VITEEE 2009 Paper

Momentum, p = m.v ⇒ v = $\frac{p}{m}$ Kinetic energy, KE = $\frac{1}{2} m v^{2} = \frac{1}{2} m \left( \frac{p^{2}}{m^{2}} \right) = \frac{1}{2m} p^{2}$ or, $KE \propto p^{2} \; \left(\because \frac{1}{2m} = \text{constant}\right)$ Hence, the graph between KE and $p^{2}$ will be linear. Now, kinetic energy KE $= \frac{1}{2} m v^{2}$ The velocity component at point P, $v_{y} = (u \sin\theta - gt)$ and $v_{x} = (u \cos\theta)$ Resultant velocity at point P, $\vec{v} = v_{y} \hat{j} + v_{x} \hat{i}$ $= (u \sin\theta - gt) \hat{j} + u \cos\theta \hat{i}$ $|\vec{v}| = \sqrt{(u \cos\theta)^{2} + (u \sin\theta - gt)^{2}}$ $= \sqrt{u^{2} \cos^{2}\theta + u^{2} \sin^{2}\theta + g^{2}t^{2} - 2ugt \sin\theta}$ $= \sqrt{u^{2} (\cos^{2}\theta + \sin^{2}\theta) + g^{2}t^{2} - 2ugt \sin\theta}$ ∴ $KE = \frac{1}{2} m (u^{2} + g^{2}t^{2} - 2ugt \sin\theta)$ i.e., $KE \propto t^{2}$ Hence, graph will be parabolic intercept on y-axis. Hence, the graph between KE and t. Now, in case of height $KE = \frac{1}{2} m (v^{2}) \text{ and } v^{2} = (u^{2} - 2gy)$ ∴ $KE = \frac{1}{2} m (u^{2} - 2gy)$ $KE = -mgy + \frac{1}{2} \mu^{2}$ Intercept on y-axis = $\frac{1}{2} \mu^{2}$ Now, $KE = \frac{1}{2} m v^{2}$