Question

A man throws a ball of mass $m$ from a moving plank ( $A$ ). The ball has a horizontal component of velocity $u=3\sqrt{34}m{s}^{-1}$ with respect to ground towards another man standing on stationary plank ( $B$ ), as shown in the figure. The combined mass of plank $A$ with man as well as that of plank $B$ with the other man is $2m$ each. the x-y plane represents the horizontal plane which is frictionless. If initially, plank $A$ was moving with speed $u\hat{i}$ with respect to ground, then the relative velocity of plank $B$ (in $m{s}^{-1}$ ) with respect to plank $A$ (after the man on plank $B$ catches the ball) minus $10m{s}^{-1}$ is

Answer 1

Answer: 7

By conservation of momentum,
$3mu\hat{i}=m(ucos{\left(37\right)}^o\hat{i}+usin{\left(37\right)}^o\hat{j})+2m\overrightarrow{V_A}$
$=\overrightarrow{V_A}=\frac{11u}{10}\hat{i}-\frac{3u}{10}\hat{j}$
Similarly $=\overrightarrow{V_B}=\frac{4u}{15}\hat{i}+\frac{3u}{10}\hat{j}$
$\therefore \left|V_{BA}\right|=17m{s}^{-1}$