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, 0.230m on a side and with mass 0.720kg , that pivots on a horizontal axis along its top edge. The board is struck face-on at its center by a bullet with mass 1.80g that is traveling at 370m/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

apply the law of conserevation of momentum as m1u1 + m2u2 = MV

V = m2v2/( m1 + m2) R

V =   (0.0018 * 370) /(0.0018 + 0.72 ) /(0.23/2)

V = 8.023 m/s

ang speed W = 8.023 /2pi = 1.277 re/sec

-------------------------

apply the law of conservation of energy as

(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 = [(0.0018*370) / (0.72 + 0.0018)}^2 / (2*9.8)

h = 0.0434 m

to find H

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

Sqrt(h) = mb vb / (mb+mt) Sqrt

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

vb = 715 m/s

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