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

1.2.2 · D3 · Physics › Newton's Laws & Dynamics › Newton's second law — F = ma (net force), impulse-momentum f

Ye page parent note Newton's Second Law ka full case-map hai. Yahan koi nayi theory nahi seekhni — hum woh do formulas lenge jo tumne pehle se dekhi hain aur unhe har tarah ke possible inputs pe thokenge: forces har direction mein point karti hain, ek moment jab net force zero ho, ek sign-flip wala collision, ek limit jahan time shrink hokar almost zero ho jaata hai, ek poora word problem, aur ek nasty exam twist.

Scenario matrix

Newton's Second Law jo kuch bhi throw kar sakta hai woh in cells mein sort hota hai. Do tools hain:

Force form: $F_{net} = m a$ (mass constant)
Impulse form: $J = F Δ t = Δ p = m v_{f} - m v_{i}$

#	Cell class	Isme kya unusual hai	Example jo ise hit karta hai
A	Saari forces ek hi direction mein (simple sum)	sirf magnitudes add karo	Ex 1
B	Opposing forces, positive net	subtraction, net ka sign	Ex 2
C	Zero net force (degenerate)	$a = 0$ par forces exist karti hain	Ex 3
D	Two-dimensional forces (sabhi quadrants)	vectors, sirf $\pm$ nahi	Ex 4 (figure)
E	Sign-flip collision (velocity reverse ho jaati hai)	$Δ p$ tumhari guess se bada hoga	Ex 5
F	Same $Δ p$ , time ko stretch karo (limiting behaviour)	force $\to \infty$ as $Δ t \to 0$	Ex 6 (figure)
G	Real-world word problem	tumhe numbers khud build karne honge	Ex 7
H	Exam twist: variable mass	$F = ma$ jhooth bolta hai; use karo $\frac{d p}{d t}$	Ex 8

Neeche, har worked example ko us cell ke saath tag kiya gaya hai jise woh fill karta hai.

Building blocks (taaki koi bhi symbol unearned na ho)

Worked Examples

Worked example Ex 3 — Cell C: zero net force (degenerate)

Ek car seedhi road pe constant $25 m/s$ se cruise karti hai. Engine use $600 N$ se aage drive karta hai. Total resistive force kya hai, aur $a$ kya hai?

Forecast: "Yeh move kar rahi hai, toh zaroor forward net force hoga," right? Dekho yeh trap kaise spring karta hai.

Constant velocity ka matlab hai $a = 0$ . Yeh step kyun? Acceleration velocity ka rate of change hai; unchanging velocity ⇒ rate $= 0$ .
Toh $F_{net} = ma = m \cdot 0 = 0$ kisi bhi mass $m$ ke liye. Yeh step kyun? $F = ma$ mein $a = 0$ daalne se net zero ho jaata hai — mass chahe kuch bhi ho, kyunki kisi bhi cheez ko zero se multiply karo toh zero aata hai.
Toh resistance drive ko balance karta hai: resistive force $= 600 N$ backward. Yeh step kyun? Agar net zero hai aur drive $+ 600$ hai, toh doosri force $- 600$ honi chahiye.

Verify: $600 + (- 600) = 0 = F_{net}$ . ✓ Yeh First Law hai jo Second ke andar zinda hai: zero net force ⇒ velocity frozen. Speed irrelevant hai.

Worked example Ex 4 — Cell D: two-dimensional forces ek

different quadrant mein Frictionless ice pe ek $5 kg$ ke puck ko ek saath do forces feel hoti hain: $F_{x} = - 30 N$ (yaani west) aur $F_{y} = + 40 N$ (north). $2 s$ mein deliver hua impulse, acceleration ki magnitude, aur uski direction find karo. (Yeh quadrant II mein land karta hai: negative $x$ , positive $y$ — toh humein $arctan$ sign trap se bachna hoga.)

