Exercises — Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

• $v_2-v_1$ = speed of separation — how fast the two bodies fly apart after impact.
• $u_1-u_2$ = speed of approach — how fast they close in before impact.
• Range: $0\le e\le 1$ for an ordinary passive collision. Both quantities are positive (that is why the indices are swapped), so $e\ge 0$; and separation cannot exceed approach without an internal energy source, so $e\le 1$. (A superelastic collision with a hidden energy source can give $e>1$ — see the edge-case note above.)

$e=1$ is a perfectly elastic collision — separation speed equals approach speed. Besides momentum (always conserved), kinetic energy is also conserved. See Elastic Collisions.

They move off together → separation speed $=0$ → $e=0$. This is a perfectly inelastic (perfectly plastic) collision. See Perfectly Inelastic Collisions.

Approach $=u_1-u_2=8-2=6$ m/s. Separation $=v_2-v_1=7-4=3$ m/s. $e=\frac{3}{6}=0.5$ Positive and less than 1 → a partially elastic collision. ✓

Height depends on speed squared ($h=v^2/2g$), so the height ratio is $e^2$ and we need a square root: $e=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{0.8}{1.8}}=\sqrt{0.4444}=0.667$

The approach speed correctly uses the signed velocities: $u_1-u_2=5-(-3)=8\text{ m/s}$ That is the physical closing speed (each moving toward the other adds up). Then $e=\frac{\text{separation}}{\text{approach}}=\frac{4}{8}=0.5$

Momentum: $2(6)+4(0)=2v_1+4v_2\Rightarrow 12=2v_1+4v_2\Rightarrow v_1+2v_2=6$. Restitution (they must separate at half the closing speed): $v_2-v_1=e(u_1-u_2)=0.5(6-0)=3$. From restitution: $v_1=v_2-3$. Substitute: $(v_2-3)+2v_2=6\Rightarrow 3v_2=9\Rightarrow v_2=3,v_1=0.$ $\boxed{v_1=0\text{ m/s},v_2=3\text{ m/s}}$ Check: separation $=3-0=3=e\cdot 6$ ✓; momentum $2(0)+4(3)=12$ ✓. Body 1 stops dead, body 2 carries the momentum forward. See Conservation of Linear Momentum.

Momentum: $v_1+2v_2=6$ (unchanged). Restitution (elastic → separate at the full closing speed): $v_2-v_1=1(6)=6\Rightarrow v_1=v_2-6$. $(v_2-6)+2v_2=6\Rightarrow 3v_2=12\Rightarrow v_2=4,v_1=-2.$ $\boxed{v_1=-2\text{ m/s},v_2=+4\text{ m/s}}$ Body 1 bounces back (negative). Check KE:

• Before: $\frac{1}{2}(2)(6^2)=36$ J.
• After: $\frac{1}{2}(2)(2^2)+\frac{1}{2}(4)(4^2)=4+32=36$ J. ✓ Conserved, as $e=1$ demands.

Momentum: $3(4)+1(-2)=3v_1+v_2\Rightarrow 10=3v_1+v_2$. Restitution (separate at $0.8$ of the closing speed): $v_2-v_1=0.8(4-(-2))=0.8(6)=4.8\Rightarrow v_2=v_1+4.8$. Substitute: $3v_1+(v_1+4.8)=10\Rightarrow 4v_1=5.2\Rightarrow v_1=1.3$, and $v_2=6.1$. $\boxed{v_1=1.3\text{ m/s},v_2=6.1\text{ m/s}}$ Check: separation $=6.1-1.3=4.8$ ✓; momentum $3(1.3)+1(6.1)=3.9+6.1=10$ ✓.

Reading the figure: the horizontal axis is time (s), the vertical axis is height (m). The blue curve is the ball's actual up-and-down path (each hump is one flight). The red dots sit on the top of each hump — those are the peak heights $h_1,h_2,h_3,\dots$, and each red-dot value printed beside it is exactly $0.81$ times the previous one. The dashed grey line marks the $2.00$ m release height. The maths: each bounce multiplies the speed by $e$, and since $h\propto v^2$, it multiplies the height by $e^2=0.81$. So $h_n=e^{2n}{h}_0$.

• (a) $h_1=e^2{h}_0=0.81\times 2.0=1.62$ m.
• (b) $h_3=e^6{h}_0=0.9^6\times 2.0=0.531441\times 2.0=1.063$ m. Compare these numbers to the red-dot labels in the figure — they match the 1st and 3rd peaks.

Momentum: $2(5)=2v_1+3v_2\Rightarrow 10=2v_1+3v_2$. Restitution: $v_2-v_1=0.6(5)=3\Rightarrow v_2=v_1+3$. Substitute: $2v_1+3(v_1+3)=10\Rightarrow 5v_1=1\Rightarrow v_1=0.2$, $v_2=3.2$.

• KE before: $\frac{1}{2}(2)(5^2)=25$ J.
• KE after: $\frac{1}{2}(2)(0.2^2)+\frac{1}{2}(3)(3.2^2)=0.04+15.36=15.4$ J.
• Lost: $25-15.4=9.6$ J → fraction $=9.6/25=0.384$. Cross-check with the shortcut $\Delta KE=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(1-e^2)(u_1-u_2{)}^2=\frac{1}{2}\cdot \frac{6}{5}(1-0.36)(25)=0.6\times 0.64\times 25=9.6$ J ✓. See Kinetic Energy Loss in Collisions.

