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A particle of mass \mathrm{2 \mathrm{~kg}} at rest, it its acted upon the force \mathrm{\vec{F}=2 \hat{i}+3 \hat{j}}. What will be its equation of path-

Option: 1

Ellipse
 


Option: 2

Straigh line

 


Option: 3

circle

 


Option: 4

Parabola


Answers (1)

best_answer

Firstly we will have to find out x and y. Using Newton's formula-

\begin{array}{ll} \vec{F}_x=m \vec{a}_x, \,\, \, \, \, \, \, \, \, \, \, ,&\, \, \, \, \, \vec{F}_y=m \vec{a}_y \\ 2=2 a_x & \, \, \, \, \, \, 3=2 \vec{a}_y \\ \vec{a}_x=1 \mathrm{~m} / \mathrm{s}^2 & \, \, \, \, \, \, \, \vec{a}_y=3 / 2 \mathrm{~m} / \mathrm{s}^2 \end{array}

Use Newton's equation of motion - 

\mathrm{\begin{aligned} & x=u_x t+\frac{1}{2} a_x t^2 \\ & x=0 x t+\frac{1}{2} \cdot 1 t^2 \\ & x=\frac{1}{2} t^2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -(1) \end{aligned}}

\mathrm{\begin{aligned} \& \quad y & =u_y t+\frac{1}{2} a_y t^2 \\ & =0 x t+\frac{1}{2} \times \frac{3}{2} \cdot t^2 \\ y & =\frac{3}{4} t^2\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, -\text { (2) } \end{aligned}}

Equation (1) dévide (2) - 

\mathrm{ \begin{aligned} & \frac{x}{y}=\frac{1}{2} t^2 \times \frac{4}{3} t^2=\frac{2}{3} \\ & y=\frac{3}{2} x \quad(\because y=m x) \end{aligned} }
\longrightarrow straight line Ans.

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Pankaj

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