Question

A bullet losses 20% of its speed while penetrating 3 cm into a log of wood . find how much further the bullet will penetrate ? Assume constant retardation.
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Answer 1

Answer:

i tried if i am wrong u can delete this answer

let the velocity of the bullet be x

the speed lost=0.2x

the velocity=x(1-0.2)

u=x(0.8)

since the bullet is stuck into the log the final velocity is 0

the acceleration is constant so we can apply the formula of

1st case

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

0-(0.8x)²=2a(3)

-0.64x²=6a

-0.64/6=-0.1xm/s²

since the a is constant

2nd case

0-(0.8)²=2(-0.1)s

0.64/0.2x=s

s=3.2x

Answer 2

Given:

Loss in speed by 20%

Penetraion=3cm

To find:

distance covered by the bullet further

Solution:

The initial speed while penetrating the log of wood is reduced by 20%.

We will use the formula,

$v^2=u^2-2as$

$(u-\frac{20}{100}u)^2=u^2-2as$

⇒$(0.8u)^2=u^2-2as$

⇒$2as=u^2-0.64u^2$

⇒$a=\frac{(1-0.64)u^2}{6}$

⇒$a=0.06u^2$

When the bullet comes to rest, the final velocity will be zero, therefore,

$0=u^2-2as$

⇒$u^2=2$×$0.06u^2$×$s$

⇒$s=\frac{1}{0.12}$

⇒$s=8.33$

The distance covered by the bullet further=(8.33-3)cm=5.33 cm

Hence, the bullet will cover a distance of 5.33cm further into the log of wood.