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A system consists of three particles, each of mass 5.00g, located at the corners of an equilateral triangle with sides of 30.0cm. (a) Calculate the potential energy of the system. (b) Assume the particles are released simultaneously. Describe the subsequent motion of each. Will any collisions take place? Explain.

Short Answer

Expert verified

(a) Total potential energy of the system will beUTotal=-1.67×10-17J

(b) All particles will collide with each other at the center of the triangle.

Step by step solution

01

Conservation of energy

The conservation of energy says that’s the total energy in the initial condition of the object will be equal to the total energy in the final condition.

K+Uinitial=K+UfinalK=kineticenergyU=Potentialenergy

02

Given

massm=5.00gsidesofthetriangle=30.0cm

03

(a) Potential energy of the system

As described in the question the particles are located at the corners of the triangle and this triangle have equal three sides that is equilateral triangle.

The length of sides is 0.30m, so the gravitational potential energy of the system will be, if triangle is ABC,

UTotal=UAB+UBC+UCA=3UABUTotal=3-GmAmBrABUTotal=3-6.67×10-11N-m2/Kg×5.00×10-3Kg×5.00×10-3Kg0.300mUTotal=-1.67×10-14J

So the Total potential energy of the system will beUTotal=-1.67×10-14J.

04

(b) Motion of the particles

The net force of attraction felt by a particle will be at the midpoint of the other two points of the triangle. Each particle of the triangle is moving towards the center point of the triangle, so the all particles will collide with each other at the center of the triangle.

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