Two balls impact e = 0.67; work-energy principle

Impact of the two balls

Assumptions (figure garbled): the line of impact is the horizontal x-axis. Ball A moves to the right at $3$ m/s inclined $40^\circ$ above the line of impact; ball B moves to the left at $4.5$ m/s inclined $25^\circ$ below the line of impact. $m_A=6$ kg, $m_B=3$ kg, $e=0.67$.

Components (normal $n$ = along line of impact, tangential $t$):

Tangential components do not change (smooth balls): $v'_{At}=1.928$, $v'_{Bt}=-1.902$ m/s.

Normal direction: momentum and restitution.

$$6(2.298)+3(-4.078)=6v'_{An}+3v'_{Bn}\quad\Rightarrow\quad 2v'_{An}+v'_{Bn}=0.5185$$ $$v'_{Bn}-v'_{An}=e(v_{An}-v_{Bn})=0.67(2.298+4.078)=4.272$$

Solving: $v'_{An}=-1.251$ m/s and $v'_{Bn}=3.021$ m/s.

Final velocities:

$$v'_A=\sqrt{1.251^{2}+1.928^{2}}=2.30\ \text{m/s},\quad \text{at }57.0^\circ\text{ above the negative line of impact}$$ $$v'_B=\sqrt{3.021^{2}+1.902^{2}}=3.57\ \text{m/s},\quad \text{at }32.2^\circ\text{ below the positive line of impact}$$

Principle of work and energy

The work done by all forces acting on a particle as it moves from position 1 to 2 equals the change in its kinetic energy:

$$T_1+U_{1\to2}=T_2,\qquad T=\tfrac12mv^{2}$$

where $U_{1\to2}=\int\vec F\cdot d\vec r$. Work of the weight is $\pm mg\,\Delta h$, work of a spring is $\tfrac12k(x_1^2-x_2^2)$, friction does negative work. It is a scalar equation, needs no time, and connects speed with displacement. For conservative forces it becomes $T_1+V_1=T_2+V_2$.

Answer: $v'_A\approx2.30$ m/s, $v'_B\approx3.57$ m/s (for the assumed directions).