a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west…

a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? (unit = m/s) remember: right is +, left is -

a 74.9 kg person sits at rest on an icy pond holding a 2.44 kg physics book. he throws the physics book west at 8.25 m/s. what is his recoil velocity? (unit = m/s) remember: right is +, left is -

Answer

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum (P_i = 0) (since the person - book system is at rest initially). According to the law of conservation of momentum (P_i=P_f), where (P_f = m_1v_1 + m_2v_2). Let (m_1 = 2.44\space kg), (v_1=- 8.25\space m/s) (west is negative), and (m_2 = 74.9\space kg), (v_2) is the recoil velocity of the person. So, (0=m_1v_1 + m_2v_2).

Step2: Solve for (v_2)

Rearrange the equation (0=m_1v_1 + m_2v_2) to get (v_2=-\frac{m_1v_1}{m_2}). Substitute (m_1 = 2.44\space kg), (v_1=-8.25\space m/s), and (m_2 = 74.9\space kg) into the formula: (v_2=-\frac{2.44\times(- 8.25)}{74.9}) First, calculate (2.44\times(-8.25)=-20.13). Then (v_2=\frac{20.13}{74.9}) (v_2\approx0.27\space m/s)

Answer:

(0.27\space m/s)