Question

A bullet is passed through a wooden block. After a displacement of 2cm, it loses 1/3 rd of itsvelocity. The distance travelled by it in the block before coming to rest is​

Answer 1

I hope the above picture perfectly helps you mate

Answer 2

$after \: travelling \: 2cm \: it \: loses \: vel. = \frac{1}{3} \: of \: initial$

$now$

$velocity \: becomes \\ v = \: u - \frac{u}{3} = \frac{2u}{3}$

$u \: = initial \: velocity \:$

$v = final \: velocity$

$applying$

${v}^{2} - {u}^{2} = 2aS$

$( \frac{2u}{3} )^{2} - u^{2} = 2a \times 2$

$\frac{4 {u}^{2} }{9} - {u}^{2} = 4a$

$\boxed{a = \frac{ - 5 {u}^{2} }{36} }$

$again \: applying \:$

${v}^{2} - {u}^{2} = 2aS$

$this \: time \: v = 0 \\ \: because \: finally \: bullet \: comes \: to \: the \: rest$

$so$

$0 - {u}^{2} = 2( \frac{ - 5 {u}^{2} }{36} )S$

$S = \frac{36}{10} = 3.6 \: cm$