Question

The initial velocity of bullet in penetrating distance 'a' through a target is reduced by u/n
How far will the bullet proceed through the target before coming to rest ?​

Answer 1

Given:

The initial velocity of bullet in penetrating distance 'a' through a target is reduced by u/n.

To find:

Distance travelled before coming to rest.

Calculation:

Let acceleration be f ;

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

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

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

$= > 2fa = - {u}^{2} \bigg \{1 - \dfrac{1}{ {n}^{2} } \bigg \}$

$= > f = - \dfrac{{u}^{2}}{2a} \bigg \{1 - \dfrac{1}{ {n}^{2} } \bigg \}$

Now , final Velocity be zero and distance travelled be s ;

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

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

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

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

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

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

So, final answer is:

$\boxed{ \sf{s = \dfrac{a {n}^{2} }{ {n}^{2} - 1 } }}$