--- title: "Coefficient of restitution; car on track" subject: "Applied Mechanics (Dynamics)" chapter: 2 key: "semester_3/applied_mechanics_dynamics:direct-central-impact-restitution-car-track" source: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/direct-central-impact-restitution-car-track site: BCT Engineering --- # Coefficient of restitution; car on track Applied Mechanics (Dynamics) · Chapter 2: Kinetics of particles: Energy and Momentum Methods - Asked in: 2076 Chaitra, 4+4 marks - Video: [Coefficient of Restitution - Inelastic Collisions and Elastic Collisions - Physics](https://www.youtube.com/watch?v=DyfGzKOwtYU) — The Organic Chemistry Tutor - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/applied_mechanics_dynamics/direct-central-impact-restitution-car-track - Whole chapter: https://www.bctengineer.com/md/semester_3/applied_mechanics_dynamics/chapter-2 ## Question Derive the coefficient of restitution for two particles involved in direct central impact. A 30 kN car starts from rest at point 1 and moves without friction down the track shown. Determine the force exerted by the track on the car at point 2, and the minimum safe value of the radius of curvature at point 3. [Figure: track starting at point 1, 40 m above the lowest point 2 (radius of curvature $\rho_2 = 20$ m), then rising to a crest at point 3, 15 m above the level of point 2.] ## Short answer The coefficient of restitution e is the ratio of relative separation speed to relative approach speed. For the car, conservation of energy gives speed at point 2 and point 3. The track exerts an upward force of 150 kN at point 2, and the minimum safe radius of curvature at the crest 3 is 50 m. ## Answer #### Coefficient of restitution (direct central impact) Two particles A and B (masses $m_A,m_B$) move along the same line with $v_A>v_B$ and collide. **Deformation period:** they are compressed until they have a common velocity $u$. Let $P$ be the impulse of the compression force. $$ m_Av_A-\int P\,dt=m_Au,\qquad m_Bv_B+\int P\,dt=m_Bu $$ **Restitution period:** they separate under the restoring impulse $\int R\,dt$: $$ m_Au-\int R\,dt=m_Av'_A,\qquad m_Bu+\int R\,dt=m_Bv'_B $$ **Definition:** $e=\dfrac{\int R\,dt}{\int P\,dt}$. From the A equations: $\int P\,dt=m_A(v_A-u)$ and $\int R\,dt=m_A(u-v'_A)$, so $e=\dfrac{u-v'_A}{v_A-u}$. From the B equations: $e=\dfrac{v'_B-u}{u-v_B}$. Adding numerators and denominators of these equal ratios: $$ e=\frac{v'_B-v'_A}{v_A-v_B} $$ i.e. velocity of separation divided by velocity of approach; $e=1$ elastic, $e=0$ plastic. #### Car on the track $W=30$ kN, starts from rest at 1, which is 40 m above point 2. Point 3 is 15 m above point 2, so 25 m below point 1. The track is frictionless, so energy is conserved. ```text 1 o \ 3 (crest) \ _o_ \ / \ \__________ 2 / rho2 = 20 m ``` **Force at point 2 (lowest point):** $$ v_2^{2}=2g(40)=80g $$ $$ N-W=\frac{W}{g}\frac{v_2^{2}}{\rho_2}\ \Rightarrow\ N=W\left(1+\frac{80}{20}\right)=5W=5(30)=150\ \text{kN} $$ **Minimum radius at crest 3:** speed there from 1 to 3 (drop 25 m): $v_3^{2}=2g(25)=50g$. For the car to stay on the track the normal force $N_3\ge0$: $$ W-N_3=\frac{W}{g}\frac{v_3^{2}}{\rho_3}\ \Rightarrow\ \rho_3\ge\frac{v_3^{2}}{g}=\frac{50g}{g}=50\ \text{m} $$ **Answer: force of the track on the car at 2 is $150$ kN (upward); minimum safe radius of curvature at 3 is $\rho_3=50$ m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the coefficient of restitution defined in the derivation? A. Impulse of compression divided by impulse of restitution B. Ratio of restitution impulse to deformation impulse C. Ratio of final to initial kinetic energy D. Ratio of masses of the particles Answer: B. e = (integral R dt)/(integral P dt), restitution impulse over deformation impulse. (Key point: Definition of e) 2. What does the derivation give for e? A. (v_A - v_B)/(v'_B - v'_A) B. (v'_A + v'_B)/(v_A + v_B) C. (v'_B - v'_A)/(v_A - v_B), separation speed over approach speed D. m_A v_A/(m_B v_B) Answer: C. Combining the A and B equations gives e = relative velocity of separation over relative velocity of approach. (Key point: Result of derivation) 3. Why is N at the bottom point 2 equal to 5W? A. N = W + (W/g)(v^2/rho) with v^2 = 2g(40) and rho = 20 m, so N = W(1 + 4) B. N = W only C. N = W - (W/g)(v^2/rho) D. N = 2W because the car falls 40 m Answer: A. At the lowest point N - W = (W/g)v^2/rho, with v^2 = 80g, rho = 20, so N = 5(30) = 150 kN. (Key point: Force at point 2) 4. What condition gives the minimum safe radius at crest 3? A. N3 = W B. N3 = 2W C. The car's speed is zero D. N3 >= 0, so rho3 >= v3^2/g = 50 m Answer: D. At the crest W - N3 = (W/g)v3^2/rho; requiring N3 >= 0 with v3^2 = 2g(25) gives rho3 >= 50 m. 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