Three uniform spheres are fixed at the positions shown in the diagram. Assume th

ID: 1447231 • Letter: T

Question

Three uniform spheres are fixed at the positions shown in the diagram. Assume they are completely isolated and there are no other masses nearby.

(a) What is the magnitude of the force on a 0.20 kg particle placed at the origin (Point P)? What is the direction of this force?

(b) If the 0.20 kg particle is placed at (x,y) = (-500 m, 400 m) and released from rest, what will its speed be when it reaches the origin?

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A)
Gravitational force = G m1 m2 / d^2

force on 0.20 kg

due to upper 1kg :

F1 = (6.67 x 10^-11 x 1 x 0.20 ) / (0.5^2) (j) = 5.336 x 10^-11 j N

due to 2 kg :

F2 =[(6.67 x 10^-11 x 2 x 0.20) /(0.5^2 + 0.5^2) ] ( cos45i + sin45j)

F2 = (3.773i + 3.773j) x 10^-11 N

due to 1 kg on x axis.

F3 = (6.67 x 10^-11 x 1 x 0.20 ) / (0.5^2) (i) = 5.336 x 10^-11 i N

Fnet = F1 + F2 + F3 = (9.109i + 9.109j) x 10^-11 N

magnitude = sqrt(9.109^2 + 9.109^2) x 10^-11 = 12.88 x 10^-11 N

direction = tan^-1(9.109/9.109) = 45 deg

b) Gravt. PE of two mass system = - G M1 m2 / d

PE at (-500, 400)

PEi = -[G(1)(0.20) /sqrt(500^2 + 399.5^2)] - [G(2)(0.20) /sqrt(500.5^2 + 399.5^2)] - [G(1)(0.20) /sqrt(500.5^2 + 400^2)]

PEi =- 8.333 x 10^-14 J

KEi = 0

when at origin,

PEf = -[G(1)(0.20) /0.50] - [G(2)(0.20) /sqrt(0.50^2 + 0.5^2)] - [G(1)(0.20) /0.5)]

= - 9.109 x 10^-11 J

KE = 0.20 v^2 / 2 = 0.10v^2

Using energy conservation,

PEi + KEi = PEf + KEf

- 8.333 x 10^-14 J + 0 = - 9.109 x 10^-11 J + 0.1v^2

v = 3.02 x 10^-5 m/s

c) PE energy of system

PE = -[G(1)(2) /0.50] - [G(1)(1) /sqrt(0.50^2 + 0.5^2)] - [G(1)(2) /0.5)]

= - -6.279 x 10^-10 J

energy required to separate them = 6.279 x 10^-10 J

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