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29 January, 05:31

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?

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  1. 29 January, 07:19
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    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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