Exercises — Elastic collisions — 2D - angle relationship

90° rule kehta hai ki dono outgoing directions perpendicular hain: ${\theta }_2=90^{\circ }-{\theta }_1=90^{\circ }-25^{\circ }=65^{\circ }.$ Kyun? Equal masses + ek at rest + elastic + dono moving $\Rightarrow$ ${\theta }_1+{\theta }_2=90^{\circ }$. Bas yahi poora trick hai.

Opening angle $={\theta }_1+{\theta }_2=90^{\circ }$. Diya gaya ${\theta }_1=40^{\circ }$: ${\theta }_2=90^{\circ }-40^{\circ }=50^{\circ }\text{ below the line}.$

Cue ball apne angle ke adjacent leg hai, struck ball opposite leg hai: $v_1=u\cos {\theta }_1=12\cos 30^{\circ }=12(0.8660)=10.39\text{m/s},$ $v_2=u\sin {\theta }_1=12\sin 30^{\circ }=12(0.5)=6.00\text{m/s}.$ Energy check: $10.39^2+6^2=107.9+36=143.9\approx 12^2=144$ ✓.

Pehle 90° rule se cue ball ka angle nikalo: ${\theta }_1=90^{\circ }-{\theta }_2=90^{\circ }-53^{\circ }=37^{\circ }.$ Phir speeds (abhi bhi ${\theta }_1$, cue ball ka angle, use karo): $v_1=20\cos 37^{\circ }=20(0.7986)=15.97\text{m/s},$ $v_2=20\sin 37^{\circ }=20(0.6018)=12.04\text{m/s}.$ Check: $15.97^2+12.04^2=255.1+145.0=400.1\approx 20^2$ ✓.

Energy (Pythagoras) seedha cue ball ki speed deta hai: $v_1=\sqrt{u^2-v_2^2}=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12\text{m/s}.$ Ab $v_2=u\sin {\theta }_1$ se struck ball ka angle: $\sin {\theta }_1=\frac{v_2}{u}=\frac{9}{15}=0.6\Rightarrow {\theta }_1=36.87^{\circ }.$ ${\theta }_2=90^{\circ }-{\theta }_1=53.13^{\circ }.$ Yeh classic $9$–$12$–$15$ (yaani $3$–$4$–$5$) triangle hai. ✓

$v_2=u\sin {\theta }_1$ aur $0^{\circ }<{\theta }_1<90^{\circ }$ (dono balls move karte hain) se: $v_2=u\sin {\theta }_1\le u\text{because }\sin {\theta }_1\le 1.$ Toh $v_2\le u$ hamesha. Maximum $v_2=u$ ke liye $\sin {\theta }_1=1$ chahiye, yaani ${\theta }_1=90^{\circ }$ — lekin iska matlab hai cue ball sideways deflect ho gayi aur $v_1=u\cos 90^{\circ }=0$: cue ball ruk gayi. Yeh exactly head-on (degenerate) case hai jahan sirf ek ball move karti hai, toh 90° "opening" define nahi hoti. Genuine glancing hit ke liye (${\theta }_1$ thoda $90^{\circ }$ se kam) struck ball $u$ ke paas pahunchti hai lekin kabhi reach nahi karti.

(a) Equal masses, target at rest, elastic $\Rightarrow$ 90° rule: ${\theta }_2=90^{\circ }-60^{\circ }=30^{\circ }.$ (b) Speeds: $v_1=u\cos 60^{\circ }=2.0\times 10^6(0.5)=1.0\times 10^6\text{m/s},$ $v_2=u\sin 60^{\circ }=2.0\times 10^6(0.8660)=1.732\times 10^6\text{m/s}.$ (c) ${\vec{v}}_1$ ko $+60^{\circ }$ par aur ${\vec{v}}_2$ ko incoming (x) axis se $-30^{\circ }$ par rakho: ${\vec{v}}_1=v_1(\cos 60^{\circ },\sin 60^{\circ })=(0.5,0.866)\times 10^6,$ ${\vec{v}}_2=v_2(\cos 30^{\circ },-\sin 30^{\circ })=(1.5,-0.866)\times 10^6.$ ${\vec{v}}_1\cdot {\vec{v}}_2=(0.5)(1.5)+(0.866)(-0.866)\times 10^{12}=(0.75-0.75)\times 10^{12}=0.✓$ Dot product zero hai, jo $90^{\circ }$ separation confirm karta hai.

