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Question: A bullet of mass 20g is fired from the rifle with a velocity of \(800\,m{s^{ - 1}}\). After passing ...

A bullet of mass 20g is fired from the rifle with a velocity of 800 ms−1800\,m{s^{ - 1}}. After passing a mud wall 100 cm100\,cm thick, velocity drops to 100 ms−1.100\,m{s^{ - 1}}. What is the average resistance of the wall ?? (Neglect friction due to air and work of gravity)\left( {{\text{Neglect friction due to air and work of gravity}}} \right).

Explanation

Solution

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=12mv2F = \dfrac{1}{2}m{v^2}
Where,
mm be the mass and vv be the final velocity.

Complete step by step solution
Given that m=20 g, m = 20\,g,\, Initial velocity u=800 ms−1u = 800\,m{s^{ - 1}} and final velocity v=100 ms−1.v = 100\,m{s^{ - 1}}.
Convert the value of mass in terms of kg.kg.
Mass m=0.02 kg.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,
12mu2−12mv2\dfrac{1}{2}m{u^2} - \dfrac{1}{2}m{v^2}
Simplify the above equation we get
12m(u2−v2)\dfrac{1}{2}m\left( {{u^2} - {v^2}} \right).
Substitute all the known values in the above equation, we get
12×0.02×(8002−1002)\dfrac{1}{2} \times 0.02 \times \left( {{{800}^2} - {{100}^2}} \right)
Performing the algebraic operations in the above equation, we get
12×0.02×(800+100)(800−100)\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
12×0.02×900×700\dfrac{1}{2} \times 0.02 \times 900 \times 700
W=6300 JW = 6300\,J
Now, we have to using the work energy formula, we get
W=F×SW = F \times S
S=100 cm or S=1 m.S = 100\,cm\,or\,S = 1\,m.
Convert the above equation, in terms of force, we get
Force F=WSForce\,F = \dfrac{W}{S}
Substituting the known values in the work energy equation, we get
Force F=63001Force\,F = \dfrac{{6300}}{1}
Force F=6300 N.Force\,F = 6300\,N.

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