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(D) By the principle of conservation of energy,the total mechanical energy at the initial position equals the total mechanical energy at the final pos

Vedclass · Accepted Answer

$\frac{7GMm}{R^3}$

Vedclass · Answer

(D) By the principle of conservation of energy,the total mechanical energy at the initial position equals the total mechanical energy at the final position (where the ball comes to rest momentarily).Initial energy $E_i = K_i + U_i = 0 - \frac{GMm}{2R}$.Final energy $E_f = K_f + U_f = 0 + U_{grav, depth} + U_{spring}$.The gravitational potential energy at a depth $d = R/2$ (i.e.,at distance $r = R/2$ from the center) is $U_{grav} = -\frac{GMm}{2R^3}(3R^2 - r^2) = -\frac{GMm}{2R^3}(3R^2 - (R/2)^2) = -\frac{GMm}{2R^3}(3R^2 - R^2/4) = -\frac{GMm}{2R^3}(\frac{11R^2}{4}) = -\frac{11GMm}{8R}$.The spring is compressed by $x = R/2$,so $U_{spring} = \frac{1}{2}K(R/2)^2 = \frac{KR^2}{8}$.Equating $E_i = E_f$: $-\frac{GMm}{2R} = -\frac{11GMm}{8R} + \frac{KR^2}{8}$.$\frac{KR^2}{8} = \frac{11GMm}{8R} - \frac{4GMm}{8R} = \frac{7GMm}{8R}$.Therefore,$K = \frac{7GMm}{R^3}$.

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