In outer space a rock with mass 8 kg, and velocity < 3500, -2800, 2800 > m/s, st
ID: 1352797 • Letter: I
Question
In outer space a rock with mass 8 kg, and velocity < 3500, -2800, 2800 > m/s, struck a rock with mass 17 kg and velocity < 300, -320, 200 > m/s. After the collision, the 8kg rock's velocity is < 3200, -2000, 3300 > m/s.
What is the final velocity of the 17 kg rock?
What is the change in the internal energy of the rocks?
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:
Explanation / Answer
m1 = 8 kg
V1i = initial velocity before collision for 8 kg = 3500 i^ - 2800 j^ + 2800 k^
V1f = final velocity after collision for 8 kg = 3200 i^ - 2000 j^ + 3300 k^
m2 = 17 kg
V2i = initial velocity before collision for 17 kg = 300 i^ - 320 j^ + 200 k^
V2f = final velocity after collision for 17 kg
using conservation of momemtum :
m1 v1i + m2 v2i = m1 v1f + m2 v2f
8 (3500 i^ - 2800 j^ + 2800 k^) + 17 (300 i^ - 320 j^ + 200 k^) = 8 (3200 i^ - 2000 j^ + 3300 k^) + 17 v2f
28000 i^ - 22400 j^ + 22400 k^ + 5100 i^ - 5440 j^ + 3400 k^ = 25600 i^ - 16000 j^ + 26400 k^ + 17 V2f
(28000 - 5100 - 25600) i^ + (-22400 - 5440 + 16000) j^ + (22400 + 3400 - 26400) = 17 V2f
V2f = -158.82 i^ - 696.5 j^ - 35.3 k^
V1i = 3500 i^ - 2800 j^ + 2800 k^ = sqrt (35002 + (-2800)2 + (2800)2) = 5284.88 m/s
V1f = 3200 i^ - 2000 j^ + 3300 k^ = sqrt (32002 + (-2000)2 + (3300)2) = 5012.98 m/s
V2i = 300 i^ - 320 j^ + 200 k^ = sqrt (3002 + (-320)2 + (200)2) = 482.1 m/s
V2f = -158.82 i^ - 696.5 j^ - 35.3 k^ = sqrt ((-158.82)2 + (- 696.5)2 + (- 35.3)2) = 715.25 m/s
change in internal energy = final KE - initial KE
change in internal energy = (0.5) (m1 v21f + m2 v22f - m1 v21i - m2 v22i )
change in internal energy = (0.5) (8 (5012.98)2 + 17(715.25)2 - 8 (5284.88)2 - 17(482.1)2) = - 8.83 x 106 J
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