Question

A rifle bullet loses 1/20th of its velocity in passing through a plank. Assuming constant resistive force, the least number of such planks required just to stop the bullet is :

Answer 1

When a bullet is passed through the plank, the plank will provide some resistive force. Due to that resistive force the velocity of the bullet will decrease to some extent. Now if we place few other planks then kinetic energy keeps on reducing and finally the bullet comes to rest. We use the work energy theorem to solve this.
Formula used:
$W = F.x = \Delta KE$

Complete answer:
The units of kinetic energy are joules. The units of work done are also joules. The relation between kinetic energy and work done can be understood by the work-energy theorem.
Work energy theorem states that work done by all the forces combined will result in change in kinetic energy
$W = F.x = \Delta KE$
Where ‘W’ is work done by all the forces and ‘x’ is displacement and ‘F’ is the force and $\Delta KE$ is change in kinetic energy.
Here since we were not given the value of resistive force and we need the work done, we don’t substitute the work done by the plant with its force and displacement.
After passing through one plank the change in its kinetic energy will be
$\Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v_i}^2$
\eqalign{ & \Rightarrow {v_f} = \left( {{v_i} - \dfrac{{{v_i}}}{{20}}} \right) \cr & \Rightarrow {v_f} = \dfrac{{19{v_i}}}{{20}} \cr}
$\Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v_i}^2$
\eqalign{ & \Rightarrow \Delta KE = \dfrac{1}{2}m{\left( {\dfrac{{19{v_i}}}{{20}}} \right)^2} - \dfrac{1}{2}m{v_i}^2 \cr & \Rightarrow \Delta KE = \dfrac{1}{2}m{v_i}^2\left( {\dfrac{{ - 39}}{{400}}} \right) \cr}
Negative sign indicates that the energy is lost i.e plank did the negative work.
Let number of planks like such require be ‘n’. work done by one plank is $\dfrac{1}{2}m{v_i}^2\left( {\dfrac{{ - 39}}{{400}}} \right)$
Hence work done by the ‘n’ planks will be $\dfrac{1}{2}nm{v_i}^2\left( {\dfrac{{ - 39}}{{400}}} \right)$
Finally the bullet should come to rest. That means the final kinetic energy of the bullet must become zero. Hence apply work energy theorem here
$\Delta KE = \dfrac{1}{2}m{v_f}^2 - \dfrac{1}{2}m{v_i}^2$
\eqalign{ & \Rightarrow \Delta KE = 0 - \dfrac{1}{2}m{v_i}^2 \cr & \Rightarrow \Delta KE = - \dfrac{1}{2}m{v_i}^2 \cr & \Rightarrow \dfrac{1}{2}nm{v_i}^2\left( {\dfrac{{ - 39}}{{400}}} \right) = - \dfrac{1}{2}m{v_i}^2 \cr & \Rightarrow n = \dfrac{{400}}{{39}} \cr & \Rightarrow n = 10.25 \cr & \Rightarrow n \simeq 11 \cr}
Hence the least number of planks required will be 11.

Note:
The least number of planks required will be 11. Even though we got the fractional value 10.25, that means the planks needed must be more than 10. Hence the nearest greater value will be 11. If the thickness of the plank is given, then we can calculate the average resistive force produced by the plank on the bullet.