Answer
The speed of the ball just after the collision is $~~2.47~m/s$
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 (9-67) to find $v_{1f}$, the velocity of the ball just after the collision:
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$v_{1f} = \left(\frac{0.500~kg-2.50~kg}{0.500~kg+2.50~kg}\right)~(3.70~m/s)$
$v_{1f} = (-\frac{2}{3})~(3.70~m/s)$
$v_{1f} = -2.47~m/s$
The speed of the ball just after the collision is $~~2.47~m/s$
