In outer space a rock with mass 8 kg, and velocity < 3300, -2500, 3200 > m/s, st

Question

In outer space a rock with mass 8 kg, and velocity < 3300, -2500, 3200 > m/s, struck a rock with mass 14 kg and velocity < 270, -290, 240 > m/s. After the collision, the 8 kg rock's velocity is < 2900, -2000, 3700 > m/s.

In outer space a rock with mass 8 kg, and velocity m/s, struck a rock with mass 14 kg and velocity m/s. After the collision, the 8 kg rock's velocity is m/s. What is the final velocity of the 14 kg rock? In outer space a rock with mass 8 kg, anf = m/s What is the change in the internal energy of the rocks? ?Einternal = J

Explanation / Answer

Using momentum conservation ,

initial momentum = final momntum

8 x (3300i - 2500j + 3200k) + 14 (270i - 290j + 240k) = 8(2900i - 2000j + 3700k) + 14v

v   = 498.57i - 575.71j - 45.71   or < 498.57 , - 575.71, -45.71 >

initial speed of 8kg = sqrt(3300^2 + 2500^2 + 3200^2) = 5232.60 m/s

similarly initial speed of 14kg = 463.25 m/s

final speed of 8kg = 5108.82 m/s

Final speed of 14 kg = 762.96 m/s

Initial Energy = m1v1^2 /2 + m2v2^2 /2

= 8 x 5232.60^2 /2 + 14 x 463.25^2 /2 = 111022615 J

Final energy = 8 x 5108.2^2 /2 + 14 x762.96^2 /2

       = 108474922.9 J

change = final - initial = 2547692.1 J or 2.55 x 10^6 J

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