Exercises — Linear momentum p = mv

KYA karein: magnitudes ko $p=mv$ mein plug karo. KYUN: object ka ek hi mass hai aur ek hi speed, kuch add nahi karna. $p=mv=12\times 1.5=18\text{kg⋅m/s}$ Direction (clearly batao): momentum hamesha $\vec{v}$ ke along point karta hai kyunki $m>0$ direction kabhi nahi paltata. 1D sign notation mein, agar hum rolling direction ko $+$ lein, toh momentum hai $p=+18\text{kg⋅m/s}$ (positive value = jis taraf roll kar rahi hai usi taraf point karna).

KYUN rearrange karein: hume $p$ aur $m$ pata hai, $v$ chahiye, toh $p=mv$ ko invert karke $v=p/m$ banao. $v=\frac{p}{m}=\frac{6\times 10^{-6}}{2\times 10^{-6}}=3\text{m/s}$

KYUN sign: velocity ek vector hai; "left" jabki right $+$ hai matlab $v=-25\text{ m/s}$. $p=mv=0.16\times (-25)=-4.0\text{kg⋅m/s}$ Minus sign hi direction hai — ye signed value $+$ axis ke along $\vec{p}$ ka component hai. 1D mein, bas itna hi kisi vector ko chahiye.

KYUN rearrange karein: hume $p$ aur $v$ diya gaya hai lekin $m$ chahiye, toh definition $p=mv$ ko $v$ se divide karke invert karo, milega $m=p/v$. $m=\frac{p}{v}=\frac{2.4}{8}=0.30\text{kg}$

KAUN sa tool aur KYUN: hume $KE$ diya gaya hai aur $p$ (magnitude) chahiye, aur wo bridge jo $v$ calculate kiye bina kaam karta hai wo hai $KE=\frac{p^2}{2m}$. $p$ ke liye solve karo: $p^2=2mKE⟹p=\sqrt{2mKE}=\sqrt{2\times 0.5\times 16}=\sqrt{16}=4\text{kg⋅m/s}$ KYUN sirf positive root: $p^2=16$ solve karne par $p=\pm 4$ milta hai, lekin $p$ yahan ek magnitude hai — momentum arrow ki length — aur lengths kabhi negative nahi hoti. Toh hum $-4$ hata dete hain aur $p=+4\text{kg⋅m/s}$ rakhte hain. (Agar direction poochi jaati, toh hum use alag se attach karte; magnitude khud positive rehti hai.)

KYUN signs ke saath add karein: total momentum vector sum ${\vec{p}}_A+{\vec{p}}_B$ hai; 1D mein ye ek ordinary sum hai jab signs set ho jaayein (right $=+$). $p_A=3(+4)=+12,p_B=5(-2)=-10$ $p_{\text{tot}}=12+(-10)=+2\text{kg⋅m/s (right direction mein)}$

Step 1 — components (KYUN): $m$ ek positive scalar hai, toh ye har velocity component ko same tarah scale karta hai. $p_x=0.30\times 6=1.8\text{kg⋅m/s},p_y=0.30\times 8=2.4\text{kg⋅m/s}$ Step 2 — magnitude (KYUN Pythagoras): $p_x$ aur $p_y$ perpendicular hain, toh ye ek right triangle ki do legs banate hain jiska hypotenuse $|\vec{p}|$ hai (figure mein red triangle dekho). $|\vec{p}|=\sqrt{1.8^2+2.4^2}=\sqrt{3.24+5.76}=\sqrt{9}=3.0\text{kg⋅m/s}$ Step 3 — direction (KYUN arctan): angle ki steepness $\tan \theta =\frac{\text{opposite}}{\text{adjacent}}=\frac{p_y}{p_x}$ se capture hoti hai; angle recover karne ke liye hum poochte hain "kaunsa angle ye tan rakhta hai?" — wo hai $\arctan$. $\theta =\arctan \left(\frac{2.4}{1.8}\right)=\arctan (1.333)\approx 53.1^{\circ }$ Dono components positive hain → quadrant I, toh raw $\arctan$ value already sahi hai.

