A target in a shooting gallery consists of a vertical square wooden board, 0.230

Question

A target in a shooting gallery consists of a vertical square wooden board, 0.230 m on a side and with mass 0. 730 kg, that pivots on a horizontal axis along its top edge. The board is struck face-on al its center by a bullet with mass 1.70 g that is traveling at 305 m/s and that remains embedded in the board. What maximum height above the equilibrium posit ion does the center of the board reach before s tarting to swing down again? What minimum bullet speed would be required for the board to swing all the way over after impact?

Explanation / Answer

b) Because the bullet sticks in the target we know kinetic energy of the bullet is not conserved. Compute the initial (immediately after the bullet strikes) kinetic energy of the target / bullet system. That kinetic energy is converted to gravitational potential energy as the center of mass of the target rises above it intial postion. The equation is

(1/2) mv^2 = mgh, g =9.807 m/s^2
(1/2) (mb + mt) vt^2 = (mb + mt) g h, where h=height that ctr mass rises.

h = vt^2 / 2 g
vt = (mb vb) / (mb + mt) (from above)

h = [(mb vb) / (mb + mt)]^2 / 2 g
h = [(0.0017*305) / (0.73 + 0.0017)] ^2 / 2*9.807

h = 0.02559366 m

To make target swing over, h = 2R = 0.2 m
Rearrage eqn for h, to get eqn for velocity of bullet vb

h = [(mb vb) / (mb + mt)]^2 / 2 g
Sqrt(h) = mb vb / (mb+mt) Sqrt(2g)
vb = Sqrt(2gh) [(mb+mt)/mb]

vb = sqrt( 3.922) (430.4117647) = 1.98*430.4117647 = 852.2152941 m/s

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