Answer
7.8 percent of the energy is lost.
Work Step by Step
Since the gold has an initial velocity of 0, the equation for the speed of the gold after the collision simplifies to:
$v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}$
$v_{2f}=\frac{2(4)}{4+197}v_{1i}$
$v_{2f}=.0398v_{1i}$
We square this value since it is squared in the kinetic energy equation: $.0398^2=.00158$
We know that the second particle is 49.25 times more massive, so it follows:
$=.00158\times49.25=.078$
7.8 percent of the energy is lost.
