Answer:
a
The speed of the quarterback backward is [tex]v_q = 0.08 \ m/s[/tex]
b
Known are
[tex]m_Q , m_B , (v_{Bx})_i (v_{Qx})_f, (v_{Bx})_f[/tex]
Unknown
[tex](v_{Qx})_f[/tex]
Explanation:
From the question we are told that
The mass of the quarterback is [tex]m_Q = 80 \ kg[/tex]
The mass of the ball is [tex]m_B = 0.43 \ kg[/tex]
The speed of the ball is [tex]v_{B x}= 15 \ m/s[/tex]
The law of momentum conservation can be mathematically represented as
[tex]m_Q u_{Qx} + m_Bu_{Bx} = - m_{Q} v_{Qx} + m_B v_{Bx}[/tex]
Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball
So
[tex]m_Q v_{Qx} = m_B v_ {Bx}[/tex]
=> [tex]v_{Qx} = \frac{m_Bv_{Bx}}{m_Q}[/tex]
substituting values
[tex]v_q = \frac{0.43 * 15}{80}[/tex]
[tex]v_q = 0.08 \ m/s[/tex]
