Exercises — Elastic collisions — 2D - angle relationship

The 90° rule says the two outgoing directions are perpendicular: ${\theta }_2=90^{\circ }-{\theta }_1=90^{\circ }-25^{\circ }=65^{\circ }.$ Why? Equal masses + one at rest + elastic + both moving $\Rightarrow$ ${\theta }_1+{\theta }_2=90^{\circ }$. That's the whole trick.

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

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

First get the cue ball's angle from the 90° rule: ${\theta }_1=90^{\circ }-{\theta }_2=90^{\circ }-53^{\circ }=37^{\circ }.$ Then the speeds (still using ${\theta }_1$, the cue ball's angle): $v_1=20\cos 37^{\circ }=20(0.7986)=15.97\text{m/s},$ $v_2=20\sin 37^{\circ }=20(0.6018)=12.04\text{m/s}.$ Check: $15.97^2+12.04^2=255.1+145.0=400.1\approx 20^2$ ✓.

Energy (Pythagoras) gives the cue ball's speed directly: $v_1=\sqrt{u^2-v_2^2}=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12\text{m/s}.$ Now the struck ball's angle from $v_2=u\sin {\theta }_1$: $\sin {\theta }_1=\frac{v_2}{u}=\frac{9}{15}=0.6\Rightarrow {\theta }_1=36.87^{\circ }.$ ${\theta }_2=90^{\circ }-{\theta }_1=53.13^{\circ }.$ This is the classic $9$–$12$–$15$ (i.e. $3$–$4$–$5$) triangle. ✓

From $v_2=u\sin {\theta }_1$ and $0^{\circ }<{\theta }_1<90^{\circ }$ (both balls move): $v_2=u\sin {\theta }_1\le u\text{because }\sin {\theta }_1\le 1.$ So $v_2\le u$ always. The maximum $v_2=u$ would require $\sin {\theta }_1=1$, i.e. ${\theta }_1=90^{\circ }$ — but that means the cue ball is deflected sideways and $v_1=u\cos 90^{\circ }=0$: the cue ball has stopped. That is exactly the head-on (degenerate) case where only one ball moves, so the 90° "opening" isn't defined. For a genuine glancing hit (${\theta }_1$ just under $90^{\circ }$) the struck ball approaches but never reaches $u$.

(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) Put ${\vec{v}}_1$ at $+60^{\circ }$ and ${\vec{v}}_2$ at $-30^{\circ }$ from the incoming (x) axis: ${\vec{v}}_1=v_1(\cos 60^{\circ },\sin 60^{\circ })=(0.5,0.866)\times 10^6,$ ${\vec{v}}_2=v_2(\cos 30^{\circ },-\sin 30^{\circ })=(1.5,-0.866)\times 10^6.$ ${\vec{v}}_1\cdot {\vec{v}}_2=(0.5)(1.5)+(0.866)(-0.866)\times 10^{12}=(0.75-0.75)\times 10^{12}=0.✓$ The dot product is zero, confirming the $90^{\circ }$ separation.

Multiply the energy relation $u^2=v_1^2+v_2^2$ by $\frac{1}{2}m$: $\frac{1}{2}m{u}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2,$ which is exactly Kinetic Energy conservation — no energy lost (elastic). The fraction carried by the struck ball is $f=\frac{\frac{1}{2}m{v}_2^2}{\frac{1}{2}m{u}^2}=\frac{v_2^2}{u^2}={\sin }^2{\theta }_1.$ At ${\theta }_1=45^{\circ }$: $f={\sin }^{2}45^{\circ }=(0.7071{)}^2=0.5$. Exactly half the energy goes to the struck ball, half stays with the cue ball — the symmetric case.

For unequal masses the parent's Step 4 leaves a non-zero remainder. Squaring $m_1\vec{u}=m_1{\vec{v}}_1+m_2{\vec{v}}_2$ and subtracting energy gives ${\vec{v}}_1\cdot {\vec{v}}_2=\frac{m_1-m_2}{2m_2}(\text{positive quantity}\sim v_1{v}_2).$ Here $m_1=2m>m_2=m$, so $m_1-m_2>0$, making ${\vec{v}}_1\cdot {\vec{v}}_2>0$. A positive dot product means the angle between ${\vec{v}}_1$ and ${\vec{v}}_2$ is less than $90^{\circ }$. So: ${\theta }_1+{\theta }_2<90^{\circ }\Rightarrow {\theta }_2<90^{\circ }-{\theta }_1=70^{\circ }.$ Physical reading: the heavy ball barely deflects and keeps ploughing forward; the light target squirts out ahead, so both cluster toward the forward direction — a narrow V, opening angle under $90^{\circ }$. (Heavy-hits-light $\Rightarrow$ opening $<90^{\circ }$; light-hits-heavy $\Rightarrow$ opening $>90^{\circ }$, can even backscatter.)

(a) Energy / Pythagoras: $u=\sqrt{v_1^2+v_2^2}=\sqrt{8^2+15^2}=\sqrt{64+225}=\sqrt{289}=17\text{m/s}$ (the $8$–$15$–$17$ triple). (b) Cue-ball angle from $v_1=u\cos {\theta }_1$: $\cos {\theta }_1=\frac{8}{17}=0.4706\Rightarrow {\theta }_1=61.93^{\circ }.$ Struck-ball angle from $v_2=u\sin {\theta }_1$ (check): $\sin {\theta }_1=15/17=0.8824$, consistent. Then ${\theta }_2=\arcsin \left(\frac{v_1}{u}\right)=\arcsin \frac{8}{17}=28.07^{\circ }.$ (c) ${\theta }_1+{\theta }_2=61.93^{\circ }+28.07^{\circ }=90.00^{\circ }$ ✓. The right angle is recovered from raw speed data alone.

• Given only the two outgoing speeds (equal mass, elastic), how do you get $u$? ::: $u=\sqrt{v_1^2+v_2^2}$ (Pythagoras).
• Struck ball's energy fraction in terms of ${\theta }_1$? ::: ${\sin }^2{\theta }_1$.
• Sign of ${\vec{v}}_1\cdot {\vec{v}}_2$ when heavy hits light? ::: Positive $\Rightarrow$ opening angle $<90^{\circ }$.
• Max possible struck-ball speed, and what it costs? ::: Approaches $u$ as ${\theta }_1\to 90^{\circ }$, but then the cue ball stops (degenerate head-on).