Matching Kinematic Scenarios with Dynamical Motion Characteristics

Analysis of Situations:

1. (A) $v = c_1\sqrt{c_2 - x^2} \implies v^2 = c_1^2(c_2 - x^2)$. Differentiating with respect to time:
   $2v\frac{dv}{dt} = -2c_1^2 x \frac{dx}{dt} \implies a = -c_1^2 x$
   Since acceleration is directly proportional to displacement and directed towards the origin, the motion is Simple Harmonic Motion (SHM). Thus, $(\text{A}) \rightarrow (\text{p})$.

2. (B) $v = -kx \implies \frac{dx}{dt} = -kx \implies x(t) = x_0 e^{-kt}$.
   The velocity is $v(t) = -k x_0 e^{-kt}$. Since $e^{-kt} > 0$ for all finite $t$, the velocity never changes sign, meaning the object does not change its direction ($q$). Furthermore, the speed $|v|$ decays exponentially towards zero, so its kinetic energy keeps on decreasing ($r$). Thus, $(\text{B}) \rightarrow (\text{q}, \text{r})$.

3. (C) In the non-inertial frame of reference of the accelerating elevator, a constant pseudo force $ma$ acts downwards in addition to gravity $mg$. The effective equilibrium position shifts downwards by $\Delta x_0 = \frac{m(g+a)}{k}$. Since the spring-mass system is released from rest, it executes SHM about this new equilibrium position with frequency $\omega = \sqrt{k/m}$. Thus, $(\text{C}) \rightarrow (\text{p})$.

4. (D) Escape velocity from the Earth is $v_{\text{esc}} = \sqrt{\frac{2GM_e}{R_e}}$. The projection velocity is $v = 2\sqrt{\frac{GM_e}{R_e}} = \sqrt{2} v_{\text{esc}} > v_{\text{esc}}$. The object escapes the gravitational field completely, moving strictly outwards to infinity without turning back. Hence, it does not change its direction ($q$), and as it climbs against gravity, its speed and kinetic energy keep decreasing ($r$). Thus, $(\text{D}) \rightarrow (\text{q}, \text{r})$.

Matching:
$(\text{A}) \rightarrow (\text{p}),\; (\text{B}) \rightarrow (\text{q}, \text{r}),\; (\text{C}) \rightarrow (\text{p}),\; (\text{D}) \rightarrow (\text{q}, \text{r})$

Correct Answer: C