocen A ad B; VA'= 1,13 m/s Ve"=0-3304 m/s Ve" = 0.7496 me B15 0 Pafter second collision 88 че 1 8kg. B/W B and c ] M:VA + Mb. VB" = mavat move VA!", VB." is the velouty of Aand B after 3rd collisiom blw A and B. 3 8X1.13 # 480.2306 = 84" + 4 Ve!!! 7 8Vqall 407" = 9.9624 - oces Vg" - VAL 30;3= Ug"-VA" Very_va" = 0.2698 VALI VG 1.13–0:2306 7. V"VA" = 0.2698 By Eq? (s) and (6); we get Veni = 1.01 ms od Vall = 0.74 01 mis O; No collision with '80, VC" = Ve" )= 0.7496 mis CS$canned with CamScanner

c LA - Up!"=0.740)MB NOM=10) ms VC"=0.7496 mis after 3rd collising ( Blw0 AcmdB] 8kg. O bere Now, Võ > Vell! So, fourth collision will occurs between B amd c. Thus By momentum conservcetim blw Band c. mbi ve + move" = move" + move Vig'""!, Vang 1 velocity of B and c opter 4thcollision Benveen B mod c. 4X107 + 8X0.7496 = 4.V!!! + B VC "Y Aval+8V6") = 10.0368 - niea vietu Vg7 0-3 = Ve"_V3"Y_ Veu Ve "=0.07812 – Up" vil 7.01-0.7496 By Eq? (7) and (8); we get V.") = 0.7843 mis, ad Ve" = 0.8624 mia. No collision with a 80, V VA! | 0.740 mig VA!".0.740)M18 215" 5 0,7843 mp3 Ve"'= 0.8624718 1 (Bj 2 godic) 829 : B C after 4th colligio .8kg 4kg : Vc > Vig > V4 80, No more collision occung further. CS$canned with CamScanner

9) Velocity of c, after A hits B and B hits c. 10:Vetontyg That is, velocity of C zter Second collision Blw Band C. Ve" = 0.74-96 m3] Answer b) Velocity gia getes hits B for second time :- . That 18, velocity g 1, after 3rd collision blw A and B. | Val" = 0.740) mis Jansman 2 c) % loss of energy in the system. > initial energy of system = Le monde et me vem tot mave Ej = 2.32) +0+0 Ngave so → E; = 165 > fimal Energy g system (EJ) = f mo va ty ma van de oven + 6 = $58. (0.40€) +$ 14.20-7893) 3-5-komponen Ef = 63965 4 4. los g energy = Ei Ek x100 = 1676.98 xloo 71% loss og chergy = 60-025 40 TI IARY toto Answer CSscanned with CamScanner,