Question

A bullet loses $(\frac{1}{n}{)}^{th}$ 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)

$\frac{n^2}{2n-1}$

Let the bullet be traveling at a speed v before entering the first plank.

If it loses $(\frac{1}{n}{)}^{th}$ velocity its new velocity will become $v-\frac{v}{n}$

Kinetic energy lost $\Delta E=\frac{1}{2}m{v}^2-\frac{1}{2}m{(v-\frac{v}{n})}^2=\frac{1}{2}m{v}^2(1-{\left(\frac{n-1}{n}\right)}^2)$

Now, let N be the number of planks after which the velocity of the bullet becomes 0.
Thus we get

$N\times \text{(Energy lost in one plank)}=\text{Total Energy}$
or

$N\times \Delta E=\frac{1}{2}m{v}^2$

$\frac{m{v}^2}{2\Delta E}=N=\frac{1}{1-(\frac{n-1}{n}{)}^2}$

$N=\frac{n^2}{2n-1}$

The velocity will become 0 after passing through $\frac{n^2}{2n-1}$ planks