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Question: When a bullet is fired at a target, its velocity gets decreased by half after penetrating \(30\,cm\)...

When a bullet is fired at a target, its velocity gets decreased by half after penetrating 30cm30\,cm into it. The additional thickness it will penetrate before coming to rest will be:
A. 3030
B. 1010
C. 4040
D. 5050

Explanation

Solution

When a bullet is fired and it starts to penetrate in a target, then it will cover some distance with negative acceleration and its final velocity will be zero and we will determine the additional thickness bullet will penetrate by using newton’s equation of motion v2u2=2aS{v^2} - {u^2} = 2aS.

Complete step by step answer:
Let us assume in first 30cm30\,cm penetration the initial velocity of bullet was uu and final velocity behalf of initial velocity as it’s given so we can write as v=u2v = \dfrac{u}{2}
S=0.3mS = 0.3\,m
Now, putting these values in equation v2u2=2aS{v^2} - {u^2} = 2aS we get,
u24=u2+2a(0.3)\dfrac{{{u^2}}}{4} = {u^2} + 2a(0.3)
3u24=0.6a(i)\Rightarrow \dfrac{{ - 3{u^2}}}{4} = 0.6a \to (i)

Now, let us assume the bullet further penetrate through distance say SS' and final velocity in this case will be zero as bullet will came to rest and initial velocity will be u2\dfrac{u}{2} , putting these values in equation v2u2=2aS{v^2} - {u^2} = 2aS we get,
u24=2aS(ii)\dfrac{{ - {u^2}}}{4} = 2aS' \to (ii)
From equations (i)and(ii)(i)and(ii) we can write as:
S=0.1mS' = 0.1\,m
S=10cm\therefore S' = 10\,cm
So, the thickness to which bullet will further penetrate the target is S=10cmS' = 10cm

Hence, the correct option is B.

Note: It’s important to remember that, the final velocity with which first 30cm30cm penetration happen will became the initial velocity of final penetration of 10cm10cm and other two equations of motion as v=u+atv = u + at , S=ut+12at2S = ut + \dfrac{1}{2}a{t^2} . These three equations are called newton’s equation of motion and whole classical mechanics of motion can be understand through these equations.>