Question

A small ball of mass $m$ starts from rest from point $A(b,c)$ on a smooth slope which is a parabola. The normal force that the ground exerts at the instant, the ball arrives at lowest point $(0,0)$ is

• A. $mg\left(\frac{b^2+4c^2}{b^2}\right)$
• B. $\frac{4mg{c}^2}{b^2}$
• C. $mg$
• D. $3mg$

Answer 1

The correct option is A $mg\left(\frac{b^2+4c^2}{b^2}\right)$

From the figure,
$mgc=\frac{1}{2}m{v}^2$
$\Rrightarrow v^2=2gc$
As $x^2=4Ay$
$\therefore b^2=4Ac$
$\Rightarrow A=\frac{b^2}{4c}$
$\therefore x^2=\frac{4b^2}{4c}\times y$
i.e., $x^2=\frac{b^2}{c}y$
$\therefore y=\frac{c}{b^2}{x}^2$
$\therefore \frac{1}{r}=\frac{\frac{d^{2}y}{d{x}^2}}{{[1+{\left(\frac{dy}{dx}\right)}^2]}^{3/2}}=\frac{2c}{\frac{b^2}{{[1+{\left(\frac{2c}{b^2}x\right)}^2]}^{3/2}}}$
At $x=0,y=0$
$\therefore \frac{1}{r}=\frac{2c}{b^2}$ or $r=\frac{b^2}{2c}$
Now, $N-mg=\frac{m\times 2gc}{b^2}\times 2c$
$\therefore N=\frac{4mg{c}^2}{b^2}+mg$
$\therefore N=mg\left(\frac{4c^2+b^2}{b^2}\right)$