A frog sits on the end of a long board of length `L = 5 m`. The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed (in `m//s`), i.e., relative to ground `'v'` that allows the frog to do the trick? The board and the frog have equal masses.

Options

A. `2sqrt(5)ms^(-1)`
B. `5 ms^(-1)`
C. `5sqrt(2)ms^(-1)`
D. `10sqrt(2)ms^(-1)`

Answer

C

Explanation

Let the speed of the board be u and from jumps with angle of inclination to the board `theta`, then from conservation of momentum in horizontal direction.
`mvcostheta - mu = 0`
`u=vcostheta`..........(i)
Let distance moved by board be x
So, `L-x = ut`......................(ii) and `x = vcos thetat`....................(iii)
Solving above equation, we get
Also, `x = (v^(2)sing2theta)/(g)`
`rArr (L/2) = (v^(2)sin 2theta)/(g)` `rArr v` = `sqrt(gL)/(2sin 2theta)`
Hence, v should be minimum for `sin 2theta =1`(i.e., maximum)
`rArr v_(min)`= `(sqrt(gL)/(2) = sqrt(10 xx 10/2) = sqrt(50) = 5sqrt(2) ms^9-1)`