While perched on an elevated site, a peregrine falcon spots a flying pigeon. The falcon dives, reaching a speed of 90 m/s (200 mi/h). The falcon hits its prey with its feet, stunning or killing it, then swoops back around to catch it in mid-air. Assume that the falcon has a mass of 0.60 kg and hits a 0.20-kg pigeon almost head-on. The falcon’s speed after the collision is 60 m/s in the same direction.
(a) Determine the final speed of the pigeon immediately after the hit.
(b) Determine the internal energy produced by the collision.
(c) Why does the falcon strike its prey with its feet and not head-on?

A) 90 m/s

B) 540 J

C) This is because the falcons claws or talons is capable of exerting a greater Impulse on the pigeon

Explanation:

A) At the instant before collision i.e. The initial state, the falcon is travelling with a velocity of;

V_fiy = 90 m/s

and the pigeon with no component of velocity;

V_piy = 0 m/s

If we apply the generalized Impulse momentum equation to this question, we'll get;

p_fi + p_pi + J_onf + J_onp = p_ff + p_pf

Where;

p_fi = initial momentum by falcon

p_pi = initial momentum by pigeon

J_onf = Impulse on falcon

J_onp = Impulse on pigeon

p_ff = final momentum by pigeon

p_pf = final momentum by pigeon

Resolving, we have;

(m_f•v_fiy) + (m_p•v_piy) + Jy = (m_f•v_ffy) + (m_p•v_pfy)

Expanding, we have;

relevant values are;

m_f = 0.6 kg

m_p =0.2 kg

v_fiy = 90 m/s

v_piy = 0 m/s

Jy = 0

v_ffy = 60 m/s

Plugging in these values;

(0.6 x 90) + (0.2 x 0) + 0 = (0.6 x 60) + (0.2v_pfy)

54 = 36 + (0.2v_pfy)

54 - 36 = (0.2v_pfy)

(0.2v_pfy) = 18

(v_pfy) = 18/0.2 = 90m/s

B) If we apply the generalized worl energy principle to this question, we will arrive at;

U_i + W = U_f

Thus,

K_i + U_i,int + W = K_f + U_f,int

Where ;

K_i = initial kinetic energy

U_i,int = initial internal energy

W = Work

K_f = final kinetic energy

U_f,int = Final internal energy

Now, at initial state, internal energy is zero and also W is zero.

Thus, we'll now have;

K_i + 0 + 0 = K_f + U_f,int

K_i = K_f + U_f,int

Expanding, we get;

(1/2)[(m_f•(v_fiy)²) + (m_p•(v_piy)²)] = (1/2)[(m_f•(v_ffy)²) + (m_p•(v_pfy)²)] + U_f,int

Multiply through by 2;

[(m_f•(v_fiy)²) + (m_p•(v_piy)²)] =[(m_f•(v_ffy)²) + (m_p•(v_pfy)²)] + 2U_f,int

Plugging in the relevant values to obtain ;

(0.6 x 90²) + 0 = (0.6 x 60²) + (0.2 x 90²) + 2U_f,int

4860 = 2160 + 1620 + 2U_f,int

4860 = 3780 + 2U_f,int

2U_f,int = 4860 - 3780

2U_f,int = 1080

U_f,int = 1080/2 = 540J

C) This is because the falcons claws or talons is capable of exerting a greater Impulse on the pigeon

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