Cray CIn outer Space A Rock v x y WiA Homework x C we can use The Textb x rk19-S
ID: 1552367 • Letter: C
Question
Cray CIn outer Space A Rock v x y WiA Homework x C we can use The Textb x rk19-SP17 C Secure h ps://www.webassign.n /web/Student/Assignment-Responses/submit?dep 5321262 Q6 My Notes 6. -13 points M 4 10.6.017 In outer space a rock with mass 9 kg, and velocity 3800, 2600, 3600 m/s, struck a rock with mass 14 kg and velocity 290, 290, 200 m/s. After the collision, the 9 kg rock's velocity is 3400, 1800, 4000 m/s. What is the final velocity of the 14 kg rock? what is the change in the internal energy of the rocks? AEi nternal Which of the following statements about Q (transfer of energy into the system because of a temperature difference between system and surroundings are correct? (Ignore heat transfer by radiation Check all that apply 0 because there are no significant objects in the surroundings. 2 nternal of the rocks AK of the rocks. 0 because the duration of the collision was very short. Additional Materials Section 10.6 Submit Answer Save Progress 6 6 points I Previous Answers MI4 10.6.020. My Notes 27 AM sk me anything 3/30/2017Explanation / Answer
m1 = 9 kg, u1 = 3800 i -2600 j +3600 k
m2 = 14 kg, u2 = 290 i -290 j +290 k
v1 = 3400 i -1800 j +4000 k
From conservation of momentum along x direction
m1u1x +m2u2x = m1v1x +m2v2x
9*3800 +14*290 = 9*3400 +14*v2x
v2x = 547 m/s
From conservation of momentum along y direction
9*-2600 +14*-290= 9*-1800+14*v2y
v2y = -804.3 m/s
From conservation of momentum along z direction
9*3600+14*290= 9*4000+14*v2z
v2z = 32.86 m/s
v2 = <547, -804.3, 32.86>
(b) DE = Ki - Kf
DE = 0.5*9*(3800^2+2600^2+3600^2) +0.5*14*(290^2+290^2+290^2) -0.5*9(3400^2+1800^2+4000^2)-0.5*14*(547^2+804.3^2+32.86^2)
DE = 1.026*10^7 J
(c) Q = DE
Q = DK