--- title: "Two balls impact e = 0.67; work-energy principle" subject: "Applied Mechanics (Dynamics)" chapter: 2 key: "semester_3/applied_mechanics_dynamics:two-balls-e067-work-energy" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-balls-e067-work-energy site: BCT Engineering --- # Two balls impact e = 0.67; work-energy principle Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods - Asked in: 2079 Bhadra, 6+2 marks - Video: [Impact: Coefficient of Restitution (learn to solve any problem)](https://www.youtube.com/watch?v=uzfeib3MkF8) — Question Solutions - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/two-balls-e067-work-energy - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2 ## Question Two frictionless balls ($m_A = 6$ kg, $m_B = 3$ kg) strike each other as shown in figure. The coefficient of restitution between the balls is $e = 0.67$. Find the velocities of A and B after the impact if initial velocity are $v_A = 3$ m/s and $v_B = 4.5$ m/s. Explain the principle of work and energy with governing equation. [Figure: ball A with $v_A$ at $40^\circ$ to the horizontal along the line of centres A to B; ball B with $v_B$ along the line of centres at $25^\circ$ to the horizontal.] ## Short answer Work-energy: work done equals change in kinetic energy (T1+U1→2=T2). For impact, tangential components unchanged; normal from momentum and e=0.67 yields v'_A≈2.30 m/s (57° above negative line), v'_B≈3.57 m/s (32° below positive line). ## Answer #### 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 $t$): | Ball | $v_n$ (m/s) | $v_t$ (m/s) | |---|---|---| | A | $3\cos40^\circ=2.298$ | $3\sin40^\circ=1.928$ | | B | $-4.5\cos25^\circ=-4.078$ | $-4.5\sin25^\circ=-1.902$ | **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).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What happens to velocity components perpendicular to the line of impact (smooth balls)? A. They are reversed B. They are unchanged C. They become zero D. They are multiplied by e Answer: B. No impulse acts tangentially on smooth balls. (Key point: Tangential components) 2. Which relation uses e along the line of impact? A. v'_Bn - v'_An = e (v_An - v_Bn) B. v'_Bn + v'_An = e (v_An + v_Bn) C. v'_Bn - v'_An = v_An - v_Bn D. v'_An = e v_An Answer: A. Relative speed of separation equals e times relative speed of approach. (Key point: Restitution equation) 3. Along with restitution, which equation gives the normal velocities? A. Conservation of momentum along the line of impact B. Conservation of kinetic energy C. Momentum along the tangent D. Work-energy Answer: A. Total momentum along the line of impact is conserved: 6(2.298) + 3(-4.078) = 6v'_An + 3v'_Bn. (Key point: Second equation) 4. What does the principle of work and energy state? A. T1 + U(1 to 2) = T2, work done equals change in kinetic energy B. T1 = T2 always C. U = m v D. Impulse equals change in kinetic energy Answer: A. It is a scalar relation linking speed and displacement without time. 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