Question

22.
A bullet fired into a fixed target loses half of its velocity after
penetrating 3 cm. How much further it will penetrate before
coming to rest assuming that it faces constant resistance to
motion
[AIEEE 2005]
(a) 1.5 cm
(b) 1.0 cm
(c) 3.0 cm
(d) 2.0 cm​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Left\:penetration=1\:cm}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies First \: penetrating \: length(s_{1}) = 3 \: cm \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies left \: Penetration \: length \: before \: it \: comes \: to \: rest ( s_{2} =?$

• According to given question :

$\tt \circ \: Let \: Initial \: velocity = v\:m/s \\ \\ \tt \circ \:Left \: velocity \: after \: s_{1} \: penetration = \frac{v}{2} \:m/s \\ \\ \tt \circ \: s_{1} = \frac{3}{100} = 0.03 \: m \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies \bigg(\frac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1} \\ \\ \tt: \implies \frac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 \\ \\ \tt: \implies \frac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a \\ \\ \tt: \implies \frac{ - 3{v}^{2} }{4} = 0.06 \times a \\ \\ \tt: \implies a = \frac{ - 3 {v}^{2} }{4 \times 0.06} \\\\ \tt: \implies a = \frac{ - 25 {v}^{2} }{2}\:m/s^{2} - - - - - (1)$

$\tt\circ\:Initial\:velocity=v\:m/s \\\\\tt \circ \: Final \: velocity = 0 \: m/s\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {v}^{2} = {u}^{2} + 2as \\ \\ \tt: \implies {0}^{2} = {v}^{2} + 2 \times \frac{ - 25 {v}^{2} }{2} \times s \\ \\ \tt: \implies - {v}^{2} = - 25 {v}^{2} \times s \\ \\ \tt: \implies s = \frac{ - {v}^{2} }{ - 25 {v}^{2} } \\ \\ \tt: \implies s = \frac{1}{25} \\ \\ \green{\tt: \implies s = 0.04 \: m} \\ \\ \bold{For \: left \: penetration(s_{2})} \\ \tt: \implies s = s_{1} + s_{2} \\ \\ \tt: \implies 0.04 = 0.03 + s_{2} \\ \\ \tt: \implies s_{2} = 0.04 - 0.03 \\ \\ \green{\tt: \implies s_{2} = 0.01 \: m = 1 \: cm} \\ \\ \green{\tt \therefore Left \: penetration \: before \: it \: come \: to \: rest \: is \: 1 \: cm}$

Answer 2

$\orange{\bigstar}$ Answer $\green{\bigstar}$

For 1st 3 cm :

Initial velocity = u cm/s

Final velocity = $\rm \dfrac{u}{2}$ cm/s

Distance , s = 3 cm

Acceleration = a cm/s²

Apply 3rd equation of motion ,

$\to \rm \left(\dfrac{u}{2}\right)^2-u^2=2a(3)\\\\\to \rm \dfrac{u^2}{4}-u^2=6a\\\\\to \rm -\dfrac{3u^2}{4}=6a\\\\\to \rm a=-\dfrac{u^2}{8}$

For remaining distance :

Initial velocity = $\rm \dfrac{u}{2}$ cm/s

Final velocity = 0 m/s

Acceleration = $\rm -\dfrac{u^2}{8}\ cm/s^2$

Distance = x cm

Apply 3rd equation of motion ,

$\to \rm v^2-u^2=2as\\\\\to \rm (0)^2-\left(\dfrac{u}{2}\right)^2=2\left(-\dfrac{u^2}{8}\right)x\\\\\to \rm 0-\dfrac{u^2}{4}=-\dfrac{u^2}{4}x\\\\\to \rm x\ \times \cancel{\left(-\dfrac{u^2}{4}\right)}=\cancel{-\dfrac{u^2}{4}}\\\\\to \rm x=1\ cm\ \; \bigstar$

So , It will penetrate 1 cm before coming to rest

Option (b) is correct