(A)→(P) In SHM, v = ω√a2−x2 which resembles v = c1√c2−x2 and hence the object executes simple harmonic motion. (B)→(Q), (R) We have v = - kx =
dx
dt
⇒ ∫
dx
x
= - k ∫ dt That is , ln (x) = - kt ⇒ x = e−kt Therefore, v =
dx
dt
= - ke−kt Now, K.E. =
1
2
mv2 =
1
2
mk2e−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 √
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√
GMe
Re
> Escape speed = √
2GMe
Re
we conclude that the object cannot return to Earth once again.