The imprint was made on the window by a sizeable bird—I’m presuming a pigeon—but there was no trace of the bird when I went outside to check. Hopefully, it was just stunned and managed to fly away after a visit to pigeon urgent care.

Question 1: Estimate (i) the momentum and (ii) the kinetic energy of the bird prior to hitting the window.

Question 2: Estimate (i) the force with which it hit the window, (ii) the pressure on the window during impact, and (iii) the wingspan of the bird, given that the wingtip-to-wingtip horizontal distance on the window was about 18 in. (∼46 cm).

Solutions to Question 1: (i) I estimate the weight of a typical adult pigeon (assuming it was a pigeon) to be m ≈ ½ kg, and speed in flight v ≈ 40 km/h or ∼10 m/s. Hence, the momentum was approximately 5 kg · m/s.

(ii) The kinetic energy ½mv2 ≈ 25 N·m.

Solutions to Question 2: (i) I think that 1 s is too long, and 0.025 s is too short, so the geometric mean is 0.2 s. If so, then the rate of change of momentum is (5 – 0)/0.2 = 25 N.

Note that I’ve assumed the bird suffers an inelastic collision; i.e., it doesn’t bounce back when it hits the window. If that were not the case, the force on the window (and the bird) would be greater.

(ii) I estimate the impact area (ignoring the slender wing cross sections) to be approximately 100 cm2 = 10−2 m2, so the pressure would have been, under these assumptions, equivalent to 25 × 102 = 2500 N/m2, or 2500 Pa. This is like a 250-kg person (500+ lb) standing on the glass if it was oriented horizontally.

(iii) Under the assumption that the arc of the wingtips from vertical to horizontal is semicircular and noting that the angle between the wings is close to 90°, simple trigonometry shows that
where x = 18 in. and R is the radius of the semicircle (i.e., the length of the wing). Hence, the wingspan 2R=x225.5in.(64cm), which is about right for an adult pigeon!