Question

Initially, mass one (2.80 kg) has a velocity of 5.90 m/s and mass two (2.00 kg) is at rest. After they collide, mass one emerges at an angle theta = 31.0 degrees. What is the speed of mass one after the collision if the collision is completely elastic? (Note, there are actually two possible answers two this problem, choose the solution which has m1 going as fast as possible.) What is the angle phi, between mass two's velocity and the initial velocity of mass one? (Give your answer as a positive number in degrees.) What is the final speed of mass two after the collision?

Answer 1

m1 = 2.80kg ļ Vo = 5.90m 15 m2 = 2kg (at rest initially) v= 95mm > consonation of KE C collision is elastic ) men 1+ 1 = + , vo? + 0 = Lm 1, 2 + 1 m. 2. Vz? - mi yo? = m. v, 2 + m2 V 2² - 0. using conservation of linear momentum ferst along x-axis - Mi vo to = mivLos & + m2 V2 ios - & along y-axis 7 milli sono = M₂ V₂ Sunid -

fueritte Larango €9" @ >> Ve - mive 1958 m2 V (60 - 4 Noces squarury 4 odsung eq?? ☺ f@ => [mive Sine ) 2 + cm, Vo - , Leo) 2 = mv, . V,? = cm, visino )? + c m Vip - mi vplase) ? putting this value in cano mivo 2 = mir,? ++ [Cmimi done.)? + (mvo-mv, oso)? - Mi Mz vo? - m, mv,² = m visin?o + m² Vo² + m² v 2 cos²0 -2 m," VO V 1906 Nero mi = 2.8089 Vo = 5.90 m 15 m2 = 2kg = 310 -> (2.80) (@) (590)- (2.80)62) vi? - (2.80) > 1,? sur3) " + (2.8)2 (509)? + (2.8)? ? 2 31" - 2(2.8)(540) X M 60531" 194.936 - 5.662 = 2:07967 1? + 272.9104 + 5.764, 2. - 79. 2982 6 V ) 13.43967"? - 79.29826V, +177.9944) - O > Vis 79.29826 + (79.2982672 - +677,9711) (1343%) 2 X 13.43967

Vi= 79.2982 6 1 45.78 66 2 X 13.43967 Vis 40 65 36 & 1.2467 ::V, (max) = 406536 mis : from o = m, no? = mui2 + M₂ V ₂ 2 =) mi (vo? - v,?) = m. Uz ² - V2 = m (vo? 1,2) after collision V2 = 402913 - speed of mass 2 mis. ... from © => m, visno = m2 V2 End = sino a me vieno m) V. Sin 8 = 0.781929405 5 $ = 51.437 6 °