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

• $v_2-v_1$ = speed of separation — impact ke baad dono bodies kitni tezi se alag hoti hain.
• $u_1-u_2$ = speed of approach — impact se pehle woh kitni tezi se paas aa rahi thi.
• Range: ek ordinary passive collision ke liye $0\le e\le 1$. Dono quantities positive hain (isliye indices swap kiye gaye hain), to $e\ge 0$; aur separation bina internal energy source ke approach se zyada nahin ho sakti, isliye $e\le 1$. (Kisi hidden energy source wali superelastic collision mein $e>1$ aa sakta hai — upar edge-case note dekhein.)

$e=1$ ek perfectly elastic collision hai — separation speed approach speed ke barabar hoti hai. Momentum (hamesha conserved) ke alawa, kinetic energy bhi conserved hoti hai. Dekho Elastic Collisions.

Woh saath milke chalte hain → separation speed $=0$ → $e=0$. Yeh ek perfectly inelastic (perfectly plastic) collision hai. Dekho 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 aur 1 se kam → ek partially elastic collision. ✓

Height speed ke square par depend karti hai ($h=v^2/2g$), isliye height ratio $e^2$ hai aur hume square root chahiye: $e=\sqrt{\frac{h_2}{h_1}}=\sqrt{\frac{0.8}{1.8}}=\sqrt{0.4444}=0.667$

Approach speed sahi se signed velocities use karti hai: $u_1-u_2=5-(-3)=8\text{ m/s}$ Yeh physical closing speed hai (ek doosre ki taraf aane wali dono ek doosri add ho jaati hain). Phir $e=\frac{\text{separation}}{\text{approach}}=\frac{4}{8}=0.5$

Momentum: $2(6)+4(0)=2v_1+4v_2\Rightarrow 12=2v_1+4v_2\Rightarrow v_1+2v_2=6$. Restitution (inhe closing speed ka aadha alag hona chahiye): $v_2-v_1=e(u_1-u_2)=0.5(6-0)=3$. Restitution se: $v_1=v_2-3$. Substitute karo: $(v_2-3)+2v_2=6\Rightarrow 3v_2=9\Rightarrow v_2=3,v_1=0.$ $\boxed{v_1=0\text{ m/s},v_2=3\text{ m/s}}$ Check: separation $=3-0=3=e\cdot 6$ ✓; momentum $2(0)+4(3)=12$ ✓. Body 1 bilkul ruk jaata hai, body 2 momentum aage le jaata hai. Dekho Conservation of Linear Momentum.

Momentum: $v_1+2v_2=6$ (unchanged). Restitution (elastic → full closing speed par alag ho): $v_2-v_1=1(6)=6\Rightarrow v_1=v_2-6$. $(v_2-6)+2v_2=6\Rightarrow 3v_2=12\Rightarrow v_2=4,v_1=-2.$ $\boxed{v_1=-2\text{ m/s},v_2=+4\text{ m/s}}$ Body 1 bounce back karta hai (negative). KE check karo:

• 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, jaisa $e=1$ demand karta hai.

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

Figure padhna: horizontal axis time (s) hai, vertical axis height (m) hai. Blue curve ball ka actual up-and-down path hai (har hump ek flight hai). Red dots har hump ke top par baithe hain — woh peak heights $h_1,h_2,h_3,\dots$ hain, aur unke paas printed har red-dot value bilkul pichle wali ki $0.81$ guni hai. Dashed grey line $2.00$ m release height mark karti hai. Maths: har bounce speed ko $e$ se multiply karta hai, aur kyunki $h\propto v^2$, height ko $e^2=0.81$ se multiply karta hai. To $h_n=e^{2n}{h}_0$.

• (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. In numbers ko figure ke red-dot labels se compare karo — woh 1st aur 3rd peaks se match karte hain.

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 $=9.6/25=0.384$. Shortcut se cross-check: $\Delta KE=\frac{1}{2}\frac{m_1{m}_2}{m_1+m_2}(1-e^2)(u_1-u_2{)}^2=\frac{1}{2}\cdot \frac{6}{5}(1-0.36)(25)=0.6\times 0.64\times 25=9.6$ J ✓. Dekho Kinetic Energy Loss in Collisions.

Figure padhna: axes do velocity components hain — horizontal (x) rightward, vertical (y) upward, dono m/s mein, same scale par drawn (ek square grid). Grey band floor hai; red dot impact point hai. Orange arrow incoming velocity $(6,-8)$ hai — yeh neeche-daayein point karta hai aur steep hai kyunki downward part ($8$), sideways part ($6$) se bada hai. Green arrow outgoing velocity $(6,+4)$ hai — same sideways length, lekin upward part aadha ho gaya, isliye yeh shallower hai. Blue dashed vertical line impact normal hai; $e$ sirf us line ke saath velocity wale piece ko scale karta hai. Maths. Restitution sirf impact normal ke saath act karta hai (yahan vertical). Ek smooth floor koi horizontal force nahin lagaata, isliye $u_x$ unchanged rehta hai.

