Two balls of mass 0.5 kg and 0.8 kg are connected by a low-mass spring. This dev
ID: 1524408 • Letter: T
Question
Two balls of mass 0.5 kg and 0.8 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 0.5 kg ball has a velocity ‹ 5, 6, 3 › m/s and the 0.8 kg ball has a velocity ‹ 4, 2, 6 › m/s.
(a) At this instant, what is the total momentum of the device?
[p with arrow] total =
5.7,1.4,6.3
[Correct: Your answer is correct.] kg · m/s
(b) What is the net gravitational (vector) force exerted by the Earth on the device?
[F with arrow] grav =
0,12.74,0
[Correct: Your answer is correct.] N
(c) At a time 0.08 seconds later, what is the total momentum of the device?
[p with arrow] total = ??
Answer the question part c
Explanation / Answer
Part C:
First, Find the total momentum at 0 seconds:
p = (m1)(v1) + (m2)(v2)
p = (0.5 kg)<5, -6, 3> + (0.8 kg)<4, 2, 6> m/s
p = <2.5, -3, 1.5> + <3.2, 1.6, 4.8> kg-m/s
p = <5.7, -1.4, 6.3> kg-m/s
Use Newton's second law to find change in momentum:
F = p/t
<0, -12.74, -0N> = p / (0.08 s)
p = <0, -1, 0> kg-m/s
Momentum at 0.08s:
p + p =<5.7, -1.4, 6.3> + <0, -1, 0> kg-m/s
p + p = <5.7, -2.4, 6.3> kg-m/s