Answer
The speed of the block just after the collision is $~~1.23~m/s$
Work Step by Step
The mass of the steel ball is $m_1 = 0.500~kg$
The mass of the steel block is $m_2 = 2.50~kg$
We can use conservation of energy to find $v_{1i}$, the velocity of the ball just before the collision:
$\frac{1}{2}m_1v_{1i}^2 = m_1~gh$
$v_{1i}^2 = 2gh$
$v_{1i} = \sqrt{2gh}$
$v_{1i} = \sqrt{(2)(9.8~m/s^2)(0.70~m)}$
$v_{1i} = 3.70~m/s$
We can use Equation (9-68) to find $v_{2f}$, the velocity of the block just after the collision:
$v_{2f} = \frac{2m_1}{m_1+m_2}~v_{1i}$
$v_{2f} = \left[\frac{(2)(0.500~kg)}{0.500~kg+2.50~kg}\right]~(3.70~m/s)$
$v_{2f} = (\frac{1}{3})~(3.70~m/s)$
$v_{2f} = 1.23~m/s$
The speed of the block just after the collision is $~~1.23~m/s$
