Question

V = ? 2M An object of mass 3M, moving in the +x direction at speed vo, breaks into two pieces of mass M and 2M as shown in the figure. If 0 = 64.00 and 62 = 21.0°, determine the final velocities vi and v2 of the resulting pieces in terms of vo. 3M 2 M VI = VO U2 = VO

Answer 1

Let us conroden the diagram. Let us Ear con sucher the queq torbranitoare Co-ordinate axt at O. Lets break their momention about the perpe two perpen diculer component

va P₂ onde 3M 0 Joi P, ooo Pisud y M -3MVO 3M No Initial momention of the resten before Collisionsor won Pia momentum of nu 3 MY Vo ... this directe es a along x axis. There is no ential momenton along yaxu. After collesion, the man was clevered en to two B- parts momentur of Pe 3 Pa MV Momentur of P2 = 2M V2 men 2 M it Now, from the above chogram, not momentum along yaxes Py- & Pisu dio P₂ m2, to and watts Pinnoi and P2 zer O2 are oppositely no we subtract them at directed above ex prerren

Pay 31 Agen , from the conservation of immention, Pya No i netral monnentur mmention] 20 about us y arn 0 P, sode = P2 zu OL 22N V2 zu02 Mv, sid, vindi 2v2 suda 2 V2 O sadz sud sen (21) er (64) 0-9972 1 arer, I again Agaver net senal momentum about a PY P, cond, & Pe cord, (from degrar) from the conservation of momentirm about yoon X axe, we have Px= Pi [Insteal momenten about & axin Pi de P, Conde te conde? MV, cond, t2M V Conde Mv. 0^, OA & card. + 2 conda = 3. 0.797 cond, t2 conda - 3 vo o. Va (0.397 end + 2 cond) = 300

Paralel 3 vo On, V2 = 0-797 cond 2 es or 30 0-797 con (CW) + 2 692 (21) = 1.353 Volu (requered value) Again, from a from (F), we have on 1972 Fisvassern 0-797 X 1.333 V. v 1.078 vo Amy (Required value) 2
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