KinematicsMedium
Question
An object is projected with kinetic energy K from a point A at an angle $60^{\circ}$ with the horizontal. The ratio of the difference in kinetic energies points $B$ and C to that at point A (see figure), in the absence of air friction is :
Options
A.$1:2$
B.$2:3$
C.$1:4$
D.$3:4$
Solution
$(KE)_{A} = K = \frac{1}{2}{mu}^{2}$
$$\begin{matrix} (KE)_{B} & \ = \frac{K}{4} = \frac{1}{2}\text{ }m\left( \frac{u}{2} \right)^{2} = \frac{K}{4}\ \left( u_{B} = ucos60^{\circ} = \frac{u}{2} \right) \\ (KE)_{C} & \ = K \\ \text{~}\text{Ratio}\text{~} & \ = \frac{K - K/4}{\text{ }K} \\ & \ = \frac{3\text{ }K/4}{\text{ }K} = \frac{3}{4} \end{matrix}$$
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