Answer
The required mass of the smaller ball is $0.21~kg$
Work Step by Step
We can use conservation of energy to find the speed after falling from a height of $1.8~m$:
$\frac{1}{2}m_1v^2 = m_1~gh$
$v^2 = 2gh$
$v = \sqrt{2gh}$
$v = \sqrt{(2)(9.8~m/s^2)(1.8~m)}$
$v = 5.94~m/s$
At the moment of the collision, we can let the velocity of the larger ball be $v_{1i} = 5.94~m/s$ and the velocity of the smaller ball be $v_{2i} = -5.94~m/s$
We can use Equation (9-75) to find the mass $m$ such that $v_{1f} = 0$:
$v_{1f} = \frac{M-m}{M+m}~v_{1i}+\frac{2m}{M+m}~v_{2i}$
$0 = \frac{M-m}{M+m}~(5.94~m/s)+\frac{2m}{M+m}~(-5.94~m/s)$
$\frac{2m}{M+m}~(5.94~m/s) = \frac{M-m}{M+m}~(5.94~m/s)$
$2m= M-m$
$3m = M$
$m = \frac{M}{3}$
$m = \frac{0.63~kg}{3}$
$m = 0.21~kg$
The required mass of the smaller ball is $0.21~kg$
