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. $-b{u}^2/a^2$
• B. $-u^2/b$
• C. $-a{u}^2/b^2$
• 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
$\frac{x^2}{a^2}+\frac{dx}{dt}+\frac{y}{b^2},\frac{dy}{dt}=0\dots \left(i\right)$
Again, differentiating with time t, we have
$\frac{1}{a^2}\left(x\frac{d^{2}x}{d{t}^2}+{\left(\frac{dx}{dt}\right)}^2\right)+\frac{1}{b_2}$
$\left(y\frac{d^{2}y}{d{t}^2}+{\left(\frac{dy}{dt}\right)}^2\right)=0\dots \left(ii\right)$
Now, given $u_x=uat\left(0,b\right)$
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
$\frac{1}{a^2}{\left(\frac{dx}{dt}\right)}^2+\frac{1}{b^2}\left(b\frac{d^{2}y}{d{t}^2}+{\left(\frac{dy}{dt}\right)}^2\right)=0$
$\frac{1}{a^2}.u^2+\frac{1}{b^2}.b\left(a_y\right)=0\left[\because u_y=0\right]$
$\Rightarrow a_y=-\frac{b}{a^2}{u}^2$