Physics M.C.Q (1 Marks)

Let $x_A$ and $x_B$ be the extensions produced in springs $A$ and $B,$ respectively.
Restoring force on spring $A, F = k_Ax_A...(1)$
Restoring force on spring $B, F = k_Bx_B...(2)$
From $(1)$ and $(2),$ we get:
$k_Ax_A= k_Bx_B$
It is given that $k_A= 2k_B$
$\therefore\ \text{x}_\text{B}=2\text{x}_\text{A}$
Energy stored in spring $A:$
$\text{E}=\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\ \dots(3)$
Energy stored in spring $B:$
$\text{E}'=\frac{1}{2}\text{k}_\text{B}\text{x}_\text{B}^2=\frac{1}{2}\Big(\frac{\text{k}_\text{A}}{2}\Big)(2\text{x}_\text{A})^2$
$\therefore\ \text{E}'=2\times\Big(\frac{1}{2}\text{k}_\text{A}\text{x}_\text{A}^2\Big)=2\text{E} [$From $(3)]$

As the stone falls under the gravitational force, which is a conservative force,
the total energy of the stone remains the same at every point during its motion.
From the conservation of energy, we have:
Initial energy of the stone $=$ final energy of the stone
i.e., $(\text { K.E. })_{\mathrm{i}}+(\text { P.E. })_{\mathrm{i}}=(\text { K.E. })_{\mathrm{f}}+(\text { P.E. })_{\mathrm{f}}$
$=\frac{1}{2}\text{mv}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$
$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$
From the above expression, we can say that the maximum speed With which stone hits the ground does not depend on the initial direction.