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Work

Question

A particle is projected at $60^{\circ}$ to the horizontal with kinetic energy K. The kinetic energy at the position when particle makes $30^{\circ}$ with horizontal is ___K

Moderate

Solution

$\begin{array}{l} K = \frac{1}{2} m u^{2} \\ \tan α = \frac{v_{y}}{v_{x}} \Rightarrow \tan 30 = \frac{v_{y}}{v_{x}} \Rightarrow v_{y} = \frac{v_{x}}{\sqrt{3}} \\ {KE}_{f} = \frac{1}{2} m (v_{x}^{2} + v_{y}^{2}) = \frac{1}{2} m (u_{x}^{2} + {(\frac{v_{x}}{\sqrt{3}})}^{2}) \\ {KE}_{f} = \frac{1}{2} m \frac{4}{3} u^{2} \cos^{2} 60 = k \times \frac{4}{3} \times \frac{1}{4} \\ = 0.33 K \end{array}$

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