Page:Elementary Text-book of Physics (Anthony, 1897).djvu/53

§ 35] $$v_{2}.$$ We then nave $$m_{1}(u_{1} - V) = m_{2}(V - u_{2})$$ and $$m_{1}(V - v_{1}) = m_{2}(v_{2} - V)$$. That these equations may both be true we must have $$\frac{V - v_{1}}{u_{1} - V} = \frac{v_{2} - V}{V - u_{2}} = e$$, an experimental constant, called the coefficient of restitution. The coeificient $$e$$ depends upon the elasticity of the bodies and their mode of impact. It has been shown by experiment to be always less than unity. From these equations we deduce Combining this equation with the equation for the velocity of the centre of mass, we obtain for the velocities $$v_{1}$$ and $$v_{2}$$ after impact the equations

The kinetic energy before impact equals $$\tfrac{1}{2}m_{1}u^2_{1} + \tfrac{1}{2}m_{2}u^2_{2}$$. The kinetic energy after impact equals $$\tfrac{1}{2}m_{1}v^2_{1} + \tfrac{1}{2}m_{2}v^2_{2}$$. Substituting in this last expression the values just obtained for $$v_{1}$$ and $$v_{2}$$ and reducing, we obtain for the kinetic energy after impact By subtracting this from the kinetic energy before impact we find that the loss of kinetic energy by impact is  If the bodies are such that $$e = 0$$, or such that the velocities after impact are both equal to the velocity of the centre of mass, they are called inelastic bodies; the kinetic energy lost by their collision is $$\frac{m_{1}m_{2}}{m_{1} + m_{2}} \cdot \frac{(u_{1} - u_{2})^2}{2} \cdot$$. If, on the other hand, $$e = 0$$, so that the velocities after impact relative to the centre of mass are equal to those before impact but of opposite sign, the bodies are called perfectly elastic bodies. In this case no kinetic energy is lost by