Question

A bullet on entering a target with an initial velocity u, loses u/n of its velocity after penetrating a distance x into the target, how much further will it penetrate?
​

Answer 1

Given that,

Initial velocity = u

Final velocity $v=\dfrac{u}{n}$

Penetrate distance = x

We need to calculate the acceleration

Using third equation of motion

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

Put the value into the formula

$\dfrac{u^2}{n^2}-u^2=2a\times x$

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

We need to calculate the penetrate distance

Using third equation of motion

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

Put the value into the formula

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

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

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

Hence, The penetrate distance will be $\dfrac{n^2 x}{n^2-1}$

Answer 2

Initial velocity = u

Final velocity v=\dfrac{u}{n}v=

n

u

Penetrate distance = x

We need to calculate the acceleration

Using third equation of motion

v^2-u^2=2asv

2

−u

2

=2as

Put the value into the formula

\dfrac{u^2}{n^2}-u^2=2a\times x

n

2

u

2

−u

2

=2a×x

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

2x

u

2

(

n

2

1

−1)

We need to calculate the penetrate distance

Using third equation of motion

v^2-u^2=2asv

2

−u

2

=2as

Put the value into the formula

0-u^2=2\times\dfrac{u^2(\dfrac{1}{n^2}-1)}{2x}\times s0−u

2

=2×

2x

u

2

(

n

2

1

−1)

×s

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

1−

n

2

1

x

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

n

2

−1

n

2

x

Hence, The penetrate distance will be \dfrac{n^2 x}{n^2-1}

n

2

−1

n

2

x