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 $10 \, m \, s^{- 1}$ is

Answer 1

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