A target in a shooting gallery consists of a vertical square wooden board, 0.240

Question

A target in a shooting gallery consists of a vertical square wooden board, 0.240 m on a side and with mass 0.770 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 2.00 g that is traveling at 330 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 bullet speed would be required for the board to swing all the way over after impact?

Explanation / Answer

using law of conservation of momentumn

Li = Lf

m*v*r = (I_bullet +I_block)*w

0.002*330*(0.24/2) = [(0.002*0.12*0.12) + ((1/3)*0.77*(0.24*0.24))]*w

w = 5.34 rad/s

B) using conservation of energy

K1+U1 = K2+u2

K1 = U2

[0.5*0.002*(0.12*0.12)*5.34*5.34] + (0.5*(1/3)*0.77*0.24*0.24) = (0.002+0.77)*9.81*h

h = 0.00103 m = 1.03*10^-3 m

for h = l

wf' = sqrt[((mb+m)*g*L)/(0.5*((mb/4)+(m/3))*L^2

wf' = sqrt[((0.002+0.77)*9.81*0.24) / [(0.5*((0.002/4)+(0.77/3))*(0.24*0.24)]= 15.66 rad/s

minimum speed is v' = [2*(Mb/4)+(m/3)*L]*wf' / (mb) = [2*((0.002/4)+(0.77/3))*0.24]*14/(0.002) = 864 m/s

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