• Horizontal after: $v_x=u_x=6$ m/s.
• Vertical after: $v_y=e{u}_y=0.5\times 8=4$ m/s (ab upward).
• Rebound speed: $|v|=\sqrt{6^2+4^2}=\sqrt{52}=7.211$ m/s.
• Horizontal se angle: $\theta =\arctan \left(\frac{v_y}{v_x}\right)=\arctan \left(\frac{4}{6}\right)=33.69^{\circ }$. Green arrow orange wale se shallower hai bilkul isliye kyunki vertical part chhota ho gaya jabki horizontal part waisa hi raha — yeh $e$ ka poora geometric effect hai. Dekho Projectile Motion.

Heights: $h_0$, phir up-and-down $h_1,h_2,\dots$ jahan $h_n=e^{2n}{h}_0$. Pehla drop $h_0$ hai. Har baad wala bounce ek up + down of $2h_n$ contribute karta hai. $D=h_0+2{\sum }_{n=1}^{\infty }{h}_n=h_0+2h_0{\sum }_{n=1}^{\infty }(e^2{)}^n=h_0+2h_0\frac{e^2}{1-e^2}.$ (Infinite sum ${\sum }_{n=1}^{\infty }{r}^n=\frac{r}{1-r}$ for $0<r<1$; yahan $r=e^2=0.64$.) $\frac{e^2}{1-e^2}=\frac{0.64}{0.36}=1.7778$ ke saath: $D=1.25+2(1.25)(1.7778)=1.25+4.4444=5.694\text{ m}.$ Equivalent closed form: $D=h_0\frac{1+e^2}{1-e^2}=1.25\times \frac{1.64}{0.36}=5.694$ m ✓.

Kyun time $e^n$ se scale karta hai ($e^{2n}$ se nahin). Peak $h_n$ se ek flight up-and-down ka time $2\sqrt{2h_n/g}$ hai — yeh $\sqrt{h_n}$ ke proportional hai. Lekin $h_n=e^{2n}{h}_0$, isliye $\sqrt{h_n}=\sqrt{e^{2n}{h}_0}=e^n\sqrt{h_0}$. Square root height ke $e^{2n}$ ko time ke $e^n$ mein badal deta hai. Yahi key hai: heights har bounce par $e^2$ se shrink hoti hain, times har bounce par $e$ se shrink hote hain. Pehla fall: $t_0=\sqrt{2h_0/g}=\sqrt{2(1.25)/10}=\sqrt{0.25}=0.5$ s. Bounce $n$ ke baad, up-then-down time hai $2\sqrt{2h_n/g}=2t_0{e}^n$. $T=t_0+{\sum }_{n=1}^{\infty }2t_0{e}^n=t_0+2t_0\frac{e}{1-e}=t_0\left(1+\frac{2e}{1-e}\right)=t_0\frac{1+e}{1-e}.$ (Phir se ${\sum }_{n=1}^{\infty }{r}^n=\frac{r}{1-r}$, ab $r=e=0.8$ ke saath.) $T=0.5\times \frac{1.8}{0.2}=0.5\times 9=4.5\text{ s}.$

$(1-e^2)$ kahan se aata hai. Kisi bhi 1-D collision mein kinetic energy lost ko relative velocity use karke likha ja sakta hai. Approach relative speed hai $u_{\text{rel}}=u_1-u_2$; separation relative speed hai $v_2-v_1=e{u}_{\text{rel}}$. Ek standard result (centre-of-mass frame mein prove kiya gaya, jahan saari KE relative motion mein stored hoti hai) yeh hai: $\Delta KE=\frac{1}{2}\mu (u_{\text{rel}}^2-(e{u}_{\text{rel}}{)}^2)=\frac{1}{2}\mu (1-e^2)u_{\text{rel}}^2,\mu =\frac{m_1{m}_2}{m_1+m_2}.$ $(1-e^2)$ isliye aata hai kyunki separation speed, approach speed ka $e$ guna hai, aur energy speed के square ke saath jaati hai, isliye relative KE ka bacha hua fraction $1-e^2$ hai. Equal masses ke liye $\mu =m/2$ aur $u_{\text{rel}}=u_1=10$, jabki initial KE hai $\frac{1}{2}m{u}_1^2$. To lost ka fraction hai $f=\frac{\frac{1}{2}\cdot \frac{m}{2}(1-e^2)u_1^2}{\frac{1}{2}m{u}_1^2}=\frac{1-e^2}{2}.$ $f=0.36$ set karo: $\frac{1-e^2}{2}=0.36\Rightarrow 1-e^2=0.72\Rightarrow e^2=0.28\Rightarrow e=\sqrt{0.28}=0.529$. Ab velocities. Equal masses ke liye, momentum deta hai $v_1+v_2=10$; restitution deta hai $v_2-v_1=e(10)=5.29$. $v_2=\frac{10+5.29}{2}=7.646,v_1=\frac{10-5.29}{2}=\mathrm{2.354.}$ $\boxed{e=0.529,v_1=2.354\text{ m/s},v_2=7.646\text{ m/s}}$ KE check: after $=\frac{1}{2}m(2.354^2+7.646^2)=\frac{1}{2}m(5.541+58.46)=32.0m$; before $=50m$; lost $=18m=0.36\times 50m$ ✓.