Solve the following :


A force $\vec{F}=\vec{v} \times \vec{A}$ is exerted on a particle in addition to the force of gravity, where $\vec{v}$ is the velocity of the particle and $\vec{A}$ is a constant vector in the horizontal direction. With what minimum speed a particle of mass $m$ be projected so that it continues to move undeflected with a constant velocity?


$\vec{F}=\vec{v} \times \vec{A}$

$\vec{A}=A \hat{\imath}$

$\left|\begin{array}{llr}\hat{\imath} & \hat{\jmath} & \hat{k} \\ V_{x} & V_{y} & V_{z} \\ A & 0 & 0\end{array}\right|=\hat{\jmath}\left(-V_{2} A\right)+\hat{k}\left(-V_{y} A\right)$

$=V_{z} A \hat{\jmath}-V_{y} A \hat{k}$

$\widehat{F}_{g}=m g(-\widehat{j})$

To go undeflected $F_{\text {net }}=0$


$\left(-m g+V_{2} A\right)-V_{g} A=0$

$=>V_{g}=0 V_{2} A=m g$

$V_{2}=V \sin \theta$

$V_{2}=V \sin \theta$

$V \sin \theta=\frac{m g}{A}$

$V=\frac{m g}{A \sin \theta}$

$\circlearrowright V=V_{\min }$ if $\sin ^{\theta}>1 \Rightarrow{ }^{\theta}>>90^{\circ}$

o i.e $\mathrm{V}_{\min }=\frac{m g}{A}$ along $x$ - axis


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now