Physics, asked by jpjitendra5321, 1 year ago

A bullet of mass 25 g moving with a velocity of 200 m/s is stopped within 5 cm of the target. the average resistance offered by the target is

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Answered by tinu21
1
i hope its help you
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Answered by nalinsingh
5

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 !

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