A board able to pivot along a horizontal axis along its top edge is shot with a

Question

A board able to pivot along a horizontal axis along its top edge is shot with a bullet. The board is square and is 0.25 m on a side, and has a mass of 0.750 kg. The board is struck at its center by the bullet at an angle of phi = 15 degree relative to the surface of the board as shown below. The bullet has mass 1.90 g and is traveling at a speed of 360 m/s before the collision. The bullet remains embedded in the board. (a) What is the angular speed of the board (with embedded bullet) just after the bullet's impact? (b) What maximum height above the equilibrium position does the center of the board, also the CM of the board-bullet object, reach before starting to swing back down? (c) What is the maximum height of the CM of the board-bullet object if the bullet instead strikes the board at l = 0.200 m. The board still has the same total mass and still strikes at an angle of phi = 15 degree relative to the surface of the board

Explanation / Answer

1. (a) applying angular momentum conservation,

m v l cos(phi) + 0 = I w + (m l^2)(w)

{ for board, I = M L^2 / 3}

putting in values,

(0.0019 x 360 x 0.125 x cos15) = (0.750 x 0.25^2 / 3)w + (0.0019 x 0.125^2) w

0.08259 = 0.015625 w + (2.969 x 10^-5) w

w = 5.28 rad/s .......Ans

(b) Now applying energy conservation,

( I w^2 / 2) + (m (wl)^2)/ 2 = (m + M) g h

(0.015625)(5.28^2) / 2 + (0.0019 x (5.28 x 0.125)^2) /2 = (0.750 + 0.0019)(9.8)(h)

0.2178 + 0.000414 = 7.36862 h

h = 0.030 m Or 3 cm ........Ans

(c) (0.0019 x 360 x 0.2 x cos15) = (0.750 x 0.25^2 / 3)w + (0.0019 x 0.2^2) w

0.132 = 0.015625 w + (7.6 x 10^-5) w

w = 8.41 rad/s

after that applying energy conservation,

(0.015625)(8.41^2) / 2 + (0.0019 x (8.41x 0.2)^2) /2 = (0.750 + 0.0019)(9.8)(h)

0.5526 + 0.002688 = 7.36862 h

h = 0.075 m Or 7.5 cm

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