WELL ,  i have a shortcut for this

let   vel   deacrease  by v/n  every time it hit the plank

then no. of plank =   n^2/ 2n - 1

so ans=  400/39=10....something

which means more than 10 planks=11

plz rate if  useful

hey abhi!!

how did u apply that fornulae of no . of planks!!

WELL THAT PROBLEM HAVE LOTS OF MIXED CONCEPT of work  energy  and little  bit of integration 

and that  formula can be derived

its  also there in some of  books

ans) let the intial velocity of bullet = u thickness of plank be X . on passing thru a plank the velocity of bullet b'coms 19u/20 .



using v² - u² = 2as = (19u/20)2 - u2  = 2aX ---(1) . n  let the bullet pass thru N planks n then come t rest . here total no of planks = N+1.. includin the first 1 .. therefore v get 

02 - u2 =  2a(N+1)X  ----(2)

solvin(1) v get  -39u² = 800aX , (2) => -u² = 2a(N+1)X 

v get N = 10(aprox) n total no of planks = 11 .

rate if useful