A solid wooden door,90 cm wide by 2.0 m tall,has a mass of 35 kg.It is open and at rest.A small 500-g ball is thrown perpendicular to the door with a speed of 20 m/s and hits the door 60 cm from the hinged side,causing it to begin turning.The ball rebounds along the same line with a speed of 16.0 m/s relative to the ground.How much energy is lost during this collision?
A)30 J
B)16 J
C)15 J
D)13 J
E)4.8 J

Question

Quizplus · Accepted Answer

The energy lost during the collision is calculated by finding the difference in kinetic energy before and after the collision. The initial kinetic energy of the ball is (1/2)mv^2 = (1/2)(0.5 kg)(20 m/s)^2 = 100 J. The final kinetic energy of the ball is (1/2)(0.5 kg)(16 m/s)^2 = 64 J. The energy lost is 100 J - 64 J = 36 J. However, the rotational kinetic energy of the door must also be considered. The moment of inertia of the door about the hinge is (1/3)ML^2 = (1/3)(35 kg)(0.9 m)^2 = 9.45 kg·m^2. The angular velocity ω can be found using conservation of angular momentum. The initial angular momentum is (0.5 kg)(20 m/s)(0.6 m) = 6 kg·m^2/s. The final angular momentum of the ball is (0.5 kg)(16 m/s)(0.6 m) = 4.8 kg·m^2/s. The change in angular momentum is 6 - 4.8 = 1.2 kg·m^2/s, which is equal to Iω. Solving for ω gives ω = 1.2 / 9.45 = 0.127 rad/s. The rotational kinetic energy of the door is (1/2)Iω^2 = (1/2)(9.45)(0.127)^2 = 0.076 J. The total energy after the collision is 64 J (ball) + 0.076 J (door) = 64.076 J. The energy lost is 100 J - 64.076 J = 35.924 J, which rounds to 36 J. However, considering the choices, the closest is 30 J.

A Solid Wooden Door,90 Cm Wide by 2