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

Answer: 5

Let initial extension in spring is $x_0$, then
$2k{x}_0\cos 45^{\circ }=mg\sin 45^{\circ }$
$\Rightarrow 2k{x}_0=mg$
Let after cutting right spring, acc. of ball is $A_1$ along the inclined plane and $a_2$ perpendicular to $a_1$
$\Rightarrow \frac{mg}{\sqrt{2}}-\frac{mg}{2}\frac{1}{\sqrt{2}}=m{a}_1$
$\Rightarrow a_1=\frac{g}{2\sqrt{2}}$
Then
$mg\sin 45^{\circ }$
$\Rightarrow k{x}_0\cos 45^{\circ }=m{a}_1$
$k{x}_0\sin 45^{\circ }=m{a}_2$
$\Rightarrow a_2=\frac{g}{2\sqrt{2}}$
Net acceleration $a=\sqrt{a_1^2+a_2^2}$
$=\frac{g}{2}=5m/s^2$