Question

A bullet of mass 20g is fired from the rifle with a velocity of $800\,m{s^{ - 1}}$. After passing a mud wall $100\,cm$ thick, velocity drops to $100\,m{s^{ - 1}}.$ What is the average resistance of the wall $?$$\left( {{\text{Neglect friction due to air and work of gravity}}} \right)$.

Answer 1

Hint In the question, mass, initial velocity and final velocity is given. By substituting the values in the equation of work-energy we get the value of the average resistance of the wall.
Formula used
The expression for finding the average resistance of the wall is
Force $F = \dfrac{1}{2}m{v^2}$
Where,
$m$ be the mass and $v$ be the final velocity.

Complete step by step solution
Given that $m = 20\,g,\,$ Initial velocity $u = 800\,m{s^{ - 1}}$ and final velocity $v = 100\,m{s^{ - 1}}.$
Convert the value of mass in terms of $kg.$
Mass $m = 0.02\,kg.$
Now, we have to find the loss of kinetic energy we get
Kinetic energy in initial velocity – Kinetic energy in Final velocity
So,
$\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{v^2}$
Simplify the above equation we get
$\dfrac{1}{2}m\left( {{u^2} - {v^2}} \right)$.
Substitute all the known values in the above equation, we get
$\dfrac{1}{2} \times 0.02 \times \left( {{{800}^2} - {{100}^2}} \right)$
Performing the algebraic operations in the above equation, we get
$\dfrac{1}{2} \times 0.02 \times \left( {800 + 100} \right)\left( {800 - 100} \right)$
Performing the arithmetic operations in the above equation, we get
$\dfrac{1}{2} \times 0.02 \times 900 \times 700$
$W = 6300\,J$
Now, we have to using the work energy formula, we get
$W = F \times S$
$S = 100\,cm\,or\,S = 1\,m.$
Convert the above equation, in terms of force, we get
\[Force\,F = \dfrac{W}{S}\]
Substituting the known values in the work energy equation, we get
\[Force\,F = \dfrac{{6300}}{1}\]
$Force\,F = 6300\,N.$

Therefore, the average resistance of the wall is $6300\,N.$

Note: In the question, we neglect the friction due to the air and gravity. By using the algebraic operations in the equation of kinetic energy we get the value of work, then substitute the values in the equation of work-energy we get the value of the resistance force of the wall.