#### 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.