Question

A bullet loses 1/10th of it's velocity while passing through a plank. What is the least number of planks required to stop the bullet?

Answer 1

Explanation:

Let number of block be n . and initial velocity v then

$\: v \times ( \frac{9}{10} ) {}^{n} = 0 \\ so \: \: n \to \: \infty$

I think more information should be given . or there should be a least velocity to passing through plank , bullet must have.

Answer 2

The least number of planks required to just stop the bullet is 6

Explanation:

Bullet looses 1/10th of its velocity while passing through plank

Let the resistance offered by the plank be a

if the initial velocity of bullet be u and final velocity be v and the depth of plank d

Then

By the second equation of motion

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

$v^2=u^2-2ad$

But

$v=u-\frac{1}{n}u$

$\implies v=u(1-\frac{1}{10})=\frac{9u}{10}$

Therefore,

$u^2(\frac{9u}{10})^2=u^2-2ad$

$\implies 2ad=u^2[1-(\frac{9}{10})^2]$

$\implies 2ad=u^2[1-(1-\frac{81}{100}]$

$\implies 2ad=u^2[\frac{19}{100}]$  .............. (1)

Let there are N planks

Then, after passing through these N planks, the velocity of bullet will be zero

Thus,

$0^2=u^2-2aNd$

$\implies 2aNd=u^2$  ................... (2)

Dividing eq (1) by eq (2)

$\frac{1}{N}=\frac{19}{100}$

$\implies N=\frac{100}{19}=5.26$

But the no. of planks must be whole number

Therefore, N = 6

Hope this answer is helpful.

Know More:

Q: A bullet leaving the muzzle of a rifle barrel with a  velocity v penetrates a plank and loses one fifth of its  velocity. It then strikes second plank, which it just  penetrates through. Find the ratio of the thickness of  the planks supposing average resistance to the penetration is same in both the cases.

Click Here: https://brainly.in/question/6771967

Q: A rifle bullets loses 1/20 of its velocity in passing through a plank exerts a constant retarding force the least no. of such planks required just to stop the bullet.

Click Here: https://brainly.in/question/1512084