Question

A bullet is moving with velocity \[800m{{s}^{-1}}\] strikes two wooden plates of width $x_1$ and $x_2$ and in passing through each of them loses \[200m{{s}^{-1}}\] of its velocity. Assuming the resistance of the plated to be uniform, the ratio $x_1$/$x_2$ is:
A. 15/13
B. 9/7
C. 7/5
D. 5/3

Answer 1

Hint: When the bullet hits (or collides) the wooden plates, the velocity changes on interactions. We use the equation of motion to find which relation of distance covered and velocities are used. At the initial point its velocity is\[800m{{s}^{-1}}\]and it gets lesser at the interacting points.
Formula used:
We use the 3rd equation of motion:
${{u}^{2}}-{{v}^{2}}=2aS$

Complete answer:
When a bullet collides with wooden plates it undergoes inelastic collision along a straight line.
In the inelastic collisions, the velocities change and corresponding kinetic energy changes due to the effect of frictional forces.
The frictional forces come into play when the bullet strikes with wooden plates.
Also, the momentum in inelastic collisions is conserved.
Now, we use the 3rd equation of motion, in which the body starts from rest with initial velocity and when it stops, it is having final velocity.
In the question it is mentioned that, if the bullet strikes two plates then the final velocity changes correspondingly.
The 3rd equation of motion is given by:
${{u}^{2}}-{{v}^{2}}=2aS$
Given, u= \[800m{{s}^{-1}}\]and when it strike/collide with plate of width $x_1$ it loses\[200m{{s}^{-1}}\] velocity, i.e. \[{{v}_{1}}=\text{ }600m{{s}^{-1}}\]
Also, for width $x_2$ it again lost\[200m{{s}^{-1}}\], then
\[{{v}_{2}}=\text{ }400m{{s}^{-1}}\]
 (i) We will use the 3rd equation of motion:
$\begin{align}
& {{u}^{2}}-v_{1}^{2}=2a\times {{x}_{1}} \\
& {{\left( 800 \right)}^{2}}-\text{ }{{\left( 600 \right)}^{2}}=\text{ }2a\times{{x}_{1}} \\
& 640000-360000=2a\times{{x}_{1}} \\
& 280000=\text{ }2a\times \text{ }{{x}_{1}} \\
\end{align}$
(ii) Again using above equation of motion:
$\begin{align}
  & {{u}^{2}}-v_{2}^{2}=2a\times S \\
 & \begin{array}{{35}{l}}
   {{\left( 600 \right)}^{2}}-\text{ }{{\left( 400 \right)}^{2}}=\text{ }2a\times {{x}_{2}} \\
   360000-160000=\text{ }2a\times {{x}_{2}} \\
\end{array} \\
 & 200000=\text{ }2a\times {{x}_{2}} \\
\end{align}$
 Dividing equations obtained, i.e. (i) by (ii)
\[\begin{array}{{35}{l}}
   2a\times \text{ }{{x}_{1}}/\text{ }2a\times \text{ }{{x}_{2}}=\text{ }280000/200000. \\
   {{x}_{1}}/{{x}_{2}}=\text{ }28/20 \\
   {{x}_{1}}/{{x}_{2}}=\text{ }7/5. \\
\end{array}\]
So, the ratio of width of plates is given by 7:5

Hence, option C is the correct answer.

Note:
When a bullet strikes with the two wooden plates, then velocity at various points change. We use the equation of motions to find the ratio of the two plates, which have relation between the displacements and velocities.