Question

A bullet loses 1/n if its velocity while penetrating a distance x into the target. The further distance travelled before coming to rest?

Answer 1

Given:

A bullet loses 1/n if its velocity while penetrating a distance x into the target.

To find:

Distance travelled further before coming to rest.

Calculation:

Let acceleration be a ;

${v}^{2} = {u}^{2} + 2ax$

$= > { (\dfrac{u}{n}) }^{2} = {u}^{2} + 2ax$

$= > \dfrac{ {u}^{2} }{ {n}^{2} } = {u}^{2} + 2ax$

$= > \dfrac{ {u}^{2} }{ {n}^{2} } - {u}^{2} = 2ax$

$= > {u}^{2} \bigg( \dfrac{1}{ {n}^{2} } - 1 \bigg) = 2ax$

$= > - {u}^{2} \bigg( \dfrac{ {n}^{2} - 1}{ {n}^{2} } \bigg) = 2ax$

$= > a = - \dfrac{ {u}^{2} }{2x} \bigg( \dfrac{ {n}^{2} - 1}{ {n}^{2} } \bigg)$

Now , final Velocity be 0 , and further distance be s ;

${0}^{2} = {v}^{2} + 2as$

$= > 0 = \dfrac{ {u}^{2} }{ {n}^{2} } - \bigg \{2 \times \dfrac{ {u}^{2} }{2x} \bigg( \dfrac{ {n}^{2} - 1}{ {n}^{2} } \bigg) \times s \bigg \}$

$= > \dfrac{ {u}^{2} }{ {n}^{2} } = \bigg \{2 \times \dfrac{ {u}^{2} }{2x} \bigg( \dfrac{ {n}^{2} - 1}{ {n}^{2} } \bigg) \times s \bigg \}$

$= > \: s = \dfrac{x}{ {n}^{2} - 1}$

So, final answer is:

$\boxed{ \bold{ \red{ \huge{ \: s = \dfrac{x}{ {n}^{2} - 1} }}}}$