Newton's laws of motion

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Question

Figure shows a ball of mas s 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/s²) of ball?

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Solution

Let initial extension in spring is x₀, 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 A₁ along the inclined plane and a₂ perpendicular to a₁
$\begin{array}{l} \Rightarrow \frac{m g}{\sqrt{2}} - \frac{m g}{2} \frac{1}{\sqrt{2}} = m a_{1} \\ \Rightarrow a_{1} = \frac{g}{2 \sqrt{2}} \end{array}$
Then
$\begin{array}{l} m g \sin 45^{\circ} \Rightarrow k x_{0} \cos 45^{\circ} = m a_{1} \\ k x_{0} \sin 45^{0} = m a_{2} \Rightarrow a_{2} = \frac{g}{2 \sqrt{2}} \end{array}$
$Net acceleration a = \sqrt{a_{1}^{2} + a_{2}^{2}} = \frac{g}{2} = 5 m / s^{2}$

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Newton's laws of motion

Remember concepts with our Masterclasses.

Figure shows a ball of mas s 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/s2) of ball?

Ready to Test Your Skills?

free study material

Figure shows a ball of mas s 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/s²) of ball?