Answer:
His recoil velocity is 0.269 m/s to the East
Explanation:
This question is easily solved by using the law of conservation of linear momentum.
The formula for the momentum is
[tex]Mo = mv[/tex]
, where m is the mass and v its speed.
The person + the book are at rest which means their momentum is
[tex]Mo_0=0[/tex]
After the book is released, they both start to move and their combined momentum is
[tex]Mo_f=m_pv_p+m_bv_b[/tex]
Where [tex]m_p, v_p[/tex] are the mass and speed of the person respectively and [tex]m_b, v_b[/tex] are the mass and speed of the book
Knowing that
[tex]m_p=74.9 Kg, m_b=2.44 Kg, v_b=-8.25m/s[/tex] (positive speed is assumed to the right or East), and the total momentum is zero:
[tex]m_pv_p+m_bv_b=0[/tex] =>
[tex]v_p=-\frac{m_bv_b}{m_p} =-\frac{2.44 (-8.25)}{74.9}[/tex]
[tex]v_p=0.269 m/s[/tex]
Since the sign of [tex]v_p[/tex] is positive, it's directed to the East
