Answer
$733.36m/s$
Work Step by Step
We apply the law of conservation of energy as follows:
$T_1+V_1=T_2+V_2$eq(1)
$T_1=\frac{1}{2}(4+0.002)v^2=2.001v^2$
$V_1=(4+0.002)\times 9.81\times 0=0$
$T_2=\frac{1}{2}(4+0.002)(0)^2=0$
and $V_2=(4+0.002)\times 9.81\times (1.25-1.25cos 60^{\circ})=0.2688J$
This simplifies to:
$v=0.3665m/s$
Now we apply the law of conservation of momentum
$m_bv_b+m_w v_w=(m_b+m_w)v$
We plug in the known values to obtain:
$0.002v_b+4(0)=(4+0.002)(0.3665)$
This simplifies to:
$v_b=733.36m/s$
