Q.Two identical balls, each of mass $m$, undergo a perfectly inelastic, head-on collision, one moving with speed $v$ and the other initially at rest. Show that exactly half of the initial kinetic energy is lost in the collision, and state what becomes of the lost energy.

Given: two identical balls, each of mass $m$; ball $1$ moves at speed $v$, ball $2$ is initially at rest; the collision is perfectly inelastic, so the two balls move off together afterward at one common speed $v'$.

Common final velocity. By momentum conservation: $mv + m(0) = (m+m)v' \quad\Longrightarrow\quad v' = \frac{v}{2}$

Kinetic energy before: $K_i = \frac{1}{2}mv^2 + 0 = \frac{1}{2}mv^2$

Kinetic energy after: $K_f = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 = \frac{1}{2}(2m)\frac{v^2}{4} = \frac{mv^2}{4}$

Fraction lost. The kinetic energy lost is $K_i - K_f = \frac{1}{2}mv^2 - \frac{1}{4}mv^2 = \frac{1}{4}mv^2$ and comparing this to the initial kinetic energy, $\frac{K_i-K_f}{K_i} = \frac{\tfrac14mv^2}{\tfrac12mv^2} = \frac{1}{2}$ confirming that exactly half of the initial kinetic energy is lost in this particular (equal-mass, one-at-rest) perfectly inelastic collision. …

Find the common final velocity from momentum conservation ($v'=v/2$ for equal masses, one initially at rest), compute $K_i$ and …

• Assuming momentum conservation alone implies kinetic energy is also conserved, and concluding (wrongly) that no energy is lost at all. …