Question

Figure shows a ball of mass $m$ connected with two ideal springs of force constant $k$, kept in equilibrium on a smooth incline, suddenly right spring is cut. What is magnitude of instantaneous acceleration (in $m / s ^{2}$ ) of ball?

Answer 1

Let initial extension in spring is $x_{0}$, then
$2 k x_{0} \cos 45^{\circ}=m g \sin 45^{\circ}$
$\Rightarrow 2 k x_{0}=m g$
Let after cutting right spring, acc. of ball is $A_{1}$ along the inclined plane and $a_{2}$ perpendicular to $a_{1}$
$\Rightarrow \frac{m g}{\sqrt{2}}-\frac{m g}{2} \frac{1}{\sqrt{2}}=m a_{1}$
$\Rightarrow a_{1}=\frac{g}{2 \sqrt{2}}$
Then
$m g \sin 45^{\circ}$
$\Rightarrow k x_{0} \cos 45^{\circ}=m a_{1}$
$k x_{0} \sin 45^{\circ}=m a_{2}$
$\Rightarrow a_{2}=\frac{g}{2 \sqrt{2}}$
Net acceleration $a=\sqrt{a_{1}^{2}+a_{2}^{2}}$
$=\frac{g}{2}=5\, m / s ^{2}$