Question

An insect sits on the end of a long board of length $5m$. The board rests on a frictionless horizontal table. The insect wants to jump to the opposite end of the board. What is the minimum take-off speed (in m/s) of insect relative to ground, that allows the insect to do the trick? The board and the insect have equal masses. $(g=10m/s^2)$.

Answer 1

Let insect jump with speed $V$ relative to ground and at an angle $\theta$ with horizontal.

The velocity of insect has two components.

$v\sin \theta ⟹$verticle

$v\cos \theta ⟹$horizontal relative to ground.

Then horizontal speed of insect relative to bound is $(v\cos \theta +v\cos \theta )$ by momentum conservation. Board speed is also same in opposite direction. Due to same mass horizontal speed is same then horizontal velocity=$2v\cos \theta$.

Time of flight of insect=$\frac{2v\sin \theta }{g}⟶\left(1\right)$

In this time insect cover $5$m.

Then, $t_1=\frac{2m}{2v\cos \theta }\frac{2v\sin \theta }{g}=\frac{5}{2v\cos \theta }2v^2\sin \theta \cos \theta =\frac{5g}{2}{v}^2\sin 2\theta =\frac{5\times 10}{2}=25$

$V$ should b minimum then $\sin 2\theta ⟹$maximum.

$2\theta =90$

$\theta =45°$

$v^2=25$

$v=5$m/s