Question

A bullet looses ${\left(\frac{1}{n}\right)}^{th}$ of its velocity passing through one plank. The number of such planks that are required to stop the bullet can be :

• A. $\frac{n^2}{2n-1}$
• B. $\frac{2n^2}{n-1}$
• C. Infinite
• D. n

Answer 1

Answer: (A)

Let the bullet be traveling at a speed v before entering the first plank.
If it loses ${\left(\frac{1}{n}\right)}^{\text{th }}$ velocity its new velocity will become $v-\frac{v}{n}$
Kinetic energy lost $\Delta E=\frac{1}{2}m{v}^2-\frac{1}{2}m{\left(v-\frac{v}{n}\right)}^2=\frac{1}{2}m{v}^2\left(1-{\left(\frac{n-1}{n}\right)}^2\right)$
Now, let $N$ be the number of planks after which the velocity of the bullet becomes 0 .
Thus we get
$N\times ($ Energy lost in one plank $)=$ Total Energy or
$N\times \Delta E=\frac{1}{2}m{v}^2$
$\frac{m{v}^2}{2\Delta E}=N=\frac{1}{1-{\left(\frac{n-1}{n}\right)}^2}$
$N=\frac{n^2}{2n-1}$
The velocity will become $o$ after passing through $\frac{n^2}{2n-1}$ planks