Question

A bullet loses $\frac{1}{20}$ of its velocity in passing through a plank. What is the least number of planks required to stop the bullet?

• A. 9
• B. 10
• C. 11
• D. 12

Answer 1

The correct option is C 11

let the width of a plank be d and no of planks be n

then final speed after one plank is $v=u-\frac{u}{20}=\frac{19u}{20}$

$v^2=u^2+2aS$

Velocity decreasing so acceleration is negative.

$(\frac{19u}{20}{)}^2=u^2-2a\times d$

$a=\frac{39u^2}{800d}$

number of planks required to stop the bullet

$V^2=U^2+2aS$

$0=u^2-2\times \frac{39u^2}{800d}\times nd$

$n=\frac{400}{39}=10.25$

$\therefore \text{There are}11$ full planks.