Answer
$v_A=1.58ft/s$, $v=0.904ft/s$
Work Step by Step
We can determine the required speed as follows:
First, we apply the conservation of momentum for car A and the boy
$m_A(v_A)_1+m_b(v_b )_1=m_A(v_1)2+m_b(v_b)2$
We plug in the known values to obtain:
$\frac{80}{32.2}(0)+\frac{60}{32.2}(0)=-\frac{80}{32.2}v_A+\frac{60}{32.2}v_b$
This simplifies to:
$v_A=0.75v_b$.eq(1)
We know that:
$4cos\theta=v_b+v_A$
$\implies 4(\frac{12}{13})=v_b+0.75v_b$
$\implies v_b=2.190ft/s$
From eq(1), we obtain:
$v_A=1.58ft/s$
Now we apply the conservation of momentum for car B and the boy
$m_B(v_B)_1+m_b(v_b)1=(m_B+m_b) v$
We plug in the known values to obtain:
$\frac{80}{32.2}(0)+\frac{60}{32.2}(2.109)=(\frac{80+60}{32.2})v$
This simplifies to:
$v=0.904ft/s$