Forecast: Answer $\frac{- 30 + 40}{5} = 2$ nahi hai. Right angles pe forces sirf add nahi hoti, aur west-north pull up-and-left point karta hai, up-and-right nahi. Figure dekho.

Do forces ko tip-to-tail draw karo (figure): ek west arrow, phir ek north arrow. Unka sum diagonal (amber) hai jo upper-left mein jaata hai. Yeh step kyun? Forces vectors hain; net tip-to-tail resultant hai, jiska length right triangle se aata hai.
Net force ki magnitude Pythagoras se: $F_{net} = (- 30)^{2} + 4 0^{2} = 900 + 1600 = 2500 = 50 N$ . Yeh step kyun? Squaring sign ko khatam kar deta hai, toh leg lengths hi hypotenuse set karte hain; resultant magnitude wahi $50 N$ hai jaisi hoti agar dono positive hote.
$2 s$ mein Impulse: $J = F_{net} Δ t$ , toh $J_{x} = (- 30) (2) = - 60 N\cdots$ , $J_{y} = (40) (2) = + 80 N\cdots$ , aur $∣ J ∣ = (- 60)^{2} + 8 0^{2} = 3600 + 6400 = 10000 = 100 N\cdots$ . Yeh step kyun? Yeh impulse form $J = F Δ t$ hai header se, real mein use kiya gaya: $2 s$ ke liye constant force puck mein $100 N\cdots$ ka momentum change stamp karta hai.
Acceleration magnitude $a = \frac{F _{net}}{m} = \frac{50}{5} = 10 m/s^{2}$ . Yeh step kyun? $F = ma$ net vector ki magnitude pe kaam karta hai.
Direction — quadrant fix: raw $arctan \frac{F _{y}}{F _{x}} = arctan \frac{40}{- 30} \approx - 53.1 3^{\circ}$ hai, jo quadrant IV mein point karta hai — galat, kyunki $tan$ har $18 0^{\circ}$ pe repeat hota hai aur quadrant II aur IV mein fark nahi bata sakta. Kyunki $F_{x} < 0, F_{y} > 0$ hai toh hum sach mein quadrant II mein hain, toh $18 0^{\circ}$ add karo: $θ = - 53.1 3^{\circ} + 18 0^{\circ} = 126.8 7^{\circ}$ east se counter-clockwise measure kiya gaya. Yeh step kyun? $arctan$ kabhi bhi $- 9 0^{\circ}$ aur $+ 9 0^{\circ}$ ke beech se bahar answer nahi deta; west-pointing $x$ -component ko $+ 18 0^{\circ}$ correction chahiye taaki angle sahi (upper-left) quadrant mein land kare.

Verify: Acceleration usi direction mein point karta hai jisme net force hai (dono sirf scalar $1/ m$ se differ karte hain), toh $a$ bhi $126.8 7^{\circ}$ pe hai — upar aur left mein, bilkul amber arrow se match karta hai. Component check: $a_{x} = - 30/5 = - 6$ , $a_{y} = 40/5 = 8$ , aur $(- 6)^{2} + 8^{2} = 36 + 64 = 100 = 10$ ✓. Impulse cross-check: $∣ J ∣ = m a Δ t = 5 \cdot 10 \cdot 2 = 100 N\cdots$ ✓.

Worked example Ex 5 — Cell E: sign-flip collision

Ek $0.15 kg$ ki ball wall se $20 m/s$ pe takraati hai aur $0.01 s$ mein opposite direction mein $30 m/s$ se rebound karti hai. Impulse aur average force find karo.

Forecast: Dhyan se — kya speed change $30 - 20 = 10$ hai, ya kuch bada? Sign decide karta hai.

