Coefficient of restitution; car on track

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.

 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.