Question

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further will it penetrate before coming to rest assuming that it faces constant resistance to its motion?​

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Answer 1

Answer:

It will penetrate more = 1 cm

Step by step explanations :

Given that,

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm.

let the initial velocity of the bullet be u

so,

it's velocity after penetrating 3 cm = u/2

let the acceleration produced by the blocks a

now we have,

initial velocity of the bullet = u

final velocity(v) = u/2

distance travelled(s) = 3 cm

acceleration = a

now,

by the equation of motion,

v² - u² = 2as

putting the values,

(u/2)² - u² = 2a(3)

6a = u²/4 - u²

a = -3u²/4 × 1/6

a = -u²/8

now,

after penetrating 3 cm ut will penetrate more,

and final velocity will zero

let the extra penetrating distance be s

here,

Initial velocity(u) = u/2

final velocity(v) = 0 cm/s

acceleration(a) = -u²/8

distance(s) = s

again,

v² - u² = 2as

0 - (u/2)² = 2(-u²/8)s

-u²/4s = -u²/4

1/s = 1

S = 1 cm

so,

It will penetrate more = 1 cm

Answer 2

Answer:

It will penetrate more = 1 cm

Step by step explanations :

let the initial velocity of the bullet be 2u

then, velocity after penetrating 3 cm = u

now,

initial velocity = u

final velocity(v) = u

penetrated distance(s) = 3 cm

acceleration = a

v² - u² = 2as

u² - (2u)² = 2a(3)

6a = u² - 4u²

a = -3u²/6

a = -u²/2

let after penetrating 3 cm it will penetrate s

and final velocity will zero

now,

Initial velocity(u) = u

final velocity(v) = 0 cm/s

acceleration(a) = -u²/8

distance(s) = s

v² - u² = 2as

0 - u² = 2(-u²/2)s

-u²s = -u²

S = 1 cm

so,

It will penetrate more = 1 cm