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3-2.Motion in Plane
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A particle moves in the $xy$ plane with a constant acceleration $'g'$ in the negative $y$-direction. Its equation of motion is $y = ax-bx^2$, where $a$ and $b$ are constants. Which of the following are correct?
AThe $x$-component of its velocity is constant.
BAt the origin, the$y$-component of its velocity is a$\sqrt {\frac{g}{{2b}}} $.
CAt the origin, its velocity makes an angle $tan^{-1}(a)$ with the $x$-axis.
DAll of the above
Solution
As in this it is given that the particle will only have the acceleration in the negative $y-$ direction so the acceleration in the $x-$direction is zero so $\quad a_{x}=0$
so the velocity in the $x$ -direction will be constant as acceleration in the $x$ -direction is
zero now $y=a x-b x^{2}$
so now differentiate the equation so we get
$\frac{\mathrm{d} y}{\mathrm{d} t}=a \frac{\mathrm{d} x}{\mathrm{d} t}-b 2 x \frac{\mathrm{d} x}{\mathrm{d} t}$
now differentiate this again $\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}}=a \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}-2 b\left(\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+x \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}\right)$
that is $a_{y}=a a_{x}-2 b\left(v_{x}^{2}+2 x a_{x}\right)$
as $a_{x}=0$ and $a_{y}=-g$
$\mathrm{so}-g=-2 b v_{x}^{2}$
so $v_{x}=\sqrt{g / 2 b}$
so as we know $v_{y}=a v_{x}-2 b x v_{x}$
as at origin means $x=0$ $v_{y}=a v_{x}$
so $v_{y}=a \sqrt{g / 2 b}$
so the velocity in the $x$ -direction will be constant as acceleration in the $x$ -direction is
zero now $y=a x-b x^{2}$
so now differentiate the equation so we get
$\frac{\mathrm{d} y}{\mathrm{d} t}=a \frac{\mathrm{d} x}{\mathrm{d} t}-b 2 x \frac{\mathrm{d} x}{\mathrm{d} t}$
now differentiate this again $\frac{\mathrm{d}^{2} y}{\mathrm{d} t^{2}}=a \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}-2 b\left(\left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{2}+x \frac{\mathrm{d}^{2} x}{\mathrm{d} t^{2}}\right)$
that is $a_{y}=a a_{x}-2 b\left(v_{x}^{2}+2 x a_{x}\right)$
as $a_{x}=0$ and $a_{y}=-g$
$\mathrm{so}-g=-2 b v_{x}^{2}$
so $v_{x}=\sqrt{g / 2 b}$
so as we know $v_{y}=a v_{x}-2 b x v_{x}$
as at origin means $x=0$ $v_{y}=a v_{x}$
so $v_{y}=a \sqrt{g / 2 b}$
Standard 11
Physics
Similar Questions
Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.
| $Column-I$ | $Column-II$ |
| $(A)$ Angle of projection | $(p)$ $20\,m$ |
| $(B)$ Angle of velocity with horizontal after $4\,s$ | $(q)$ $80\,m$ |
| $(C)$ Maximum height | $(r)$ $45^{\circ}$ |
| $(D)$ Horizontal range | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
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