Question
A 25 g bullet with a muzzle velocity of 300 m/s is fired into a board 4.0 cm thick. For a soft board, the bullet goes through the board and emerges with a speed of 50 m/s. What was the average force exerted on the bullet by the board?
Answer= 27344 N
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Answer= 27344 N
Show Work
Answers
Henry
V^2 = Vo^2 + 2a*d,
a = (V^2-Vo^2)/2d = (50^2-300^2)/0.08 = -1,093,750 m/s^2.
F = M*a = 0.025 * (-1,093,750) =
a = (V^2-Vo^2)/2d = (50^2-300^2)/0.08 = -1,093,750 m/s^2.
F = M*a = 0.025 * (-1,093,750) =