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

Question

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Answer 1

Case $I$ :

$[\frac{u}{2}]^{2} - u^{2} = 2 . a . 3$
or $- \frac{3 u ^{2}}{4} = 2 . a . 3 \Rightarrow a = - \frac{u ^{2}}{8}$
Case $I I$ :
$0 - [\frac{u}{2}]^{2} = 2 . a . x$ or $- \frac{u ^{2}}{4} = 2 [- \frac{u ^{2}}{8}] \times x$
$\Rightarrow x = 1 c m$
Alternative method : Let K be the initial energy and F be the resistive force. Then according to work-energy theorem,
$W = Δ K$
i.e., $3 F = \frac{1}{2} m v^{2} - \frac{1}{2} m [\frac{v}{2}]^{2}$
      $3 F = \frac{1}{2} m v^{2} [1 - \frac{1}{4}]$
      $3 F = \frac{3}{4} [\frac{1}{2} m v^{2}]$ ............(1)
and $F x = \frac{1}{2} m [\frac{v}{2}]^{2} - \frac{1}{2} m (0)^{2}$
i.e., $\frac{1}{4} [\frac{1}{2} m v^{2}] = F x$ .........(2)
Comparing eqns. (1) and (2)
    $F = F x$
or $x = 1 c m$

Answer 2

Case

I

:

[\frac{u}{2}]^{2} - u^{2} = 2 . a . 3

or

- \frac{3 u ^{2}}{4} = 2 . a . 3 \Rightarrow a = - \frac{u ^{2}}{8}

Case

I I

:

0 - [\frac{u}{2}]^{2} = 2 . a . x

or

- \frac{u ^{2}}{4} = 2 [- \frac{u ^{2}}{8}] \times x

\Rightarrow x = 1 c m

Alternative method : Let K be the initial energy and F be the resistive force. Then according to work-energy theorem,

W = Δ K

i.e.,

3 F = \frac{1}{2} m v^{2} - \frac{1}{2} m [\frac{v}{2}]^{2}

3 F = \frac{1}{2} m v^{2} [1 - \frac{1}{4}]

3 F = \frac{3}{4} [\frac{1}{2} m v^{2}]

............(1)
and

F x = \frac{1}{2} m [\frac{v}{2}]^{2} - \frac{1}{2} m (0)^{2}

i.e.,

\frac{1}{4} [\frac{1}{2} m v^{2}] = F x

.........(2)
Comparing eqns. (1) and (2)

F = F x

or

x = 1 c m