A mass M door is open when hit by a ball of mass m (m <M) at a point spaced a di

Question

A mass M door is open when hit by a ball of mass m (m <M) at a point spaced a distance D from the axis through the hinges, as shown in the figure below. The initial direction of the ball is horizontal and forms an angle ? with the normal to the face of the door, and after collision against the door remains. The modulus of the speed of the ball before the collision is vi and vf after the collision. The door has uniform density and its width is w. Neglect the effects of friction in the hinges during the collision.

a) Find the angular velocity of the door after the collision.

b) Find the change in kinetic energy of the system (more ball door).

Explanation / Answer

here,

mass of door is M and ball is m

a)

let the angular speed after the collison be w0

using conservation of angular momentum

m * v * D * cos(theta) = ( M * w^2/L + m * D^2) * w0

w0 = m * v * D * cos(theta)/(M * w^2/L + m * D^2)

b)

change in kinetic energy , KE = 0.5 * I * w0^2 - 0.5 * m * v^2

KE = 0.5 * ( (M * w^2/L + m * D^2)) * w0^2 - 0.5 * m * v^2

c)

w0 = m * v * D * cos(theta)/(M * w^2/3 + m * D^2)

w0 = 1.1 *27*0.62*cos(0.38) /( 35*0.73^2 /3 + 1.1 *0.62^2)

w0 = 2.58 rad/s

the angular speed is 2.58 rad/s

KE = 0.5 * ( (M * w^2/L + m * D^2)) * w0^2 - 0.5 * m * v^2

KE = 0.5 * ( 35*0.73^2 /3 + 1.1 *0.62^2) * 2.58^2 - 0.5 * 1.1 * 27^2

KE = - 378.85 J

the kinetic energy lost is 378.85 J

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