Question

Heya...
Good Mrng..
⤴⤴Refer the attachment.. ⤴⤴
No spam..
​

Answer 1

Answer: A bullet looses "1/n" velocity in passing through a plank what is the number of planks required to stop the bullet (Assume constant retardation) is n^2/(2n-1)

Explanation:

Let the velocity of the bullet be "v"

After hitting it looses 1/n of it's velocity. Hence it's final velocity becomes (v-1/n*v)

Let the thickness of the plank be "t"

Apply,

v^2 = u^2 + 2as

(v-1/n*v)^2 = v^2 +2at

v^2(1-1/n)^2=v^2+2at

v^2(1+(1/n)^2-2/n)=v^2+2at

v^2((1-2n)/n^2)=2at=a

a=(v^2(1-2n))/(2t*n^2)

This is the retardation for first hit, let the least no of planks required to stop the bullet be k

Apply,

v^2=u^2+2as

0=v^2+2a(kt)

-v^2=+2(v^2(1-2n))/(2t*n^2)*(kt)

-v^2=(v^2(1-2n))k/n^2

k=-n^2/(1-2n)

k=n^2/(2n-1)

Hence the number of planks required are n^2/(2n-1)

Answer 2

Good Morning.

Your Question:

A Bullet of velocity v loses (1/n) of its velocity when passing through a plank. What is the least number of planks required to stop the bullet?

Step by step explanation:

Given in the image

Answer:

n^/2-1

Hope it helps you