IIT JEE Advanced 2007 Paper 2

(A)→(P) In SHM, v = $\omega \sqrt{a^2-x^2}$ which resembles v = $c_1 \sqrt{c_2-x^2}$ and hence the object executes simple harmonic motion. (B)→(Q), (R) We have v = - kx = $\frac{dx}{dt}$ ⇒ ∫ $\frac{dx}{x}$ = - k ∫ dt That is , ln (x) = - kt ⇒ x = $e^{-kt}$ Therefore, v = $\frac{dx}{dt}$ = - $ke^{-kt}$ Now, K.E. = $\frac{1}{2} m v^2$ = $\frac{1}{2} m k^2 e^{-2kt}$ That is, the object does not change its direction as shown in the following graph: The kinetic energy keeps on decreasing as shown in the following graph: (C)→(P) We have T = 2π $\sqrt{\frac{m}{k}}$ ; therefore, the motion of the object is SHM. (D)→(Q), (R) As the object goes up against gravity, its speed and hence the kinetic energy goes on decreasing. Also, since V = $2\sqrt{\frac{GM_e}{R_e}}$ > Escape speed = $\sqrt{\frac{2GM_e}{R_e}}$ we conclude that the object cannot return to Earth once again.