What is the change in the internal energy of the rocks?
In outer space, a rock with a mass of 4 kg and a velocity of < 3300, -3200, 3000 > m/s struck another rock with a mass of 12 kg and a velocity of < 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 to be < 370, -436.7, 33.3 > m/s, but I'm uncertain if this is correct.
2) What is the change in the internal energy of the rocks?
3) Which of the following statements about Q (the transfer of energy into the system due to a temperature difference between the system and its 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 = ΔK of the rocks. - Q = ΔE_internal of the rocks.
1 Answers
Conservation of momentum
m1 v1i + m2 v2i = m1 v1f + m2 v2f
v2f = (m1 (v1i – v1f) + m2 v2i) / m2
2) The increase in internal energy will be the kinetic energy lost in the collision
Tf – 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.
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