A proton with mass 1.67*10^-27kg is propelled at an initialspeed of 3.00*10^5m/s directly toward a uranium nucleus 5.00maway. The proton is repelled by the uranium nucleus with aforce of magnitude F=α /x^2, where x is the separation betweenthe two objects and α =2.12*10^-26N*m^2. Assume that theuranium nucleus remains at rest.

a) What is the speed of the protonwhen it is 8.00*10^-10m from the uranium nucleus?
b) As the protonapproaches the uranium nucleus, the repulsive force slows down theproton until it comes momentarily to rest, after which the protonmoves away from the uranium nucleus. How close to the uraniumnucleus does the proton get?
c) What is the speed of the protonwhen it is again 5.00m away from the uranium nucleus?

Explanation:

F = 2.12 x 10⁻²⁶ / x²

Work done by electric field of nucleus

W = ∫ Fdx

= ∫2.12 x 10⁻²⁶ / x² dx

= 2.12 x 10⁻²⁶ ( - 1 / x )

= - 2.12 x 10⁻²⁶ ( 1/5 - 1 / 8 x 10⁻¹⁰ )

= - .265 x 10⁻¹⁶ J

1/ 2 x mv² = .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ - .265 x 10⁻¹⁶

= 7.515 x 10⁻¹⁷ - .265 x 10⁻¹⁶

=( .7515 - .265 )x 10⁻¹⁶

= .4865 x 10⁻¹⁶

.5 x 1.67 x 10⁻²⁷ x v² = .4865 x 10⁻¹⁶

v² = .5826 x 10¹¹

v² = 5.826 x 10¹⁰

v = 2.41 x 10⁵ m /s

b )

Let r be the closest distance

Potential at this point

2.12 x 10⁻²⁶ (  1 / r )

Kinetic energy

= 0

Total energy = 2.12 x 10⁻²⁶ (  1 / r )

Total energy at 5 m

= .5 x 1.67 x 10⁻²⁷ x 9 x 10¹⁰ + 0 ( potential energy at 5 m will be negligible as compared with that near the center )

= 7.515 x 10⁻¹⁷ J

So ,

2.12 x 10⁻²⁶ (  1 / r ) = 7.515 x 10⁻¹⁷

r = 2.12 x 10⁻²⁶ / 7.515 x 10⁻¹⁷

= .282 x 10⁻⁹

= 2.82 x 10⁻¹⁰ m

c ) As electric field is conservative , no dissipation of energy takes place . Hence its speed at 5m on returning back to this point will be same as

3.00 x 10⁵ m /s

the answer is a. at the equator