Exercises — Newton's second law — F = ma (net force), impulse-momentum form

KYA chahiye: known net force aur mass se acceleration $a$. KYUN $F=ma$: mass constant hai (ek solid crate), isliye simple child-form laagoo hoti hai. $a=\frac{F_{\text{net}}}{m}=\frac{20}{5}=4{\text{m/s}}^2\text{east}.$ $\vec{a}$ ki direction hamesha ${\vec{F}}_{\text{net}}$ ki direction se match karti hai.

Momentum bas mass times velocity hai: $p=mv=0.5\times 8=4\text{kg⋅m/s}.$ Units check: $\text{kg}\times \text{m/s}=\text{kg⋅m/s}$. Direction velocity ke saath same hai.

Constant force ke liye time-integral ek product mein collapse ho jaata hai: $J=F\Delta t=6\times 3=18\text{N⋅s}.$ Aur $\text{N⋅s}=\text{kg⋅m/s}$ — momentum ke saath same units, kyunki impulse hai hi momentum mein change.

Step 1 — net force. Right = positive choose karo. Forces ko signs ke saath add karo: $F_{\text{net}}=+18-6=+12\text{N (rightward)}.$ Signs kyun? Forces vectors hain; ek line par ek sign hi poori direction hai. Step 2 — mass se divide karo: $a=\frac{F_{\text{net}}}{m}=\frac{12}{3}=4{\text{m/s}}^2\text{right}.$

Step 1 — signs assign karo. Incoming direction ko positive mano: $v_i=+15$, $v_f=-12$ (ab yeh doosri taraf ja rahi hai). Step 2 — momentum mein change: $\Delta p=m(v_f-v_i)=0.20(-12-15)=0.20\times (-27)=-5.4\text{kg⋅m/s}.$ Step 3 — average force $J=F\Delta t=\Delta p$ se: $F=\frac{\Delta p}{\Delta t}=\frac{-5.4}{0.05}=-108\text{N}.$ Minus sign matlab force ball ki original motion ke opposite point karti hai (wall se door). Magnitude: $108\text{N}$.

Step 1 — forces ko vectors ki tarah add karo. Yeh perpendicular hain, isliye yeh ek right triangle ki do legs banate hain (figure dekho). Net force hypotenuse hai: $|{\vec{F}}_{\text{net}}|=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\text{N}.$ Pythagoras kyun? Perpendicular vectors tip-to-tail add hokar ek right triangle banate hain; diagonal ki length $\sqrt{a^2+b^2}$ hoti hai. Step 2 — mass se divide karo: $a=\frac{|{\vec{F}}_{\text{net}}|}{m}=\frac{10}{2}=5{\text{m/s}}^2,$ same $6\text{-}8\text{-}10$ diagonal ki taraf point karta hai (roughly $53°$ north of east).

Momentum change fixed hai — isse koi fark nahi ki tum kaise rokte ho: $\Delta p=m(v_f-v_i)=60(0-20)=-1200\text{kg⋅m/s},|\Delta p|=1200.$ (a) $F=\frac{1200}{0.10}=12,000\text{N}.$ (b) $F=\frac{1200}{0.50}=2,400\text{N}.$ Ratio: $12000/2400=5$. Time ko $5\times$ stretch karne se force $5\times$ shrink hoti hai. Yahi to crumple zones ka poora point hai.

Area kyun? Impulse hai $\int Fdt$ — time ke saath accumulated force — jo exactly hai $F$-vs-$t$ curve ke neeche ka area. Graph ek triangle hai jiska base $4\text{s}$ aur height $8\text{N}$ hai (figure dekho): $J=\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(4)(8)=16\text{N⋅s}.$ Final speed $J=\Delta p=m{v}_f-m{v}_i$ se jab $v_i=0$: $v_f=\frac{J}{m}=\frac{16}{2}=8\text{m/s}.$

Phase 1 (0–3 s): impulse $J_1=F\Delta t=10\times 3=30\text{N⋅s}$, isliye $v_3=\frac{J_1}{m}=\frac{30}{4}=7.5\text{m/s}.$ Phase 2 (3–5 s): koi force nahi ⇒ koi impulse nahi ⇒ momentum unchanged. Newton's First Law (Inertia) ke mutabiq object constant velocity par coast karta hai. $v_5=7.5\text{m/s}.$

