Question

A peregrine falcon sees a pigeon flying below it and dives to catch it. Suppose that the pigeon is flying horizontally at 10 m/s and has a mass of 200 grams; the falcon, diving straight downward, has a mass of 500 grams.

Immediately after the falcon catches the pigeon, you observe that it is no longer moving straight downward; its path instead makes an angle of 2.3? with the vertical.

How fast was the falcon moving when it caught the pigeon

Answer 1

Using momentum conservation

In vertical direction

Pix = Pfx

m1*v1x + m2*v2x = M*Vx

m1 = mass of pigeon = 200 gm = 0.2 kg

m2 = mass of falcon = 500 gm = 0.5 kg

v1x = 10 m/sec

v2x = 0 m/sec (since initially straight downward)

Vx = V*sin 2.3 deg

M = 0.2 + 0.5 = 0.7 kg

So

0.2*10 + 0.5*0 = 0.7*V*sin 2.3 deg

V = 2/(0.7*sin 2.3 deg)

V = 71.19 m/sec = final speed of falcon + pigeon

Now momentum conservation in vertical direction

Piy = Pfy

m1*v1y + m2*v2y = M*V*cos 2.3 deg

v1y = 0 m/sec

v2y = initial speed of falcon = v2

So,

0.2*0 + 0.5*v2 = 0.7*71.19*cos 2.3 deg

v2 = (0.7*71.19*cos 2.3 deg)/0.5

v2 = 99.58 m/sec straight downward

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