Question

All I want here if for you to solve for the initial velocity vectors of two masses as depicted below. Assume the collision is perfectly elastic. A ball of mass M collides with a second ball of mass 2M. The velocity of Mis measured to be 5 m/s at 30 degrees post collision. The velocity of 2M is measured to be 4 m/s at -15 degrees. That's all you get. Submit your response for the initial velocities of M and 2M.

Answer 1

Answer: TE D = 5 m/s Us (2MU2 150 0 V2=4 mis 2m Given that collision is perfectly elastic „s o both law will be apply 1. Momentum's of conservation. 2. Kinetic energy Conservation. First momentum conses vation along X-axis Must Muz = Mvicos 3 6 + 2m Na COSIS (Before) (after) mult 2 m 2 = mx 5 x 53 x 5 x 53 2 mx 4 Cos 15 + 2 + 22 = 553 (1) + 8 Cosis 2 Along y-dign: mv, sinzo 2m x Na sinis Mx 5 x Sinzo 2m x 4 x Sinis' 8 x Sin 15' 2.5 – 20.07. Conserved. from Kinetic energy conservation I'mu?+ 12.2 Ilmuit 1 + 2 m 2 2 2 1.2m =

u? + 202 - 25+ 2x16 2 2 uit 22 25+32 ² 57 (2) from e grain +2u2 - 12.05 U 12.05 - 202 2 2 = 145.20+ y u2 40.202 put this into ear (2) 4 2² – 48.2 at 145.20 +2= 57 62 - 48.2 U2 + 88.2 = 0 U2 - 0.03 U2 + 14.750 Solve by Quadratic ean 8.03 + J18.03)?_4x14.7 U2= 2 U2= 5.20 mis from egen VIP 1.65 mis = 1.65 mis Uz - 5. 20ls (tre dirn, (tre ve dige s. e. same dire collision of balls.