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A 30-gram bullet is fired with a speed of 500 meters per second into a wall.

a. If the deceleration of the bullet is constant and it penetrates 12 centimeters into the wall, calculate the force on the bullet while it is stopping. 

b. If the deceleration of the bullet is constant and it penetrates 12 centimeters into the wall, how much time is required for the bullet to stop? 

1 Answer

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Best answer

(a) Apply work-energy theorem

Wnc = ∆ME

Wfk = ∆K (Kf – Ki) Kf = 0– fkd = – ½mvi2 – fk(0.12) = – ½ (0.030) (500)2 fk = 31250N

(b) Find find acceleration 

– fk = ma – (31250) = (0.03)a

a = – 1.04 x 106m/s2

Then use kinematics vf = vi + at

0 = 500 + (– 1.04 x 106)t

t = 4.8 x 10– 4sec

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