A particle moves in a plane along an elliptic | vedclass

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Question

(a)

Path of particle is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Differentiating with time $t$, we get

$\frac{x}{a^2} \cdot \frac{d x}{d t}+\fr

Vedclass · Accepted Answer

$-b u^2 / a^2$

Vedclass · Answer

(a)

Path of particle is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Differentiating with time $t$, we get

$\frac{x}{a^2} \cdot \frac{d x}{d t}+\frac{y}{b^2} \cdot \frac{d y}{d t}=0 \dots(i)$

Again, differentiating with time $t$, we have

$\frac{1}{a^2}\left(x, \frac{d^2 x}{d t^2}+\left(\frac{d x}{d t}ight)^2ight)+\frac{1}{b^2}$

$\left(y \frac{d^2 y}{d t^2}+\left(\frac{d y}{d t}ight)^2ight)=0 \ldots(ii)$

Now, given $u_x=u$ at $(0, b)$

So, from Eq. $(i)$, we have

$\frac{1}{b} \frac{d y}{d t}=0 \Rightarrow \frac{d y}{d t}=u_y=0$

Now at $(0, b)$ from Eq. $(ii)$, we have

$\frac{1}{a^2}\left(\frac{d x}{d t}ight)^2+\frac{1}{b^2}\left(b \frac{d^2 y}{d t^2}+\left(\frac{d y}{d t}ight)^2ight)=0$

$	ext { or } \quad \frac{1}{a^2} \cdot u^2+\frac{1}{b^2} \cdot b\left(a_yight)=0 \quad\left[\because u_y=0ight]$

$\Rightarrow \quad a_y=-\frac{b}{a^2} u^2$

A particle moves in a plane along an elliptic path given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. At point $(0, b)$, the $x$-component of velocity is $u$. The $y$-component of acceleration at this point is

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