A target in a shooting gallery consists of a vertical square wooden board, 0.250

Question

A target in a shooting gallery consists of a vertical square wooden board, 0.250 m on a side and with mass 0.750 kg, that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.90 g that is traveling at 360 m/s and that remains embedded in the board. (a) What is the angular speed of the board just after the bullet’s impact? (b) What maximum height above the equilibrium position does the center of the board reach before starting to swing down again? (c) What minimum bul- let speed would be required for the board to swing all the way over after impact?

Explanation / Answer

angular momentum before collision = angular momentum after collision

mb*v*(L/2) = (mb*(L/2)^2 + (1/3)*M*L^2)*wf

1.9*10^-3*360*(0.25/2) = [(0.0019*(0.25/2)^2 + ((1/3)*0.75*0.25*0.25)]*wf = 0.01565 * wf

wf = 5.46 rad/s

B) Apply law of conservation of energy

h = 0.5*[(mb/4)+(M/3)]*L^2*wf^2/(Mb+M)*g = (0.5*((0.0019/4)+(0.75/3))*0.25*0.25*5.46*5.46)/(0.0019+0.75) = 0.3103 m

C) wf' = sqrt((0.0019+0.75)*9.81*0.25)/(0.5*((0.0019/4)+(0.75/4))*0.25*0.25) = 17.7 rad/s

Vb = ((2*(0.0019/4)+(0.75/3)*0.25)*17.7)/0.0019 = 591.08 m/s

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