Question

A bullet fired into a fixed target loses half of its velocity after penetrating 3cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion ?

Answer 1

let assume that 3cm= 1 planck( a piece of wood having thickness of 3 cm).
the derivation of this question is very long so, the shortcut formula to solve any of this type of question is n^2/2n-1
where n is thickness of wooden piece so solution is
3×3/2×3-1=1.8

Answer 2

Explanation:

Given:-

Case :- 1

Distance travelled = 3cm.

Initial Velocity = u

Final Velocity = $\sf{\dfrac{u}{2}}$

Case :- 2

Distance travelled = x

Initial Velocity = $\sf{\dfrac{u}{2}}$

Final Velocity = 0m/s

To Find:-

The Distance bullet will penetrate before coming to rest.

Formulae used:-

v² - u² = 2as

Where,

v = Final Velocity

u = Initial Velocity

a = Accceleration

s = Distance

Now, Case 1

→ v² - u² = 2as

→ $\sf{ (\dfrac{u}{2})^2 - u^2 = 2 * a * 3}$

→ $\sf{ \dfrac{u^2}{4} - u^2 = 6a}$

→ $\sf{ \dfrac{-3u^2}{4} = 6a}$

→ $\sf{ a = \dfrac{-u^2}{8}}$

Now, In Case 2

→ v² - u² = 2as

→ $\sf{0 - (\dfrac{u}{2})^2 = 2 * \dfrac{-u^2}{8} * x}$

∗x ......using 1

→ $\sf{ \dfrac{u^2}{4} = \dfrac{-2u^2}{8} * x}$

→ $\sf{ -8u^2 = -8u^2 * x}$

→ $\sf{ \dfrac{-8u^2}{-8u^2} = x}$

→ $\sf{ x = 1cm}$

Hence, The bullet will penetrate 1cm more.