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.