"+" = ball ki incoming direction choose karo. Toh $v_{i} = + 20$ , $v_{f} = - 30$ (ab yeh doosri taraf jaati hai). Yeh step kyun? Reversal mein zaroor minus sign hoga, warna physics galat hai.
Impulse = change in momentum $J = Δ p = m (v_{f} - v_{i}) = 0.15 (- 30 - 20) = 0.15 (- 50) = - 7.5 N\cdots$ . Yeh step kyun? Impulse–momentum: $J = Δ p = m v_{f} - m v_{i}$ . $- 50$ (na ki $- 10$ ) is cell ka poora point hai.
Average force $J = F Δ t$ se: $F = \frac{J}{Δ t} = \frac{- 7.5}{0.01} = - 750 N$ . Yeh step kyun? Constant force ke liye, impulse sirf $F Δ t$ hai, toh $F = J /Δ t$ .

Verify: Minus sign kehta hai force ball ke approach ke against point karti hai — bilkul wahi jo wall karta hai. Magnitude $750 N$ . Agar tumne galti se $Δ v = 10$ use kiya hota, toh $J = - 1.5$ aur $F = - 150 N$ aata — paanch guna chhota. Rebound speed kahan se aata hai uske liye Collisions and Elasticity dekho. ✓

Worked example Ex 6 — Cell F: same

$Δ p$ , time ko stretch karo (limiting behaviour) Ek $70 kg$ ka driver $15 m/s$ pe ruk jaana chahta hai. Hard stop ( $0.02 s$ ) versus airbag stop ( $0.20 s$ ) ke force compare karo. Phir poochho: $Δ t \to 0$ hone par kya hota hai?

Forecast: Dono stops same momentum khatam karte hain. Compute karne se pehle do forces ka ratio guess karo.

Impulse needed $J = Δ p = m Δ v = 70 \times 15 = 1050 N\cdots$ . Yeh step kyun? Insaan $15$ se $0$ jaata hai, toh seatbelt/airbag ko jo impulse deliver karna hai woh $∣Δ p ∣ = m v_{i}$ hai.
Hard dashboard: $F = \frac{J}{Δ t} = \frac{1050}{0.02} = 52, 500 N$ . Yeh step kyun? $F = J /Δ t$ chhote time ke saath.
Airbag: $F = \frac{J}{Δ t} = \frac{1050}{0.20} = 5, 250 N$ . Yeh step kyun? Same impulse, das guna time.
Limiting behaviour: jab $Δ t \to 0$ , $F = \frac{J}{Δ t} \to \infty$ . Yeh step kyun? Ek finite impulse jo zero time mein deliver ho uske liye infinite force chahiye — yahi mathematical reason hai ki instantaneous collisions itni violent kyun hoti hain. Figure ka curve $F = 1050/Δ t$ upar shoot karta hai jab $Δ t$ shrink hota hai (red asymptote).

Verify: Ratio $\frac{52500}{5250} = 10$ — bilkul times ka ratio ( $0.20/0.02 = 10$ ), confirm karta hai "10× time ⇒ 1/10 force". ✓ Units: $\frac{N\cdots}{s} = N$ ✓.

Worked example Ex 7 — Cell G: real-world word problem

Ek $1200 kg$ ki car level road pe $9 s$ mein $0$ se $27 m/s$ (lagbhag $97 km/h$ ) tak speed up karti hai. (a) Iske liye kaun sa constant net force chahiye? (b) Agar air+rolling drag $400 N$ hai, toh tyres ko kaun sa forward drive force supply karna hoga?

Forecast: Do forces kaam karenge — net jo car ko accelerate karta hai, aur extra jo drag overcome karne ke liye chahiye. Guess karo kaun sa bada hai.

Acceleration $a = \frac{Δ v}{Δ t} = \frac{27 - 0}{9} = 3 m/s^{2}$ . Yeh step kyun? Constant acceleration = velocity mein change divided by jo time laga.
(a) Net force $F_{net} = ma = 1200 \times 3 = 3600 N$ . Yeh step kyun? Yeh woh leftover force hai jo actually mass ko accelerate karta hai.
(b) Drive force ko dono kaam karne hain — accelerate bhi karo aur drag ko bhi harao: $F_{drive} = F_{net} + F_{drag} = 3600 + 400 = 4000 N$ . Yeh step kyun? Net = drive − drag, toh drive = net + drag. Drag term ke liye Friction dekho.

