Question

A man of mass m is standing on one end o a plank of mass 2 m floating on a river. The other end just touches the bank of the river as shown in figure. With what minimum speedw.r.t. the plank should the man jump to getout of the river. All surfaces are smooth and plank is always in level with the bank. Given that the length of the plank is 15 m.

• A. $7m/s$
• B. $7.5m/s$
• C. $9m/s$
• D. $5m/s$

Answer 1

The correct option is C $9m/s$

When the man reaches at the other end of the plank, river bank will be at a distance 5 m from the plank.

As $m\left(15x\right)=2mx=2mx,15=3x\Rightarrow x=5$.

Therefore for maximum range w.r.t. ground

$\frac{v^2}{g}=5\Rightarrow v=7m/s$

Velocity of man w.r.t. ground

$\therefore {\overline{v}}_m=vcos{45}^o\hat{i}+vsin{45}^o\hat{j}=\frac{7}{\sqrt{2}}\hat{i}+\frac{7}{\sqrt{2}}\hat{j}$

Velocity of plank

${\overline{v}}_P=\frac{m\left(v_m{)}_x\right(-\hat{i})}{2m}=\frac{-7}{2\sqrt{2}}\hat{i}$,

${\overline{v}}_{m/}=(\frac{7}{\sqrt{2}}+\frac{7}{2\sqrt{2}})\hat{i}+\frac{7}{\sqrt{2}}\hat{j}=\frac{2l}{2\sqrt{2}}\hat{i}+\frac{7}{\sqrt{2}}\hat{j}$

$\left|{\overline{v}}_{m/p}\right|=\sqrt{{\left(\frac{21}{2\sqrt{2}}\right)}^2+{\left(\frac{7}{\sqrt{2}}\right)}^2}$

$=\sqrt{\frac{441}{8}+\frac{49}{2}}=\sqrt{\frac{441+196}{8}}=9m{s}^{-1}$