g A 0.750-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wall and rebounds with 70.0% of its initial kinetic energy. What is the magnitude of the change in momentum of the stone

  v = 1.23 m / s

the stone goes in the initial direction of the ball.

Explanation:

This is a momentum conservation problem, they indicate that after the crash its speed is 70% of its initial energy.

           K_f = 0.7 K₀

           ½ m v_f² = 0.7 ½ m v₀²

           v_f² = 0.7 v₀²

           v_f = √0.7  v₀

           v_f = √0.7   20

           v_f = 16.73 m / s

now we can use conservation the moment

initial instant. Before the crash

           p₀ = m v₀ + 0

final instant. After the crash

           p_f = m v_f + M v

           p₀ = p_f

           m v₀ = m v_f + M v

           v =  (v₀ - v_f)

           v = 0.750  (20 - 16.73) / M

            v = 2.4525 / M

To finish the exercise we must assume a mass of the stone, generally higher than the mass of the ball.

          M = 2 kg

          v = 2.4525 / 2

          v = 1.23 m / s

the stone goes in the initial direction of the ball.

your answer is, orbit around the nucleus

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j