NEET Physics Motion in A Plane Questions Solved

(2)

 $$\begin{gathered} \mathrm{Let} \mathrm{time} \mathrm{to} \mathrm{reach} \mathrm{P} \mathrm{from} \mathrm{A} \mathrm{be} \mathrm{t} \\ \mathrm{H}=\frac{1}{2}{\mathrm{gt}}^2 \\ \mathrm{t}=\sqrt{\frac{2\mathrm{H}}{\mathrm{g}}} \\ \mathrm{Time} \mathrm{to} \mathrm{reach} \mathrm{C} \mathrm{from} \mathrm{P} \mathrm{be} \mathrm{t}\text{'} \\ \frac{\mathrm{H}}{2}=\frac{1}{2}\mathrm{g}{\left(\mathrm{t}\text{'}\right)}^2 \\ \mathrm{t}\text{'}=\sqrt{\frac{\mathrm{H}}{\mathrm{g}}}= \sqrt{\frac{2\mathrm{H}}{2\mathrm{g}}}= \frac{\mathrm{t}}{\sqrt{2}} \\ \mathrm{So} \mathrm{total} \mathrm{time} \mathrm{to} \mathrm{go} \mathrm{from} \mathrm{A} \mathrm{to} \mathrm{C} \\ \mathrm{t}+\mathrm{t}\text{'}= \mathrm{t}+\frac{\mathrm{t}}{\sqrt{2}} \\ = \mathrm{t} \left[1+\frac{1}{\sqrt{2}}\right] \end{gathered}$$

$$\begin{gathered} \mathrm{v} \mathrm{sin\theta }=\mathrm{u} \mathrm{cos\theta } \\ \mathrm{v}=\mathrm{u} \mathrm{cot\theta } \end{gathered}$$

$\mathrm{Initially}$

$$\begin{gathered} \mathrm{Using} \mathrm{relative} \mathrm{motion} \mathrm{concept} \\ {\overrightarrow{\mathrm{V}}}_{\mathrm{B}, \mathrm{A}} = {\overrightarrow{\mathrm{V}}}_{\mathrm{B}}-{\overrightarrow{\mathrm{V}}}_{\mathrm{A}} \end{gathered}$$

$$\begin{gathered} \mathrm{tan\theta }=\frac{{\mathrm{V}}_{\mathrm{B}}}{{\mathrm{V}}_{\mathrm{A}}}=1 \\ \mathrm{\theta }=45° \end{gathered}$$

At shortest distance, the line dropped from A onto B should be perpendicular

(1)

$$\begin{gathered} \mathrm{Time} \mathrm{to} \mathrm{go} \mathrm{from} \mathrm{A} \mathrm{to} \mathrm{B} = \mathrm{t} \\ \mathrm{AC}= \mathrm{Horizontal} \mathrm{displacement} \mathrm{in} \mathrm{time} \mathrm{t} \\ = \left(\mathrm{ucos}60°\right) \mathrm{t} \\ \mathrm{AC}= \frac{\mathrm{ut}}{2} \\ \mathrm{In} \mathrm{right}-\mathrm{angled} ∆\mathrm{ABC} \\ \mathrm{AC} = \mathrm{AB} \cos 30° \\ \mathrm{AB} = \frac{\mathrm{AC}}{\cos 30°}= \frac{\mathrm{ut}}{\sqrt{3}} \end{gathered}$$

(A)

At any instant,

$$\begin{gathered} \mathrm{At} \mathrm{any} \mathrm{general} \mathrm{point} \mathrm{P}(\mathrm{x},\mathrm{y}) \\ \mathrm{In} \mathrm{y}-\mathrm{direction}, {\left({\mathrm{u}}_{\mathrm{y}}\right)}^2={\left(\mathrm{usin\theta }\right)}^2-2\mathrm{gy} \\ \mathrm{In} \mathrm{x}-\mathrm{direction}, {\mathrm{u}}_{\mathrm{x}}= \mathrm{ucos\theta } \end{gathered}$$

$$\begin{gathered} \mathrm{Total} \mathrm{K}.\mathrm{E}. = \frac{1}{2} {{\mathrm{mu}}_{\mathrm{y}}}^2+\frac{1}{2}{{\mathrm{mu}}_{\mathrm{x}}}^2 \\ = \frac{\mathrm{m}}{2}\left[{\left(\mathrm{usin\theta }\right)}^2-2\mathrm{gy}+{\left(\mathrm{ucos\theta }\right)}^2\right] \\ \mathrm{K}.\mathrm{E}. = \frac{\mathrm{m}}{2}\left[{\mathrm{u}}^2-2\mathrm{gy}\right] \left[\because {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\right] \\ \mathrm{is} \mathrm{of} \mathrm{the} \mathrm{form} \mathrm{y}\text{'}=\mathrm{mx}+\mathrm{c} .................\mathrm{Equation} \mathrm{of} \mathrm{Straight} \mathrm{line} \\ \mathrm{where} \mathrm{y}\text{'}=\mathrm{K}.\mathrm{E}. \\ \mathrm{x}=\mathrm{y} \end{gathered}$$

$$\begin{gathered} \mathrm{At} \mathrm{time} \mathrm{t}, \\ \mathrm{A}, \mathrm{B} \& \mathrm{C} \mathrm{are} \mathrm{on} \mathrm{a} \mathrm{straight} \mathrm{line}, \mathrm{so} \\ \mathrm{Slope} \mathrm{of} \mathrm{AC} = \mathrm{Slope} \mathrm{of} \mathrm{BC} \\ \frac{{\mathrm{V}}_2\mathrm{t}-0}{0-{\mathrm{v}}_1\mathrm{t}}=\frac{{\displaystyle \frac{\mathrm{Vt}}{\sqrt{2}}}-0}{{\displaystyle \frac{\mathrm{Vt}}{\sqrt{2}}}-{\mathrm{V}}_1\mathrm{t}} \\ \frac{{\mathrm{V}}_2}{{\mathrm{V}}_1} = \frac{\mathrm{V}}{\sqrt{2}{\mathrm{V}}_1-\mathrm{V}} \\ \mathrm{V} = \frac{\sqrt{2}{\mathrm{V}}_1{\mathrm{V}}_2}{{\mathrm{V}}_1+{\mathrm{V}}_2} \end{gathered}$$