From a large distance away a particle of mass 300 g and char

From a large distance away, a particle of mass 3.00 g and charge 15.0 mu c is fired at 20.0 m/s straight toward a second particle, originally stationary but free to move, with mass 4.95 g and charge 8.50 mu c. Both particles are constrained to move only along the x axis. At the instant of closest approach, both particles will be moving at the same velocity. Find this velocity. vector v_c = hat i m/s Find the distance of closest approach. r_c = m After the interaction, the particles will move far apart again. At this time, find the velocity of the following the 3.00-g particle vector v_1f = hat i m/s the 4.95-g particle vector v_2f = hat i m/s

Solution

given that,

m1 = 3 g

u1 = 20 m/s

q1 = 15 uC

m2 = 4.95 g

q2 = 8.50 uC

u2 = 0.

let velocity of bothy particles at closest approch = v

apply law of momentum conservation-

m1*u1 = (m1 + m2)*v

v = 3*10^(-3)*20 / (3 + 4.95)*10^(-3)

v = 7.54 m/s

(b)

let, distance of closest approch = d

now, apply law of energy conservation,

(1/2)*m1*u1^2 = k*q1*q2 / d + (1/2)*(m1 + m1)*v^2

(1/2)*0.003*20^2 = 9*10^9*15*10^(-6)*8.50*10^(-6) / d + (1/2)*(0.003 + 0.00495)*(7.54)^2

by solving this, we got-

d = 3.071 m

(c)

after their seperation, let their velocities are v1 and v2.

apply momentum conservation-

m1*u1 = m1*v1 + m2*v2 .....eq1

now, apply law of energy conservation-

(1/2)*m1*u1^2 = (1/2)*m1*v1^2 + (1/2)*m2*v2^2 ....eq2

by solving equation 1 and 2,

v1 = (m1 - m2)*u1 / (m1 + m2)

v1 = (0.003 - 0.00495)*20 / (0.003 + 0.00495)

v1 = -4.90 m/s

v2 = 2*m1*u1 / (m1 + m2)

v2 = 2*0.003*20 / (0.00795)

v2 = 15.09 m/s

answer