** 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 ½

*mv*

^{2}≈ 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 cm

^{2}= 10

^{−2}m

^{2}, so the pressure would have been, under these assumptions, equivalent to 25 × 10

^{2}= 2500 N/m

^{2}, or 2500 Pa. This is like a 250-kg person (500+ lb) standing on the glass if it was oriented horizontally.

**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**

*(iii)**x*= 18 in. and

*R*is the radius of the semicircle (i.e., the length of the wing). Hence, the wingspan $\u223c2R=x2\u224825.5in.(\u223c64cm)$, which is about right for an adult pigeon!