Spring 2009 Final Exam In deep space, three masses are held in fixed positions (

Question

Spring 2009 Final Exam In deep space, three masses are held in fixed positions (by rods light compared to the masses) at three corners of a square of side length m as shown. A fourth mass is placed at point A, and released from rest. By symmetry it accelerates straight down and reaches point B (the midpoint between the two 3.0x109 kg masses). sin(36.87degree) 0.6 cos(36.87degree) 0.8 How fast is the fourth mass moving at point B? (Note: the acceleration is not constant.) If the fourth mass is 70 kg, what is the magnitude of the net force acting on it at point A?

Explanation / Answer

r sin36.87 = 18 sin45

>>>> r = 18 sin45/sin36.87

>>>> r = 5 m

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initial potential energy = Ui

final potential energy = Uf

U = Uf - Ui

     = GM [(m1(1/r1f - 1/r1i) + m2(1/r2f - 1/r2i) + m3(1/r3f - 1/r3i)]

    = GM [(3.00e9*(1/r1f - 1/r1i) + 3.00e9*(1/r2f - 1/r2i) + 5.64e9*(1/r3f - 1/r3i)]

r1f = r2f = 18 sin45 = 3 m

r1i = r2i = r = 5 m

r3f = 3 m

r3i = 3 + 4 = 7 m

U = GM [(3.00e9*(1/r1f - 1/r1i) + 3.00e9*(1/r2f - 1/r2i) + 5.64e9*(1/r3f - 1/r3i)]

     = 6.673e-11*((3.00e9*(1/3-1/5)+3.00e9*(1/3-1/5)+5.64e9*(1/3-1/7))

     = 0.125071086*M J

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Conservation of energy:

K = U

0.5 M v2 = U

0.5 * M * v2 = 0.125071086*M

>>>> v = 0.500 m/s

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b)

F = GM [(m1/(r1i)2)*cos36.87 + (m2/(r2i)2)*cos36.87 + (m3/(r3i)2)]

   = 6.673e-11*70*((3e9/(5)2)*0.8+ (3e9/(5)2)*0.8+ (5.64e9/(7)2))

   = 6.673e-11*70*((3e9/(5*5))*0.8+ (3e9/(5*5))*0.8+ (5.64e9/(7*7)))

   = 1.43 N

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