Particle 1has a mass of m 1 = 8.00 × 10 –20 kg and a charge of 6.00 × 10 –12 C;
ID: 1597504 • Letter: P
Question
Particle 1has a mass of m 1 = 8.00 × 10 –20 kg and a charge of 6.00 × 10 –12 C; Particle 2 has a
mass of 3m1 = 24.00 × 10 –20 kg and a charge of –3.00 × 10 –12 C. The particles are initially held
r i = 0.400 m apart from each other when they are released from rest. How fast is Particle 1
moving when they are r f = 0.200 m = r i / 2 apart from each other (treat both of the particles as
point charges)?
Show work please! Thanks!
8. Particle 1 has a mass ofm 8.00 x 10 20 kg and a charge of 6.00 x 10-12 C; Particle 2 has a mass of 3m 24.00 x 10-20 kg and a charge of-3.00 x 10-12 C. The particles are initially held ri 0.400 m apart from each other when they are released from rest. How fast is Particle 1 moving when they are rf 0.200 m ri 2 apart from each other (treat both of the particles as point charges)? Later Time ri A. 7550 ms B. 2760 ms C. 3180 m/s D. 1590 ms E. 1840 m/sExplanation / Answer
The only force acting on each particle is the conservative electric force. Therefore, the total energy (kinetic energy plus electric potential energy) is conserved as the particles move apart.
For two points, A and B, along the motion, the conservation of energy is
m1*v1a^2/2 + m2*v2a^2/2 + kq1q2/ra = m1v1b^2/2 + m2*v2b^2/2 + kq1q2/rb
initially they are at rest
v1a = v2a = 0
rb = ra/2
m1*v1b^2/2 + m2*v2b^2/2 = -kq1q2/ra ...(1)
here linear momentum also conserved
m1*v1a + m2v2a = m1v1b + m2v2b
v1a = v2a = 0
v2b = -m1*v1b/m2
v2b = -1/3 * v1b
put the value in eq(1)
m1*v1b^2/2 + m2 * v1b^2/18 = -kq1q2/ra
(9m1v1b^2 + m2*v1b^2)/18 = -kq1q2/ra
v1b^2(9m1+m2) = -18*kq1q2/ra
v1b = sqrt(-18*kq1q2/ra*(9m1+m2))
v1b = 2755.6 = 2760 m/s ( B option )