Particle 1 has a mass of m 1 = 3.80 × 10 -6 kg, while particle 2 has a mass of m
ID: 1530753 • Letter: P
Question
Particle 1 has a mass of m1 = 3.80 × 10-6 kg, while particle 2 has a mass of m2 = 6.90 × 10-6 kg. Each has the same electric charge. These particles are initially held at rest, and the two-particle system has an initial electric potential energy of 0.170 J. Suddenly, the particles are released and fly apart because of the repulsive electric force that acts on each one (see the figure). The effects of the gravitational force are negligible, and no other forces act on the particles. At one instant following the release, the speed of particle 1 is measured to be v1 = 161 m/s. What is the electric potential energy at this instant?
Explanation / Answer
Apply conservation of momentum,
momentum gained by m2 = momentum geined by m1 in opposite direction
m2*v2 = m1*v1
v2 = m1*v1/m2
= 3.8*10^-6*161/(6.9*10^-6)
= 88.7 m/s
now apply conservation of energy
U1 + K1 = U2 + K2
==> U2 = U1 + K1 - K2
= U1 + 0 - ((1/2)*m1*v1^2 + (1/2)*m2*v2^2 )
= 0.17 + 0 - ( (1/2)*3.8*10^-6*161^2 + (1/2)*6.9*10^-6*88.7^2 )
= 0.0936 J <<<<<<<----------------Answer