Question

A frog sits on the end of a long board of length $L=10cm$. The board rests on frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is minimum take of speed $v$ in $m{s}^{-1}$ relative to the ground that the frog follows to do the tricks ? [Assume that the board and frog have equal masses?]

• A. $2\sqrt{5}m{s}^{-1}$
• B. $5m{s}^{-1}$
• C. $0.5\sqrt{2}m{s}^{-1}$
• D. $10\sqrt{2}m{s}^{-1}$

Answer 1

The correct option is C $0.5\sqrt{2}m{s}^{-1}$

Suppose the frog jumped with velocity u at an

angle ${45}^{\circ }$ (for maximum ranse )

$\Rightarrow$ displacement relative to ground = $\frac{u^2}{9}$

recoil of board = x

so $x+\frac{u^2}{9}=10cm=0.1meter$

to calculate x we need to apply the faut

that center of mass shared Not move.

so

$mx-\frac{m{u}^2}{9}=0$ (No motion of c.m )

$\Rightarrow x=\frac{u^2}{9}\Rightarrow \frac{u^2}{9}=5cm=0.5meter$

$\Rightarrow u^2=0.5\times 9=0.5\times 10=0.5=\frac{50}{100}$

$\Rightarrow u=\frac{\sqrt{50}}{10}=\frac{5\sqrt{2}}{10}=0.5\sqrt{2}m/s$