Question

A bullet of mass 25 g moving with a speed of 200 m/s is stored within 5 cm of the target what is the average resistance force offered by the target

Answer 1

hey mate!!

u=0
Mass =0.25kg
velocity=200m/s
displacement=0.5 m
force=mass×acceleration
now from using a=u^2-v^2/2s

a=4000m/s^2

now ,,,FORCE=Ma

0.25×40000=1 N

thankyou

Answer 2

Hey !

Given:

      Mass of bullet m = 25 g = 0.025 kg

Solution:

         Initial velocity of bullet,

                        u = 200 m/s

Final velocity, v = 0 and

      distance, s = 5 cm = 0.05 m

Relation for the acceleration

                       v² = u² - 2as

(or)                  0 = (200)² - 2a × 0.05

(or)                    a = (200)²/2 × 0.05

(or)                             = 400000 m/sec²

     ∴ Average resistance offered by the target,

                       F = m.a

                      = 0.025 × 400000

                      = 10000 N = 10 kN

GOOD LUCK !