An open door of mass M is at rest when struck by a thrown ball of mass m (m<M) a

Question

An open door of mass M is at rest when struck by a thrown ball of mass m (m<M) at any point that is a distance D from an axis through the hinges. Just before the ball strikes the door, its path is perpendicular to the door face, and, since m<M, its path is still nearly perpendicular just after the collision. The door has a uniform density and width w. Let Vi and Vf represent the initial and final speeds of the ball. Neglect friction in the hinges during the time interval of the collision.

For part a) Taking the system to be the ball and the door, explain why the total linear momentum of the system is not conserved. B) Is the angular momentum of the system about any axis conserved? If so, identify the axis. c) Use the approximation discussed above and determine an expression for the resulting angular speed, w, of the door in terms of the quantities introduced. D) Evaluate w when m=1.1kg, M=35kg, w=73cm, D=62cm, Vi=27 m/s, and Vf, 16 m/s

Explanation / Answer

a) the total linear momentum of the ball and the door system will not be conserved because some of the balls momentum is transferred to the door in the form of angular momentum.

b) the angular momentum of the door will remain conserved along the axis connecting the hinges of the door.

c) momentum loss of the ball = m (Vf + Vi) {The direction of bothe velocities are opposite, so the positive sign}

angular momentum = m (Vf+ Vi)

M((omega)^2 *w = m (Vf +Vi)

(omega) = [m(Vf+Vi)/Mw] ^(0.5)

d)

(omega) = [1.1(27+16) / (35 * 0.73) ]^(0.5) = 1.36 rad/sec

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