Energy relation $u^2=v_1^2+v_2^2$ ko $\frac{1}{2}m$ se multiply karo: $\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2,$ jo exactly Kinetic Energy conservation hai — koi energy lost nahi (elastic). Struck ball ke paas jaane wala fraction hai $f=\frac{\frac{1}{2}m{v}_2^2}{\frac{1}{2}m{u}^2}=\frac{v_2^2}{u^2}={\sin }^2{\theta }_1.$ ${\theta }_1=45^{\circ }$ par: $f={\sin }^{2}45^{\circ }=(0.7071{)}^2=0.5$. Exactly aadhi energy struck ball ko jaati hai, aadhi cue ball ke paas rehti hai — symmetric case.

Unequal masses ke liye parent ka Step 4 ek non-zero remainder chodta hai. $m_1\vec{u}=m_1{\vec{v}}_1+m_2{\vec{v}}_2$ ko square karke energy subtract karne par milta hai ${\vec{v}}_1\cdot {\vec{v}}_2=\frac{m_1-m_2}{2m_2}(\text{positive quantity}\sim v_1{v}_2).$ Yahan $m_1=2m>m_2=m$, toh $m_1-m_2>0$, jisse ${\vec{v}}_1\cdot {\vec{v}}_2>0$ banta hai. Positive dot product ka matlab hai ${\vec{v}}_1$ aur ${\vec{v}}_2$ ke beech ka angle $90^{\circ }$ se kam hai. Toh: ${\theta }_1+{\theta }_2<90^{\circ }\Rightarrow {\theta }_2<90^{\circ }-{\theta }_1=70^{\circ }.$ Physical reading: heavy ball barely deflect hoti hai aur aage badhti rehti hai; light target aage squirt ho jaata hai, toh dono forward direction ki taraf cluster karte hain — ek narrow V, opening angle $90^{\circ }$ se kam. (Heavy-hits-light $\Rightarrow$ opening $<90^{\circ }$; light-hits-heavy $\Rightarrow$ opening $>90^{\circ }$, backscatter bhi ho sakta hai.)

(a) Energy / Pythagoras: $u=\sqrt{v_1^2+v_2^2}=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\text{m/s}$ ($8$–$15$–$17$ triple). (b) $v_1=u\cos {\theta }_1$ se cue-ball angle: $\cos {\theta }_1=\frac{8}{17}=0.4706\Rightarrow {\theta }_1=61.93^{\circ }.$ Struck-ball angle $v_2=u\sin {\theta }_1$ se (check): $\sin {\theta }_1=15/17=0.8824$, consistent hai. Phir ${\theta }_2=\arcsin \left(\frac{v_1}{u}\right)=\arcsin \frac{8}{17}=28.07^{\circ }.$ (c) ${\theta }_1+{\theta }_2=61.93^{\circ }+28.07^{\circ }=90.00^{\circ }$ ✓. Right angle sirf raw speed data se recover ho gaya.

• Sirf dono outgoing speeds diye hone par (equal mass, elastic), $u$ kaise nikaalte ho? ::: $u=\sqrt{v_1^2+v_2^2}$ (Pythagoras).
• ${\theta }_1$ ke terms mein struck ball ki energy fraction? ::: ${\sin }^2{\theta }_1$.
• Heavy hits light mein ${\vec{v}}_1\cdot {\vec{v}}_2$ ka sign? ::: Positive $\Rightarrow$ opening angle $<90^{\circ }$.
• Max possible struck-ball speed, aur uski kya cost hai? ::: Approaches $u$ as ${\theta }_1\to 90^{\circ }$, lekin tab cue ball ruk jaati hai (degenerate head-on).