What is the change in the internal energy of the rocks?

Question

In outer space a rock with mass 4 kg, and velocity < 3300, -3200, 3000 > m/s, struck a rock with mass 12 kg and velocity < 270, -270, 200 > m/s. After the collision, the 4 kg rock’s velocity is < 3000, -2700, 3500 > m/s.1). What is the final velocity of the 12 kg rock?I calculated this out to be < 370, -436.7, 33.3 > m/s. I’m not sure if this is correct though. 2). What is the change in the internal energy of the rocks?3). Which of the following statements about Q (transfer of energy into the system because of a temperature difference between system and surroundings) are correct? Check all that apply: Q ~= 0 because the duration of the collision was very short.Q ~= 0 because there are no significant objects in the surroundings.Q = Delta K of the rocks.Q = Delta E_text(internal) of the rocks.

Mina Herzog · Accepted Answer

Conservation of momentumm1 v1i + m2 v2i = m1 v1f + m2 v2fv2f = (m1 (v1i – v1f) + m2 v2i) / m22) The increase in internal energy will be the kinetic energy lost in the collisionTf – Ti = 1/2 m1 (v1f^2 – v1i^2) + 1/2 m2 (v2f^2 – v2i^2)3) Not sure exactly what conventions you’re using, but I’d say Q=0, since there’s no players in the system except for the two rocks.

Lavinia Carroll

What is the change in the internal energy of the rocks?

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