Question

a bullet losses 1/n of its velocity in passing through a plank.what is the last no of plank required to stop the bullet?

Answer 1

Answer:

Explanation: let the velocity of the bullet be v. if the bullet passes through the first plank it reduces 1/n of its velocity i.e 1/n × v = v/n

since it losses this v/n velocity then the new velocity will be v - v/n = v(n-1) / n

which is the velocity of the bullet after it passes through the first plank.

now as it passes through the second plank then again it losses v/n velocity and becomes $\frac{v(n-1)}{n} - \frac{v}{n}$ = $\frac{v(n-2)}{n}$

now as it passes through the third plank it again losses v/n of its velocity and becomes $\frac{v(n-2)}{n} - \frac{v}{n} = \frac{v(n-3)}{n}$.

now lets make this as a general statement .now let the above be

$\frac{v(n-m)}{n}$ where m = 1 as it passes through the first plank

                                                   m=2  as it passes through the second plank

                                                   m=3  as it passes through the third

                                                   m = m as it passes through the m th plank

for our velocity velocity v¹ = 0

$\frac{v(n-m)}{n}$ = 0

which bringes us with n-m = 0

∴ m = n

so the velocity becomes zero and the bullet stops when it passes through the n th plank.

THUS THE SOLUTION.