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# A $60\,g$ bullet is fired through a stack of fiberboard sheets of $200\,mm$ thickness. If the bullet approaches the stack with a velocity of $600\,m/s$ ,the average resistance offered to the bullet is:A) $54\,kN$ B) $2\,kN$ C) $20.25\,kN$ D) $10\,kN$

Last updated date: 20th Jun 2024
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Hint:In order to solve this problem, we can use the third equation of motion to find the retardation. We know that according to the third equation of motion.
${v^2} = {u^2} + 2as$
Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.
Since force is the product of mass and acceleration we can get the average force of resistance by multiplying this retardation with the mass of the bullet.

Complete step by step solution:
It is given that the mass of the bullet is $60\,g$ . Let it be denoted as ${m_b}$ . Thus we can write
$\therefore {m_b} = 60g = 60 \times {10^{ - 3}}kg$
Let initial velocity of the bullet be denoted as u. It is given as $600m/s$ .
$u = 600\,m/s$
The thickness of the fibre, d of the sheet is 200mm
$d = 200mm = 200 \times {10^{ - 3}}m$
We need to find the average resistance offered by the sheet.
When the bullet strikes the fibre sheet it penetrates into the fibre sheet and finally comes to rest.
Thus we can take the final velocity as Zero.
$\therefore v = 0$
Let the retardation that is offered by the sheet be represented as $a$ .
We know that according to the third equation of motion.
${v^2} = {u^2} + 2as$ (1)
Where v is the final velocity, u is the initial velocity, a is the acceleration and s is the distance.
We have a sheet of 200mm thickness. This can be taken as the distance travelled.
Let us substitute all the values in equation 1
Then, we get
$0 = {\left( {600} \right)^2} + 2 \times a \times \left( {200 \times {{10}^{ - 3}}} \right)$ $\Rightarrow a = - 9 \times {10^5}m/{s^2}$
From Newton's second law we have
$F = ma$
Mass here is the mass of bullet ${m_b}$
On substituting the values we get
$F = - 60 \times {10^{ - 3}} \times 9 \times {10^5}$
$\Rightarrow F = - 54 \times {10^3}N$
$\therefore F = - 54\,kN$
This is the average force of resistance offered to the bullet.

So the correct answer is option A.

Note: If the time taken to penetrate the sheet before it comes to rest is given, then we can find the average resistive force as the ratio of change in momentum of the bullet to the time taken. Since distance travelled is given, we took the equation of motion involving distance to calculate the answer.