Answer
(a) The magnitude of the recoil velocity is 5.59 cm/s.
(b) The magnitude of the recoil velocity is 3.13 cm/s.
Work Step by Step
(a) Let $m_A$ be the mass of the hunter. Let $m_B$ be the mass of the bullet. Note that the initial momentum of the system is zero.
$p_0 = 0$
$m_Av_A + m_Bv_B = 0$
$m_Av_A = -m_Bv_B$
$v_A = \frac{-m_Bv_B}{m_A}$
$v_A = \frac{-(0.00420~kg)(965~m/s)}{72.5~kg}$
$v_A = -0.0559~m/s = -5.59~cm/s$
The magnitude of the recoil velocity is 5.59 cm/s.
(b) Let $m_A$ be the mass of the hunter. Let $m_B$ be the mass of the bullet. Note that the initial momentum of the system is zero.
$p_0 = 0$
$m_Av_A + m_Bv_B~cos(\theta) = 0$
$m_Av_A = -m_Bv_B~cos(\theta)$
$v_A = \frac{-m_Bv_B~cos(\theta)}{m_A}$
$v_A = \frac{-(0.00420~kg)(965~m/s)~cos(56.0^{\circ})}{72.5~kg}$
$v_A = -0.0313~m/s = -3.13~cm/s$
The magnitude of the recoil velocity is 3.13 cm/s.
