Q73 / 90 · KVPY 2011 (SB/SX) · Physics
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 $\mathrm{u}$. The y-component of acceleration at this point is-
- Correct answerA$-\mathrm{bu}^{2} / \mathrm{a}^{2}$
- B$-u^{2} / b$
- C$-\mathrm{au}^{2} / \mathrm{b}^{2}$
- D$-\mathrm{u}^{2} / \mathrm{a}$
Solution
$\begin{array}{l}
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \\
u_{x}=u \text { at }(0, b) \\
u_{y}=0 \\
\frac{2 x}{a^{2}} \frac{d x}{d t}+\frac{2 y}{b^{2}} \frac{d y}{d t}=0
\end{array}$
Again diff. w.r.t. to time
$\frac{2 x}{a^{2}} \frac{d^{2} x}{d t^{2}}+\frac{2}{a^{2}}\left(\frac{d x}{d t}\right)^{2}+\begin{array}{ll}
2 y & d^{2} y \\
b^{2} & {d t^{2}}+\frac{2}{b^{2}}\left(\begin{array}{l}
d y \\
d t
\end{array}\right)=0
\end{array}$
acceleration at $(0, b)$ is
$a_{y}=\frac{-b}{a^{2}} u^{2}$
Hence, the correct answer is option: 'A'.