Question

A rifle bullet loses 1/20th of its velocity in passing through a plank. What will be the least number of such planks required to just stop the bullet?

Answer 1

Hint: When the bullet travels through the plank the will do a work W=-Fd, where d is the thickness of the plank. Calculate the change in kinetic energy of the bullet due to this force. The work is equal to change in kinetic energy. Repeat the same procedure when the bullet passes through n planks.

Formula used:
W=Fs.
$K=\dfrac{1}{2}m{{v}^{2}}$
$W=\Delta K$
${{v}^{2}}-{{u}^{2}}=2as$

Complete answer:
Let us first understand what happens to a bullet when passing through a plank or any material. Suppose a bullet is shot from a gun. The velocity of the bullet be v. Let us assume that there are no forces influencing the motion of the bullet.
When a body is in motion, we say that it possesses some energy. This energy is called kinetic energy. The kinetic of a body of mass m and travelling with a speed v is given as $K=\dfrac{1}{2}m{{v}^{2}}$.
Therefore, the bullet also possesses some kinetic energy. Let the kinetic energy of the bullet be K.
Now suppose we placed a material in front of this bullet. We know that the bullet passed through the material. However, when the bullet is inside the material, the material opposes the motion of the bullet by applying a force on the bullet in the direction opposite to the velocity of the bullet. The direction of the applied is in the opposite direction of the bullet’s displacement. This means the material does a negative work on the bullet.
As a result the velocity of the bullet decreases as moves through the material. This means that its kinetic energy becomes less when it comes out. Therefore, we say that the bullet loses some of its initial kinetic energy.
Let us assume that the plank applies a constant force F. Let the thickness of the plank bed. Therefore, the bullet’s displacement inside the plank is d. Hence, the work done is W=-Fd.
The work done will be equal to the change in kinetic energy.
i.e $W=\Delta K$ …. (i)
Let us calculate the change in the kinetic energy of the bullet.
It is given that the velocity of the bullet loses 1/20th of its initial velocity (velocity before hitting the plank). Therefore, the velocity of the bullet after coming out is $v-\dfrac{v}{20}=\dfrac{19v}{20}$.
Let the kinetic energy after coming out of the plank be K’.
Hence, $K'=\dfrac{1}{2}m{{\left( \dfrac{19v}{20} \right)}^{2}}=m\dfrac{361{{v}^{2}}}{800}$
Therefore,
$\Delta K=K'-K$
$\Rightarrow \Delta K=m\dfrac{361{{v}^{2}}}{800}-\dfrac{1}{2}m{{v}^{2}}=-0.04875m{{v}^{2}}$
Substitute the values of W and $\Delta K$ in equation (i).
 $\Rightarrow -Fd=-0.04875m{{v}^{2}}$
$\Rightarrow Fd=0.04875m{{v}^{2}}$ …. (ii).
Now, suppose we place a number of planks together such that it acts as a single plank with thickness nd. When the bullet with the same velocity v, penetrates this combination its velocity at the end is zero.
Therefore, now the work done on the bullet is W’=-F(nd) and the change in kinetic energy is $\Delta K'=\dfrac{1}{2}m{{v}^{2}}-0=-\dfrac{1}{2}m{{v}^{2}}$.
And $W'=\Delta K'$.
$\Rightarrow -F(nd)=-\dfrac{1}{2}m{{v}^{2}}$
$\Rightarrow nFd=\dfrac{1}{2}m{{v}^{2}}$ …. (iii).
Divide equation (iii) by equation (ii).
$\Rightarrow \dfrac{nFd}{Fd}=\dfrac{\dfrac{1}{2}m{{v}^{2}}}{0.04875m{{v}^{2}}}$
$\Rightarrow n=\dfrac{1}{2\times 0.04875}=\dfrac{1}{0.0975}=10.26$
Therefore, the least value of n is 10.26. However, the number of planks has to be a whole number. Therefore, $n=11$.
Hence, the least number of planks required to stop the bullet is 11.

Note:
We can find the value of n by using the kinematic equations.
We assumed that the plank applies a constant force. Hence it will produce a constant negative acceleration of the bullet.
Let the acceleration of the bullet be -a.
Let's use the kinematic equation ${{v}^{2}}-{{u}^{2}}=2as$ for the case when there is only one plank.
In this case, u=v, $v=\dfrac{19v}{20}$, a=-a and s=d.
Therefore,
$\Rightarrow {{\left( \dfrac{19v}{20} \right)}^{2}}-{{v}^{2}}=2(-a)d$
$\Rightarrow \dfrac{361{{v}^{2}}}{400}-{{v}^{2}}=-2ad$
$\Rightarrow 0.0975{{v}^{2}}=2ad$ …. (1)
For the case when there are n planks use the same kinematic equation.
In this case, u=v, v=0, a=-a and s=nd.
Therefore,
$\Rightarrow 0-{{v}^{2}}=2(-a)nd$
$\Rightarrow {{v}^{2}}=2and$ …. (2).
Divide (1) and (2).
$\Rightarrow \dfrac{0.0975{{v}^{2}}}{{{v}^{2}}}=\dfrac{2ad}{2and}$
$\Rightarrow 0.0975=\dfrac{1}{n}$
$\Rightarrow n=\dfrac{1}{0.0975}=10.26$