Question

A bullet of mass 10 kg travelling with a velocity of 150ms-2 strikes a stationary wooden block and comes to rest
in 0.03 s. Calculate the distance
of penetration of the bullet
into the block. Also calculate
the magnitude of the force
exerted by wooden block on bullet.​

Answer 1

Answer -

Given -

• m = 10 kg
• u = 150 m/s
• v = 0 ( because it comes to rest )
• t = 0.03 s

To find -

Distance Travelled ( s )

Force exerted by block on bullet ( F )

Solution -

For calculating distance travelled, we first need to calculate acceleration

Calculating Acceleration -

• u = 150 m/s
• v = 0 m/s
• t = 0.03 s

Substituting the value in 1st equation of motion -

$\rm v = u + at$

$\rm 0 = 150 + 0.03a$

$\rm - 150 = 0.03a$

$\rm a = \frac{-150}{0.03}$

$\rm a = -5000 m/s^2$

$\underline{\rm\pink{Calculating\: Distance \:Travelled - }}$

• u = 150 m/s
• v = 0 m/s
• a = -5000 m/s^2

Substituting the value in 3rd equation of motion -

$\rm v^2 = u^2 + 2as$

$\rm 0 = (150)^2 + 2 \times (-5000) \times s$

$\rm 22500 = 10000s$

$\rm s = \frac{22500}{10000}$

$\rm s = 2.25 m$

$\boxed{\underline{\rm\red{Distance\: travelled = 2.25 m}}}$

$\underline{\rm\pink{Calculating \:Force \:exerted - }}$

• m = 10 kg
• a = -5000 kg

Substituting the value in formula -

$\rm F = ma$

$\rm F = 10 \times (-5000)$

$\rm F = -50000 N$

$\boxed{\underline{\rm\red{Retardating\: force = 5000 N}}}$

Answer 2

$\huge\underline\bold{AnSwEr,}$

Given -

m = 10 kg

u = 150 m/s

v = 0 ( because it comes to rest )

t = 0.03 s

To find -

Distance Travelled ( s )

Force exerted by block on bullet ( F )

Solution -

For calculating distance travelled, we first need to calculate acceleration

Calculating Acceleration -

u = 150 m/s

v = 0 m/s

t = 0.03 s

Substituting the value in 1st equation of motion -

$\rm v = u + at$

$\rm 0 = 150 + 0.03a$

$\rm - 150 = 0.03a$

$\rm a = \frac{-150}{0.03}$

$\rm a = -5000 m/s^2$

$\underline{\rm\pink{Calculating\: Distance \:Travelled - }}$

u = 150 m/s

v = 0 m/s

a = -5000 m/s^2

Substituting the value in 3rd equation of motion -

$\rm v^2 = u^2 + 2as$

$\rm 0 = (150)^2 + 2 \times (-5000) \times s$

$\rm 22500 = 10000s$

$\rm s = \frac{22500}{10000}$

$\rm s = 2.25 m$

$\boxed{\underline{\rm\red{Distance\: travelled = 2.25 m}}}$

$\underline{\rm\pink{Calculating \:Force \:exerted - }}$

m = 10 kg

a = -5000 kg

Substituting the value in formula -

$\rm F = ma$

$\rm F = 10 \times (-5000)$

$\rm F = -50000 N$

$\boxed{\underline{\rm\red{Retardating\: force = 5000 N}}}$