Verify: Drive se net re-compute karke check karo: $4000 - 400 = 3600 = ma$ ✓. Impulse cross-check: $J = F_{net} Δ t = 3600 \times 9 = 32400 N\cdots$ , aur $Δ p = m Δ v = 1200 \times 27 = 32400 kg\cdotm/s$ — equal, jaisa impulse–momentum demand karta hai. ✓

Worked example Ex 8 — Cell H: exam twist jahan

$F = ma$ jhooth bolta hai (variable mass) Mass $M = 500 kg$ ka ek open railcar $v = 4 m/s$ se roll karta hai. Baarish seedha neeche girti hai aur $μ = \frac{d m}{d t} = 5 kg/s$ ki rate se collect hoti hai. Koi horizontal force nahi hai. $10 s$ baad car ki speed find karo.

Forecast: Koi horizontal force nahi ⇒ " $a = 0$ ⇒ speed $4$ rehti hai"? Yeh trap hai. Dekho.

Horizontal Newton: $F_{net} = \frac{d ( m v )}{d t} = 0$ , kyunki kuch bhi horizontally push nahi karta. Yeh step kyun? Momentum form kabhi fail nahi hota, chahe mass change ho. Baarish mass add karti hai lekin koi horizontal momentum nahi (yeh seedha neeche girti hai).
Conserved horizontal momentum: $m v = constant$ , toh $m_{i} v_{i} = m_{f} v_{f}$ . Yeh step kyun? Agar $m v$ ka time-derivative zero hai, toh $m v$ kabhi nahi badalta — yeh Conservation of Linear Momentum hai.
New mass $m_{f} = M + μ t = 500 + 5 (10) = 550 kg$ . Yeh step kyun? Mass $5 kg/s$ se linearly badhti hai.
Solve for $v_{f} = \frac{m _{i} v _{i}}{m _{f}} = \frac{500 \times 4}{550} = \frac{2000}{550} \approx 3.636 m/s$ . Yeh step kyun? Jab $m$ badhta hai aur $m v$ fixed rehta hai, toh $v$ zaroor girega — naive "constant speed" answer ka bilkul ulta.

Verify: Momentum pehle $= 500 \times 4 = 2000$ ; baad mein $= 550 \times 3.636 = 2000$ ✓. $v \frac{d m}{d t}$ term jo parent ne warn kiya tha, wahi car ko slow karta hai — Variable Mass Systems & Rocket Equation dekho. ✓

Recall Poore matrix pe quick self-test

Har phrase kaunse cell mein belong karti hai? Constant velocity lekin forces present hain ::: Cell C — zero net force, $a = 0$ . Ball wall pe direction reverse karti hai ::: Cell E — sign-flip, $Δ p = m (v_{f} - v_{i})$ jahan $v_{f}$ negative hai. Do forces $9 0^{\circ}$ pe, west component ke saath ::: Cell D — Pythagorean net magnitude, $arctan$ plus $18 0^{\circ}$ quadrant II ke liye. Airbag versus dashboard ::: Cell F — same $Δ p$ , bada $Δ t$ $F$ ko shrink karta hai. Rain-filling railcar ::: Cell H — variable mass, use karo $\frac{d ( m v )}{d t}$ , $ma$ nahi.

Connections

Parent: Newton's Second Law — yeh page uska case-map hai.
Newton's First Law (Inertia) — Cell C, Second ke andar First Law hai.
Conservation of Linear Momentum — Ex 8 ke peeche ka engine.
Collisions and Elasticity — jahan rebound speeds (Ex 5) aate hain.
Friction — Ex 2, 3, 7 mein resistive term.
Variable Mass Systems & Rocket Equation — Ex 8 ka full generalisation.
Newton's Third Law — Ex 5 mein wall ki reaction force supply karta hai.