KAUN sa tool aur KYUN: hamare paas har ek ke liye $p$ (magnitude) aur $m$ hai, aur $v$ alag se nikale bina $KE$ chahiye — $KE=\frac{p^2}{2m}$ use karo. $K{E}_{\text{bullet}}=\frac{4^2}{2\times 0.02}=\frac{16}{0.04}=400\text{J}$ $K{E}_{\text{block}}=\frac{4^2}{2\times 2}=\frac{16}{4}=4\text{J}$ Bullet mein zyada energy hai, $\frac{400}{4}=100$ factor se. KYUN itna bada fark: fixed $p$ par, $KE=p^2/2m\propto 1/m$. Bullet ka mass $100\times$ kam hai, toh energy $100\times$ zyada hai. Equal "oomph," lekin wildly unequal "damage capacity."

$p=mv$ mein $v$ linear hai: $v$ teen guna → momentum $\times 3$. $KE=\frac{1}{2}m{v}^2$ mein $v$ quadratic hai: $v$ teen guna → energy $\times 3^2=9$. Toh momentum $3\times$ badhti hai, kinetic energy $9\times$ badhti hai.

Step 1 — har object ka momentum (KYUN per-axis): A purely $x$ ke along move karta hai, B purely $y$ ke along, toh har ek sirf ek hi axis par contribute karta hai. ${\vec{p}}_A:p_{Ax}=2\times 3=6,p_{Ay}=0$ ${\vec{p}}_B:p_{Bx}=0,p_{By}=1\times 8=8$ Step 2 — component by component add karo (KYUN): vector sum ka matlab hai $x$ waale saath mein aur $y$ waale saath mein, alag-alag add karo. $p_x=6+0=6,p_y=0+8=8\text{kg⋅m/s}$ Step 3 — magnitude (right triangle): dono totals perpendicular legs hain (figure mein teal aur orange arrows dekho jo plum resultant mein combine ho rahe hain). $|\vec{p}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{kg⋅m/s}$ Step 4 — direction: $\theta =\arctan \frac{p_y}{p_x}=\arctan \frac{8}{6}=\arctan (1.333)\approx 53.1^{\circ }$ north of east. Dono components positive → quadrant I, koi correction nahi chahiye.

Magnitude: $|\vec{p}|=\sqrt{(-3{)}^2+(-4{)}^2}=\sqrt{9+16}=\sqrt{25}=5\text{kg⋅m/s}$. Direction — KYUN naive $\arctan$ yahan fail karta hai: $\tan$ har $180^{\circ }$ par repeat hota hai, toh $\arctan$ sirf $-90^{\circ }$ aur $+90^{\circ }$ ke beech ka angle return karta hai (quadrants I aur IV). Hamara vector down-and-left point karta hai → quadrant III. Raw computation deta hai: $\arctan \left(\frac{-4}{-3}\right)=\arctan (1.333)\approx 53.1^{\circ }$ Ye up-and-right point karta hai — galat direction. Kyunki dono components negative hain (quadrant III), $180^{\circ }$ add karo: $\theta =53.1^{\circ }+180^{\circ }=233.1^{\circ }$ (Equivalently $-126.9^{\circ }$.) Figure mein fake $53.1^{\circ }$ arrow vs true $233.1^{\circ }$ arrow dikhaya gaya hai — same tan, bilkul opposite physical direction.

$p$ of $12\text{ kg}$ at $1.5\text{ m/s}$ ::: $+18\text{kg⋅m/s}$ — direct $p=mv$; positive kyunki ye $+$ direction mein point karta hai. $|\vec{p}|$ from $p_x=1.8,p_y=2.4$ ::: $3.0\text{kg⋅m/s}$ — perpendicular legs par Pythagoras, na ki $1.8+2.4$. $KE$ of $2\text{ kg}$ block with $p=4$ ::: $4\text{J}$ — $KE=p^2/2m$ use karo taaki $v$ pehle nahi nikalna pade. Direction of $(p_x,p_y)=(-3,-4)$ ::: $233.1^{\circ }$ — quadrant III ($p_x<0$), toh raw arctan mein $180^{\circ }$ add karo. Triple the speed → $p\times ?$, $KE\times ?$ ::: $\times 3$ aur $\times 9$ — $p$ mein $v$ linear hai, $KE$ mein quadratic ($v^2$).