Question

Much appreciated!

In outer space rock 1, with mass 4 kg and velocity < 4300, -2600, 2000 > m/s, struck rock 2, which was at rest. After the collision, rock 1's velocity is < 4100, -2000, 2400 > m/s. What is the final momentum of rock 2? 728= (800,2400, – 1600) kg : m/s Before the collision, what was the kinetic energy of rock 1? K11 = 1.2e+08 X I Before the collision, what was the kinetic energy of rock 2? K2i = 0 After the collision, what is the kinetic energy of rock 1? K17= Suppose the collision was elastic (that is, no change in kinetic energy and therefore no change in thermal or other internal energy of the rocks). In that case, after the collision, what is the kinetic energy of rock 2? K2F= On the other hand, suppose that in the collision some of the kinetic energy is converted into thermal energy of the two rocks, where AEthermal 1 + AEthermal 2 = 2.68 x 106 ). What is the final kinetic energy of rock 2? K2= In this case (some of the kinetic energy being converted to thermal energy), what was the transfer of energy Q (microscopic work) from the surroundings into the two-rock system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.)

Answer 1

Rock 1 ма Initimassam, = 4 kg a velocity w ni = (43007 -26003+2000 Å 1 m/s Final velocity a Vgs = (42001 - 2000Î +2400) m/s Initial momentum priamo = 4x (43001 - 2600 +2000 2 =(172001 - 2040099 +80002 1 kg.mls Final momentum a Pele = maaf = 4x( 4100-2000ỉ +2400 R =( 164001 - 8000 m +960of g kg. m/s resta Rock 2 Initial momentum = P2i = of totoa final momentum = Post momentum conservation gives 7) Bit = Est Bonus in - ( 172061-10400 +8600R)- (164001 - 8000 +9600Xj Powe cs Scanned with CamScanner

= 900f - 2400-1600R bet = <800, -2400,-1600 kg.mis. । २m. kinetic energy YN (17200)?+(-104001? + 1800072 } 2x4 Kai = 5.85 x 107 g К.: (Fal- - am, O - 2mz Kai to legal? Kgf = 2min (960072 16400²+ (-800072 2x4 | K95 = 5.314 31075) kaf = (kaitk2i) - kafa = ( 5.85-5.314 ) x 107 = Kef = 0,536 x 107 I 536x104 g. Scanned with CamScanner

If, AE thermal at AE thermal, 2 = 2.68x1065 Then Kas = (kait kai) - Kef - 2.68*106 = 536x109 - 268x104 268x104g 1 K2f = 2,68x100 l 68 x 10 Q = 2,68% 10°5 = AE thermal cs Scanned with CamScanner