Question

answer is A, please explain how to get that answer

8. 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. If the momentum of inertia of the door around the hinge is I=1/12Ma’, where a is the width of the door, how much energy was lost during this collision? A) 30 J B) 16 J C) 15 J D) 13 J E) 4.8 J

Answer 1

here,

mass of door , m1 = 35 kg

width , a = 90 cm = 0.9 m

the mass of ball , m2 = 500 g = 0.5 kg

initial speed of ball , u = 20 m/s

final speed of ball , v = 16 m/s

r = 60 cm = 0.6 m

let the final angular speed of the soor be w

using conservation of angular momentum

Li = Lf

m2 * u * r = I * w + m2 * v * r

0.5 * 20 * 0.6 = (m1 * a^2 /12) * w + 0.5 * 16 * 0.6

0.5 * 0.6 * ( 20 - 16) = 35 * 0.9^2 * w /12

solving for w

w = 0.508 rad/s

the kinetic energy lost in the collison , KE = KEi - KEf

KE = 0.5 * m2 * u^2 - (0.5 * I * w^2 + 0.5 * m2 * v^2)

KE = 0.5 * (0.5 * 20^2 - (35 * 0.9^2 /12) * 0.508^2 - 0.5 * 16^2) J

KE = 30 J

the kinetic energy lost in the collison is A) 30 J