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STEP I 1990 question 12 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1990 question 12 solution

Tension in spring using modulus of elasticity:

\displaystyle T_{1} = \frac{\lambda e}{l_{0}} = \frac{\lambda (2l - l\sqrt{3})}{2l} = \frac{\lambda (2 - \sqrt{3})}{2}

Tension in string by resolving vertically:

\displaystyle 2T_{2}\cos\frac{\pi}{3} = mg \Rightarrow T_{2} = mg

Tension in string by resolving horizontally: \displaystyle T_{1} = T_{2}\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}T_{2}

\displaystyle \Rightarrow \frac{\lambda (2 - \sqrt{3})}{2} = \frac{\sqrt{3}}{2}mg

\displaystyle \Rightarrow mg = \lambda \left(\frac{2}{\sqrt{3}} - 1\right)

As required.

Using elastic potential energy = \displaystyle \frac{\lambda e^{2}}{2l_{0}}, we get \displaystyle EPE = \frac{\lambda (2l - 2xl)^{2}}{4l} = \lambda l (1 - x)^{2}

Using Pythagoras, the change in height is -\sqrt{l^{2} - l^{2}x^{2}} = -l\sqrt{1 - x^{2}}

Change in gravitational potential energy = mgh = - mgl\sqrt{1-x^{2}}

Energy is conserved, so \lambda l (1-x)^{2} - mgl\sqrt{1-x^{2}} + \frac{1}{2}mv^{2} = 0

When the particle comes to rest, v = 0, so:

\displaystyle \lambda l (1-x)^{2} - mgl\sqrt{1-x^{2}} = 0

Substituting for mg, we get:

\displaystyle \lambda (1-x)^{2} - \lambda \left(\frac{2}{\sqrt{3}} - 1\right)\sqrt{1-x^{2}} = 0

Note that in 1990 calculators were allowed, so it is easy to numerically confirm that a possible solution for x is approximately 0.66:

\displaystyle 0.34^{2} - \left(\frac{2}{\sqrt{3}} - 1\right)\sqrt{1 - 0.66^{2}} \approx -0.0006 \approx 0

Solution by Dystopia.

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