Question

A particle is projected along an inclined plane as shown in figure. What is the speed of the particle when it collides at point A? $(g=10m/s^2)$

Answer 1

Given : $u=10$ m/s

Initial velocity along x direction $u_x=ucos{30}^o$ m/s

Initial velocity along y direction $u_y=usin{30}^o$ m/s

Also acceleration x and y direction $a_x=-gsin{30}^o$ $m/s^2$ and $a_y=-gcos{30}^o$ $m/s^2$

x direction : $V_x=u_x+a_{x}t$

time $t=\frac{2\times 10\sin 30}{10cos30}$, hence $t=\frac{2}{\sqrt{3}}$ s

$\therefore$ $V_x=ucos{30}^o-\left(gsin{30}^o\right)t$

OR $V_x=10\times \frac{\sqrt{3}}{2}-10\times 0.5\times \frac{2}{\sqrt{3}}$ $=$ $\frac{15-10}{\sqrt{3}}=\frac{5}{\sqrt{3}}$ $m/s$

Also $V_y=-5$ $m/s$ (solved earlier)

$\therefore$ Speed of the particle at collision $V=\sqrt{V_x^2+V_y^2}=\sqrt{(\frac{5}{\sqrt{3}}{)}^2+(-5{)}^2}$

$⟹$ $V=\sqrt{\frac{100}{3}}=\frac{10}{\sqrt{3}}$ $m/s$