## Essential University Physics: Volume 1 (4th Edition)

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.