Question

If a bullet looses  

1

th

n

of its velocity while passing through a plank, then

the number of such planks required to just stop the bullet is ___​

Answer 1

Question:-

If a bullet loses $\sf{\left(\dfrac{1}{n}\right)^{th}}$ of its velocity while passing through a plank, then  find the number of such planks required to just stop the bullet.

Answer:-

$\Large\boxed{\sf{\dfrac{n^2}{2n-1}}}$

Solution:-

Let u be the initial velocity of the bullet so that its final velocity after passing through one plank is,

$\longrightarrow\sf{v=u\left(1-\dfrac{1}{n}\right)}$

$\longrightarrow\sf{v=u\left(\dfrac{n-1}{n}\right)}$

Let F be the force exerted by the plank against the bullet and d be the width of the plank. Let m be the mass of the bullet.

Work done by the plank against bullet is, by work - energy theorem,

$\longrightarrow\sf{Fd=\dfrac{1}{2}m(v^2-u^2)}$

If M is the mass of the plank and a, the retardation of the plank,

$\longrightarrow\sf{Mad=\dfrac{1}{2}m\left(u^2\left(\dfrac{n-1}{n}\right)^2-u^2\right)}$

$\longrightarrow\sf{Mad=\dfrac{1}{2}mu^2\left(\dfrac{(n-1)^2-n^2}{n^2}\right)}$

$\longrightarrow\sf{Fd=-\dfrac{1}{2}mu^2\left(\dfrac{2n-1}{n^2}\right)\quad\quad\dots(1)}$

Now consider when the bullet comes to rest after passing through a large mass made by arranging a no. of such identical planks together. Let M' be the mass of this large plank formed.

Work done by this large plank against the bullet is, by work - energy theorem,

$\longrightarrow\sf{F'd'=\dfrac{1}{2}m(0^2-u^2)\quad\quad\dots(2)}$

But here,

• $\sf{F'=F}$

because each plank, by which the large one is made, offers same retarding force for a particular time interval during the motion of the bullet. [Remember, the mass of an infinitesimal layer of the retarding plank is not changed though the large plank is made out.]

But the distance is increased in proportional to the no. of plank, or to the mass, as,

• $\sf{d'=d\cdot\dfrac{M'}{M}}$

since the identical planks are arranged together which are against to the motion of the bullet.

Thus (2) becomes,

$\longrightarrow\sf{Fd\cdot\dfrac{M'}{M}=-\dfrac{1}{2}mu^2}$

From (1),

$\longrightarrow\sf{-\dfrac{1}{2}mu^2\left(\dfrac{2n-1}{n^2}\right)\cdot\dfrac{M'}{M}=-\dfrac{1}{2}mu^2}$

$\longrightarrow\sf{\dfrac{2n-1}{n^2}\cdot\dfrac{M'}{M}=1}$

$\longrightarrow\sf{\underline{\underline{\dfrac{M'}{M}=\dfrac{n^2}{2n-1}}}}$

Well, the ratio $\sf{\dfrac{M'}{M}}$ is the ratio of mass of the large plank to that of each plank by which it's made, which actually indicates the no. of the identical planks required to make the larger one.

Thus we can say $\sf{\dfrac{n^2}{2n-1}}$ planks are needed to make the larger plank, and hence $\sf{\dfrac{n^2}{2n-1}}$ planks are required to stop the bullet.

Therefore $\sf{\dfrac{n^2}{2n-1}}$ is the answer.