Question

A target in a shooting gallery consists of a vertical square wooden board, 0.230 m on a side and with mass 0.700 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 1.60 g that is traveling at 355 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?

Answer 1

a)
Moment of inertia of board, I_board = M*L^2/3

= 0.7*0.23^2/3

= 0.0123 kg.m^2

initial anular momentum of the bullet, I = m*v*R

= 1.6*10^-3*355*(0.23/2)

= 0.06532 kg.m^2/s

moment of inertia of bullet about pivot, I_bullet = m*R^2

= 1.6*10^-3*(0.23/2)^2

= 2.116*10^-5 kg.m^2

let w is the angular velocity of the baord.

Apply conservation of angular momentum

Initial angular momentum of the bullet = final angular momentum of the board.

0.06532 = (0.0123 + 2.116*10^-5)*w

==> w = 0.06532/(0.0123 + 2.116*10^-5)

= 5.3 rad/s <<<<<------------Answer

b) Apply conservation of energy

(M+m)*g*h = 0.5*I*w^2

h = 0.5*I*w^2/((M+m)*g)

= 0.5*(0.0123 + 2.116*10^-5)*5.3^2/((1.6*10^-3+0.7)*9.8)

= 0.025 m <<<<<------------Answer

c) h = L

(M+m)*g*L = 0.5*I*w^2

w = (M+m)*g*L/(0.5*I)

= sqrt( (1.6*10^-3+0.7)*9.8*0.23/(0.5*(0.0123 + 2.116*10^-5)) )

= 16.02 rad/s

now Apply, m*v*R = I*w

v = I*w/(m*R)

= (0.0123 + 2.116*10^-5)*16.02/(1.6*10^-3*(0.23/2))

= 1073 m/s <<<<<------------Answer