Question

A bullet loses 1/20 of its initial velocity after penetrating a plank. How many plank's required to stop the bullet?

• A. 6
• B. 9
• C. 11
• D. 13

Answer 1

Answer: (C)

Let the acceleration provided by one plank be 'a'

The thickness of plank will be 'l'

Now, $v^2-u^2=2as$

If n no. of planks are required to stop the bullet:$0-v^2=2al$

$n=\frac{-v_0^2}{2al}$...(1)

velocity of bullet after passing though one plank,$v^{\prime }=v_0-\frac{v_0}{20}=\frac{19v_0}{20}$

After passing through one plank, $v^{\prime 2}-v_0^2=2as$

$\frac{361}{400}{v}_0^2-v_0^2=2al$

$-\frac{39}{400}{v}_0^2=2al$...(2)

putting (2) in (1): $n=\frac{-v_0^2}{-39v_0^2}\times 400$

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

Minimum 11 planks are needed.