A charge of -2.86 C is fixed in place. From a horizontaldistance of 0.0580 m, a

Question

A charge of -2.86 C is fixed in place. From a horizontaldistance of 0.0580 m, a particle of mass 7.69 x 10-3 kgand charge -8.01 C is fired with an initial speed of 71.2 m/sdirectly toward the fixed charge. What is the distance of closestapproach?

I was given a work through with the answer of:
ResponseDetails: Given q = -2.45 * 10 ^ -6 C mass   m = 7.49* 10 ^ -3 kg            q ' = -9.86 * 10 ^ -6 C Speed v = 85 m / s Let the distance of the closest approach be r   At closest distance potential energy = kineticenergy                                    K q q ' / r = ( 1/ 2) m v ^ 2                                                   r = 2Kqq' / m v ^ 2 where K = coulomb constant = 8.99 * 10 ^ 9 N m ^ 2 / C ^2 plug the values    r = 2( 8.99*10 ^ 9 ) ( 2.45*10 ^ -6 ) ( 9.86* 10 ^ -6 ) / [ 7.49* 10 ^ -3 * (85 ^ 2 ) ]                            = 8.026 * 10 ^ -3 m
This answer was not correct....
ResponseDetails: Given q = -2.45 * 10 ^ -6 C mass   m = 7.49* 10 ^ -3 kg            q ' = -9.86 * 10 ^ -6 C Speed v = 85 m / s Let the distance of the closest approach be r   At closest distance potential energy = kineticenergy                                    K q q ' / r = ( 1/ 2) m v ^ 2                                                   r = 2Kqq' / m v ^ 2 where K = coulomb constant = 8.99 * 10 ^ 9 N m ^ 2 / C ^2 plug the values    r = 2( 8.99*10 ^ 9 ) ( 2.45*10 ^ -6 ) ( 9.86* 10 ^ -6 ) / [ 7.49* 10 ^ -3 * (85 ^ 2 ) ]                            = 8.026 * 10 ^ -3 m
This answer was not correct....

Explanation / Answer

Given q = -2.86 * 10 ^ -6 C mass of the particle m = 7.69* 10 ^ -3 kg Charge of the particle q ' = -8.01 * 10 ^ -6 C Initial speed of the particle is v =71.2m/s Initial distance between the particle and the charge is d=0.058m Let the distance of the closest approach be r   At the distance of closest approach the kinetic energy of theparticle beocmes zero Now initial energy of the particle isE1= K q q ' / d  + ( 1/ 2) m v ^2 Final energy of the particle is E2= K q q '/ r Then by the conservation of energyE1=E2 Thus K q q ' / d  + ( 1/ 2) m v ^ 2=K q q '/ r ----------(A) K = 8.99*109Nm2/C2 Now substitute the given values in the equation (A) and solvefor r(distance of closest approach). Now substitute the given values in the equation (A) and solvefor r(distance of closest approach).

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