Two balls of mass 0.7 kg and 0.3 kg are connected by a low-mass spring. This dev
ID: 1883855 • Letter: T
Question
Two balls of mass 0.7 kg and 0.3 kg are connected by a low-mass spring. This device is thrown through the air with low speed, so air resistance is negligible. The motion is complicated: the balls whirl around each other, and at the same time the system vibrates, with continually changing stretch of the spring. At a particular instant, the 0.7 kg ball has a velocity <2, -5, 3> m/s and the 0.3 kg ball has a velocity <3, 2, 9> m/s. What is the total momentum of the device after 0.08 seconds?
Explanation / Answer
Momentum, P = m1v1 + m2v2
P = (0.7)(2, - 5, 3) + (0.3)(3, 2, 9)
P = (1.4, - 3.5, 2.1) + (0.9, 0.6, 2.7) = (2.3, - 2.9, 4.8) kg m/s
F = mg = (0.7 + 0.3)(0, - 9.8, 0) = (0, - 9.8, 0) N
F = dP/dt
dP = (0, - 0.78, 0)
Momentum after 0.08 sec = (2.3, - 2.9, 4.8) + (0, - 0.78, 0)
= (2.3, 3.68, 4.8) kgm/s
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