Question

A 20 gram bullet moving at 300 m/s stops after penetrating 2 cm of a bone. calculate the avg force exerted by the bullet.

Answer 1

$\large{\bf{\gray{\underline{\underline{\orange{Given:}}}}}}$

➝ Mass of bullet = 20g

➝ Initial velocity = 300m/s

➝ Distance covered = 2cm

➝ Final velocity = zero

$\large{\bf{\gray{\underline{\underline{\green{To\:Find:}}}}}}$

⟶ We have to find average force exerted by bone on the bullet.

$\large{\bf{\gray{\underline{\underline{\pink{Solution:}}}}}}$

➠ We know that, Force is defined as the product of mass and acceleration.

➠ Force is a polar vector quantity having both magnitude as well as direction.

➠ SI unit : N

• $\boxed{\bf{\red{F=ma}}}$

Conversion :

⟶ 20g = 20/1000 = 0.02kg

⟶ 2cm = 2/100 = 0.02m

፨ First we have to find acceleration of bullet. ☃

Let's apply third equation of kinematics. :D

$:\implies\tt\:v^2-u^2=2as$

$:\implies\tt\:0^2-300^2=2a(0.02)$

$:\implies\tt\:a=-\dfrac{90000}{0.04}$

$:\implies\boxed{\bf{\blue{a=-2250000\:ms^{-2}}}}$

[Note : Negative sign indicates retardation.]

$:\implies\tt\:F=ma$

$:\implies\tt\:F=0.02\times (-2250000)$

$:\implies\bf\:F=-45000N$

$:\implies\boxed{\bf{\gray{|F|=45\:kN}}}\:\bigstar$

Answer 2

Given :

• Mass of bullet, m = 20 g = (20/1000) kg = 0.02 kg
• initial velocity of bullet, u = 300 m/s
• final velocity of bullet, v = 0   [since it comes to rest]
• distance covered by bullet, s = 2 cm = (2/100) m = 0.02 m

To find :

• Average force exerted by the bullet, F = ?

Knowledge required :

• Third equation of motion

      2 a s = v² - u²

[ where a is acceleration, s is distance covered, v is final velocity, u is initial velocity of body ]

• Newton's second law of motion

   F = m a

[ where F is force exerted by body, a is acceleration of body and m is mass of body ]

Solution :

Calculating acceleration of bullet

Using third equation of motion

→ 2 a s = v² - u²

→ 2 a ( 0.02 ) = ( 0 )² - ( 300 )²

→ 0.04 a = -90000

→ a = -90000/0.04

→ a = -2250000

→ a = - 2.25 × 10⁶  m/s²

Calculating force exerted by the bullet

Using Newton's second Law of motion

→ F = m a

→ F = (0.02) × (-2.25 × 10⁶ )

→ F = -45000 N

therefore,

• magnitude of the force exerted by the bullet is equal to 45000 Newton or 45 kilo Newtons.