Question

A particle of mass $2\text{kg}$ is moving along a line $4y–3x+c=0$ with a constant speed of $5\text{m/s}$. If angular momentum about origin is $80{\text{kg m}}^2{\text{s}}^{-1}$, then $c$ is

• A. $10$
• B. $-10$
• C. $20$
• D. $-40$

Answer 1

The correct option is D $-40$

Given,
$m=2\text{kg},L=80{\text{kg m}}^2{\text{s}}^{-1}$ about origin.
Speed $v=5\text{m/s}$

Since particle is moving along $4y-3x+c=0$ i.e $y=\frac{3}{4}x-\frac{c}{4}$
Slope $\tan \theta =\frac{3}{4}\therefore \theta ={37}^{\circ }$

$\therefore$ perpendicular distance from origin
$r_{\perp }=\frac{-c}{4}\sin {53}^{\circ }=\frac{4}{5}\times \frac{c}{4}=\frac{-c}{5}$

$L=mv{r}_{\perp }$
$\Rightarrow 80=2\times 5\times \left(\frac{-c}{5}\right)$
$\Rightarrow c=\frac{-80}{2}=-40$