A bullet of mass 20g moving with velocity of 200m/a strikes a target and comes to rest in 0.04 seconds Calculate
a) Force exerted by target on bullet
b) Decceleration of bullet in target
c) Distance covered by bullet in target

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Solution

$a) u = 200 m s^{- 1}, v = 0, t = 0.04 s e c o n d . e q n o f m o t i o n v = u + a t t h e r e f o r e 0 = 200 - a \times 0.04 o r a = \frac{200}{0.04} = 5000 m s^{- 2} . t h u s t h e f o r c e e x e r t e d o n t h e b u l l e t = \frac{20}{1000} \times 5000 = 100 N . b) d e a c c e l e r a t i o n = a = 5000 m s^{- 2} .$

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A bullet of mass 20g moving with velocity of 200m/a strikes a target and comes to rest in 0.04 seconds Calculate a) Force exerted by target on bullet b) Decceleration of bullet in target c) Distance covered by bullet in target

a) u = 200ms-1, v = 0, t = 0.04 second.eqn of motion v = u + attherefore 0 = 200 - a ×0.04 or a = 2000.04 = 5000 ms-2.thus the force exerted on the bullet = 201000×5000 = 100 N.b) deacceleration = a = 5000 ms-2.

A bullet of mass 20g moving with velocity of 200m/a strikes a target and comes to rest in 0.04 seconds Calculate
a) Force exerted by target on bullet
b) Decceleration of bullet in target
c) Distance covered by bullet in target