Reading the figure: the axes are the two velocity components — horizontal (x) rightward, vertical (y) upward, both in m/s, drawn to the same scale (a square grid). The grey band is the floor; the red dot is the impact point. The orange arrow is the incoming velocity $(6,-8)$ — it points down-right and is steep because the downward part ($8$) is bigger than the sideways part ($6$). The green arrow is the outgoing velocity $(6,+4)$ — same sideways length, but the upward part has shrunk to half, so it is shallower. The blue dashed vertical line is the impact normal; $e$ scales only the piece of velocity along that line. The maths. Restitution acts only along the impact normal (vertical here). A smooth floor exerts no horizontal force, so $u_x$ is unchanged.

• Horizontal after: $v_x=u_x=6$ m/s.
• Vertical after: $v_y=e{u}_y=0.5\times 8=4$ m/s (now upward).
• Rebound speed: $|v|=\sqrt{6^2+4^2}=\sqrt{52}=7.211$ m/s.
• Angle above horizontal: $\theta =\arctan \left(\frac{v_y}{v_x}\right)=\arctan \left(\frac{4}{6}\right)=33.69^{\circ }$. The green arrow is shallower than the orange one precisely because the vertical part shrank while the horizontal part stayed put — that is the whole geometric effect of $e$. See Projectile Motion.

Heights: $h_0$, then up-and-down $h_1,h_2,\dots$ with $h_n=e^{2n}{h}_0$. The first drop is $h_0$. Each subsequent bounce contributes an up + down of $2h_n$. $D=h_0+2{\sum }_{n=1}^{\infty }{h}_n=h_0+2h_0{\sum }_{n=1}^{\infty }(e^2{)}^n=h_0+2h_0\frac{e^2}{1-e^2}.$ (The infinite sum ${\sum }_{n=1}^{\infty }{r}^n=\frac{r}{1-r}$ for $0<r<1$; here $r=e^2=0.64$.) With $\frac{e^2}{1-e^2}=\frac{0.64}{0.36}=1.7778$: $D=1.25+2(1.25)(1.7778)=1.25+4.4444=5.694\text{ m}.$ Equivalent closed form: $D=h_0\frac{1+e^2}{1-e^2}=1.25\times \frac{1.64}{0.36}=5.694$ m ✓.

Why time scales as $e^n$ (not $e^{2n}$). The time for one flight up-and-down from peak $h_n$ is $2\sqrt{2h_n/g}$ — it is proportional to $\sqrt{h_n}$. But $h_n=e^{2n}{h}_0$, so $\sqrt{h_n}=\sqrt{e^{2n}{h}_0}=e^n\sqrt{h_0}$. The square root turns the $e^{2n}$ of height into $e^n$ of time. That is the key: heights shrink by $e^2$ each bounce, times shrink by $e$ each bounce. First fall: $t_0=\sqrt{2h_0/g}=\sqrt{2(1.25)/10}=\sqrt{0.25}=0.5$ s. After bounce $n$, the up-then-down time is $2\sqrt{2h_n/g}=2t_0{e}^n$. $T=t_0+{\sum }_{n=1}^{\infty }2t_0{e}^n=t_0+2t_0\frac{e}{1-e}=t_0\left(1+\frac{2e}{1-e}\right)=t_0\frac{1+e}{1-e}.$ (Again ${\sum }_{n=1}^{\infty }{r}^n=\frac{r}{1-r}$, now with $r=e=0.8$.) $T=0.5\times \frac{1.8}{0.2}=0.5\times 9=4.5\text{ s}.$

Where the $(1-e^2)$ comes from. In any 1-D collision the kinetic energy lost can be written using the relative velocity. Approach relative speed is $u_{\text{rel}}=u_1-u_2$; separation relative speed is $v_2-v_1=e{u}_{\text{rel}}$. A standard result (proved by working in the centre-of-mass frame, where all KE is stored in the relative motion) is: $\Delta KE=\frac{1}{2}\mu (u_{\text{rel}}^2-(e{u}_{\text{rel}}{)}^2)=\frac{1}{2}\mu (1-e^2)u_{\text{rel}}^2,\mu =\frac{m_1{m}_2}{m_1+m_2}.$ The $(1-e^2)$ appears because separation speed is $e$ times approach speed, and energy goes as speed squared, so the leftover fraction of relative KE is $1-e^2$. For equal masses $\mu =m/2$ and $u_{\text{rel}}=u_1=10$, while the initial KE is $\frac{1}{2}m{u}_1^2$. So the fraction lost is $f=\frac{\frac{1}{2}\cdot \frac{m}{2}(1-e^2)u_1^2}{\frac{1}{2}m{u}_1^2}=\frac{1-e^2}{2}.$ Set $f=0.36$: $\frac{1-e^2}{2}=0.36\Rightarrow 1-e^2=0.72\Rightarrow e^2=0.28\Rightarrow e=\sqrt{0.28}=0.529$. Now the velocities. For equal masses, momentum gives $v_1+v_2=10$; restitution gives $v_2-v_1=e(10)=5.29$. $v_2=\frac{10+5.29}{2}=7.646,v_1=\frac{10-5.29}{2}=\mathrm{2.354.}$ $\boxed{e=0.529,v_1=2.354\text{ m/s},v_2=7.646\text{ m/s}}$ Check KE: after $=\frac{1}{2}m(2.354^2+7.646^2)=\frac{1}{2}m(5.541+58.46)=32.0m$; before $=50m$; lost $=18m=0.36\times 50m$ ✓.