Answer to Question #304986 in Mechanics | Relativity for Ogar Precious Owog

Question

Two objects of equal masses collide head-on while moving in the opposite direction and joined together. Show that the fractional loss of original kinetic energy of the objects is given by the expression; E=1/2+V1V2/V1²+V2²

Answer 1

Explanations & Calculations

• Final speed needs to be found using the conservation of linear momentum.

$\qquad\qquad \begin{aligned} \small \to\\ \small mv_1-mv_2&=\small 2mV\\ \small V&=\small \frac{v_1-v_2}{2} \end{aligned}$

• Then the energy loss is

$\qquad\qquad \begin{aligned} \small L&=\small K.E_{inital}\\ \small \Delta L&=\small K.E_{inital}-K.E_{final}\\ &=\small \frac{1}{2}mv_1^2+\frac{1}{2}mv_2-\frac{1}{2}.2m.V^2\\ &=\small \frac{m}{2}\Big[\frac{v_1^2}{2}+\frac{v_2^2}{2}+v_1v_2\Big]\\ \end{aligned}$

• Then the fractional loss is

$\qquad\qquad \begin{aligned} \small \frac{\Delta L}{L}&=\small \frac{\frac{v_1^2}{2}+\frac{v_2^2}{2}+v_1v_2}{v_1^2+v_2^2}\\ &=\small \frac{\frac{(v_1^2+v_2^2)}{2}+v_1v_2}{(v_1^2+v_2^2)}\\ &=\small \frac{1}{2}+\frac{v_1v_2}{(v_1^2+v_2^2)} \end{aligned}$