Question

a bullet enters a wooden block and loses half of the speed at 1 centimeter,and how much distance the bullet will take to stop?​

Answer 1

$s = \frac{ {v}^{2} - {u}^{2} }{2a}$

when it travels 1cm, velocity changes from v to v/2.

$1 = \frac{ { (\frac{v}{2} )}^{2} - {v}^{2} }{2a} = \frac{ \frac{ {v}^{2} }{4} - {v}^{2} }{2a} \\ \\ \Rightarrow 1 = \frac{ \frac{ - 3 {v}^{2} }{4} }{2a} = \frac{ - 3 {v}^{2} }{8a} \\ \\ \Rightarrow \frac{ - {v}^{2} }{a} = \frac{8}{3}$

Now, velocity is v/2 and when it stops, its velocity will be 0. distance travelled will be

$s = \frac{{0}^{2} - ({ \frac{v}{2}) }^{2} }{2a} = \frac{ \frac{ - {v}^{2} }{4} }{2a} = \frac{ - {v}^{2} }{8a} \\ \\ \Rightarrow s = \frac{1}{8} \times \frac{ - {v}^{2} }{a} = \frac{1}{8} \times \frac{8}{3} \\ \\ \Rightarrow s = \frac{1}{3} \: cm$

So total distance travelled before stoping = 1 cm + 1/3 cm = 4/3 cm

Answer 2

Answer:-

$\begin{lgathered}1 = \frac{ { (\frac{v}{2} )}^{2} - {v}^{2} }{2a} = \frac{ \frac{ {v}^{2} }{4} - {v}^{2} }{2a} \\1 = \frac{ \frac{ - 3 {v}^{2} }{4} }{2a} = \frac{ - 3 {v}^{2} }{8a} \\ \frac{ - {v}^{2} }{a} = \frac{8}{3}\end{lgathered}$

The velocity is v/2

$\begin{lgathered}s = \frac{{0}^{2} - ({ \frac{v}{2}) }^{2} }{2a} = \frac{ \frac{ - {v}^{2} }{4} }{2a}= \frac{ - {v}^{2} }{8a} \\ s = \frac{1}{8} \times \frac{ - {v}^{2} }{a} = \frac{1}{8} \times \frac{8}{3} \\ s = \frac{1}{3} \: cm\end{lgathered}$

Therefore ,1 cm + 1/3 cm = 4/3 cm is the total distance travelled before stopping