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Level 2 Solutions: Motion in a Plane
Student Name: ____________________________________ Class: 11th (Physics) Subject: Physics
Level 2 Solutions
1.
A projectile is launched up an inclined plane of inclination $\alpha=30^\circ$ with initial velocity $u=30\text{ m/s}$ at an angle $\beta=30^\circ$ relative to the incline. Find the range along the incline.
Sol: $R = \frac{2u^2 \sin\beta \cos(\alpha+\beta)}{g \cos^2\alpha} = \frac{2(900)\sin 30^\circ \cos 60^\circ}{9.8 \cos^2 30^\circ} = \frac{1800(0.5)(0.5)}{9.8(0.75)} = 61.22\text{ m}$.
2.
A particle moves in x-y plane with position coordinates $x(t) = R\cos(\omega t)$ and $y(t) = R\sin(\omega t)$. Prove that velocity is perpendicular to position vector.
Sol: $\vec{r} = R\cos(\omega t)\hat{i} + R\sin(\omega t)\hat{j}$. $\vec{v} = -R\omega\sin(\omega t)\hat{i} + R\omega\cos(\omega t)\hat{j}$. $\vec{r}\cdot\vec{v} = -R^2\omega\sin(\omega t)\cos(\omega t) + R^2\omega\sin(\omega t)\cos(\omega t) = 0$. Perpendicular!
3.
Two projectiles are thrown simultaneously from the top of a tower of height $H$ with speeds $u_1$ and $u_2$ horizontally in opposite directions. Find distance between them when their velocities become perpendicular.
Sol: $\vec{v}_1 = u_1\hat{i} - gt\hat{j}$, $\vec{v}_2 = -u_2\hat{i} - gt\hat{j}$. $\vec{v}_1 \cdot \vec{v}_2 = -u_1 u_2 + g^2 t^2 = 0 \implies t = \frac{\sqrt{u_1 u_2}}{g}$. Separation $d = (u_1 + u_2)t = \frac{(u_1 + u_2)\sqrt{u_1 u_2}}{g}$.