Question

A bullet when fired at a target has its velocity decreased to 50 % after penetrating 30 cm into it.Then the additional thickness it will penetrate in cm before coming to rest

a) 10 cm b) 30 cm c) 40 cm d) 60 cm

Answer 1

40 cm ( use equation 2as = v^2 - u^2 )

Answer 2

$s=10\ cm$ is the distance penetrated by the bullet before stopping.

Explanation:

Let initial velocity of the bullet be, $2v\ cm.s^{-1}$

∴ Final velocity of the bullet after penetration of the given distance, $v\ cm.s^{-1}$

distance travelled by the bullet initially, $s=30\ cm$

Now, use the equation of motion:

$v^2=(2v)^2+2.a.s$

$-3v^2 =2\times a\times 30$

$a=-\frac{v^2}{20}\ cm.s^{-2}$

Using another eq. of motion for the bullet the time taken to come to rest:

$v=u+a.t$

$0=v-\frac{v^2}{20} \times t$

$t=\frac{20}{v} \ s$

Now distance further penetrated:

$s=u.t+\frac{1}{2} a.t^2$

$s=v\times (\frac{20}{v} )-\frac{1}{2}\times \frac{v^2}{20}\times (\frac{20}{v} )^2$

$s=10\ cm$

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TOPIC: eq. of motion

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