$F=ma$ kyun nahi? Mass change ho raha hai, isliye humein momentum form use karni chahiye. Horizontally koi external force nahi hai, isliye horizontal momentum conserved hai (yeh hai Conservation of Linear Momentum): $m_i{v}_i=m_f{v}_f.$ $200\times 6=(200+40)v_f\Rightarrow 1200=240v_f\Rightarrow v_f=5\text{m/s}.$ Car slow ho jaati hai even though kuch bhi ise peeche nahi dhakel raha — added mass same fixed momentum share karta hai. Ek naive "$a=0\Rightarrow v$ constant" deta $6\text{m/s}$ — jo product-rule term $v\frac{dm}{dt}$ ki wajah se galat hai.

Target impulse: $\Delta p=m\Delta v=1.5\times 12=18\text{N⋅s}$ (rest se). Stage 1 impulse: $J_1=9\times 1=9\text{N⋅s}$. Stage 2 ke liye remaining impulse: $J_2=18-9=9\text{N⋅s}$. Stage 2 time: $t_2=\frac{J_2}{F_2}=\frac{9}{3}=3\text{s}.$ Note karo: dono stages mein same $9\text{N⋅s}$, lekin kamzor force ko $3\times$ zyada time chahiye — "chhoti force, lamba time" utna hi momentum khareedti hai jitna "badi force, chhota time."

Gases bullet ko aage dhakelti hain aur bullet-gas pair rifle ko equal-and-opposite force se peeche dhakelti hai (Newton's Third Law), same contact time ke liye — isliye impulses equal aur opposite hain, aur total momentum zero rehta hai: $0=m_b{v}_b+m_r{v}_r.$ $0=(0.01)(400)+(2)v_r\Rightarrow v_r=-\frac{4}{2}=-2\text{m/s}.$ Minus sign = backward. Rifle $2\text{m/s}$ par recoil karti hai.

Vectors kyun matter karte hain: speed unchanged hai ($10\to 10$), phir bhi momentum vector $90°$ rotate hua, isliye $\Delta \vec{p}\ne 0$. Component by component karo. East = $+x$, north = $+y$ mano. ${\vec{p}}_i=(m\cdot 10,0)=(3,0)\text{kg⋅m/s},{\vec{p}}_f=(0,m\cdot 10)=(0,3).$ $\Delta \vec{p}={\vec{p}}_f-{\vec{p}}_i=(0-3,3-0)=(-3,+3).$ Magnitude: $|\vec{J}|=|\Delta \vec{p}|=\sqrt{(-3{)}^2+3^2}=\sqrt{18}\approx 4.24\text{N⋅s}.$ Impulse north-west point karta hai (figure mein upar-aur-baayein) — na incoming path ke saath, na outgoing path ke saath.

(a) Impulse = force ka time par integral. Kyunki $F$ constant nahi hai, humein integrate karna hoga ($F=12t$ ke neeche area, jo $t=2$ par height $24$ wala triangle hai): $J={\int }_0^{2}12tdt={\left[6t^2\right]}_0^2=6(4)-0=24\text{N⋅s}.$ (b) $\Delta v=\frac{J}{m}=\frac{24}{3}=8\text{m/s}.$ (c) Average force = impulse divided by the time it acted: $\bar{F}=\frac{J}{\Delta t}=\frac{24}{2}=12\text{N}.$ Sanity check: $F$ linearly $0$ se $24\text{N}$ tak gaya, isliye uska average hai $\frac{0+24}{2}=12\text{N}$. ✓

(a) Work–energy (Work-Energy Theorem): force ka space-integral kinetic energy mein change ke barabar hai. $W=Fd=5\times 4=20\text{J}=\frac{1}{2}m{v}_f^2-0.$ $v_f=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2\times 20}{2}}=\sqrt{20}\approx 4.47\text{m/s}.$ (b) Impulse = momentum mein change: $J=m{v}_f-0=2\times 4.472\approx 8.94\text{N⋅s}.$ Contact time $J=F\Delta t$ se: $\Delta t=\frac{J}{F}=\frac{8.944}{5}\approx 1.79\text{s}.$ Sabak: impulse force ko time par integrate karta hai; work force ko distance par integrate karta hai. Same force, do alag accumulations — ek deta hai momentum, doosra energy.