Question

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

• A. $-bu^{2}/a^{2}$
• B. $-u^{2}/b$
• C. $-au^{2}/b^{2}$
• D. $-u^{2}/a$

Answer 1

Answer: (A)

Path of particle is
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$
Differentiating with time t, we get
$\frac{x^{2}}{a^{2}}+\frac{dx}{dt}+\frac{y}{b^{2}}, \frac{dy}{dt}=0 \ldots\left(i\right)$
Again, differentiating with time t, we have
$\frac{1}{a^{2}}\left(x\frac{d^{2}x}{dt^{2}}+\left(\frac{dx}{dt}\right)^{2}\right) +\frac{1}{b_{2}}$
$\left(y \frac{d^{2}y}{dt^{2}}+\left(\frac{dy}{dt}\right)^{2}\right)=0\ldots\left(ii\right)$
Now, given $u_{x }= u at \left(0, b\right)$
So, from Eq. $\left(i\right),$ we have
$\frac{1}{b} \frac{dy}{dt}=0\Rightarrow \frac{dy}{dt}=u_{y}=0$
Now at $\left(0, b\right)$ from Eq. $\left(ii\right),$ we have
$\frac{1}{a^{2}}\left(\frac{dx}{dt}\right)^{2}+\frac{1}{b^{2}} \left(b \frac{d^{2}y}{dt^{2}}+\left(\frac{dy}{dt}\right)^{2}\right)=0$
$\frac{1}{a^{2}}.u^{2}+\frac{1}{b^{2}}.b \left(a_{y}\right)=0 \left[\because u_{y}=0\right] $
$\Rightarrow a_{y}=-\frac{b}{a^{2}}u^{2}$