Motion Laws
Key Concepts
| # | Concept | Explanation |
|---|---|---|
| 1 | Newton’s 1st Law (Inertia) | A body stays at rest or in uniform motion unless an external force acts. |
| 2 | Newton’s 2nd Law | F = ma; acceleration ∝ net force, inversely ∝ mass. |
| 3 | Newton’s 3rd Law | Every action has an equal & opposite reaction acting on different bodies. |
| 4 | Momentum (p) | p = mv; vector quantity; conserved in isolated system. |
| 5 | Impulse (J) | J = FΔt = Δp; large force acting for short time. |
| 6 | Conservation of Momentum | Total momentum before collision = total momentum after collision (no external force). |
| 7 | Recoil Velocity | Gun-bullet system: 0 = mgvg + mbvb ⇒ vg = –(mb/mg)vb. |
| 8 | Friction Types | Static > limiting > kinetic; always opposes relative motion. |
15 Practice MCQs
1. A 2 kg stone & 10 kg bench are both at rest. Which needs more force to start motion?
A. Stone B. Bench C. Same D. Depends on surface **Answer:** B **Solution:** F = ma; same a needs F ∝ m. **Shortcut:** Heavier ⇒ more inertia ⇒ more force. **Tag:** Newton-I2. A 0.05 kg bullet leaves a 5 kg rifle at 400 m/s. Recoil speed of rifle is nearly
A. 4 m/s B. 0.4 m/s C. 40 m/s D. 0.04 m/s **Answer:** A **Solution:** 0 = 5v + 0.05×400 ⇒ v = –4 m/s (magnitude 4). **Shortcut:** vrecoil = (mbullet/mgun) vbullet. **Tag:** Conservation of momentum3. A 500 g ball changes velocity from 10 m/s East to 15 m/s West in 0.01 s. Average force is
A. 250 N B. 500 N C. 1250 N D. 2500 N **Answer:** C **Solution:** Δv = –15 –10 = –25 m/s; Δp = 0.5×(–25)= –12.5 kg·m/s; F = Δp/Δt = –12.5/0.01 = –1250 N (magnitude 1250 N). **Shortcut:** F = 2mv/Δt when direction reverses. **Tag:** Impulse4. When a fast-moving car stops suddenly, passengers lurch forward due to
A. Inertia of motion B. Inertia of rest C. Friction D. Gravity **Answer:** A **Solution:** Body tends to keep its state of motion. **Shortcut:** “Lurch forward” ⇒ motion inertia. **Tag:** Newton-I5. A cricket ball is pushed 2 m by a恒定 20 N force. Work done is
A. 10 J B. 20 J C. 40 J D. 400 J **Answer:** C **Solution:** W = Fs = 20×2 = 40 J. **Shortcut:** W = F × distance (when force ∥ displacement). **Tag:** Work-energy6. Action-reaction forces
A. Cancel each other on same body B. Act on different bodies C. Are always perpendicular D. Can be unequal **Answer:** B **Solution:** Newton-III pairs act on two interacting bodies. **Shortcut:** “Different bodies” ⇒ never cancel. **Tag:** Newton-III7. A 4 kg object accelerates at 3 m/s². Net force is
A. 7 N B. 12 N C. 1.33 N D. 0.75 N **Answer:** B **Solution:** F = ma = 4×3 = 12 N. **Shortcut:** F = ma (direct multiplication). **Tag:** Newton-II8. A 20 g bullet at 300 m/s stops in 0.1 s on hitting wall. Average resistance force is
A. 30 N B. 60 N C. 300 N D. 600 N **Answer:** B **Solution:** Δp = 0.02×300 = 6 kg·m/s; F = 6/0.1 = 60 N. **Shortcut:** F = mv/t (grams→kg!). **Tag:** Impulse9. Two ice skaters push each other; 60 kg moves 2 m/s left, 40 kg moves right at
A. 2 m/s B. 3 m/s C. 4 m/s D. 5 m/s **Answer:** B **Solution:** 0 = 60×2 – 40v ⇒ v = 3 m/s. **Shortcut:** v2 = (m1/m2)v1. **Tag:** Conservation of momentum10. A horse pulls a cart; the cart pulls horse with equal force. Why does system move?
A. Forces cancel B. Forces act on different bodies C. Frictionless ground D. Horse heavier **Answer:** B **Solution:** Forward force on cart > ground friction on cart. **Shortcut:** Motion because pair acts on different bodies. **Tag:** Newton-III11. A 1000 kg car decelerates from 20 m/s to rest in 4 s. Braking force is
A. 2000 N B. 4000 N C. 5000 N D. 8000 N **Answer:** C **Solution:** a = (0–20)/4 = –5 m/s²; F = 1000×5 = 5000 N. **Shortcut:** F = m(Δv/t). **Tag:** Newton-II12. A ball thrown upward has at top point
A. Zero velocity, zero acceleration B. Zero velocity, acceleration g downward C. Non-zero velocity, zero acceleration D. Non-zero velocity, non-zero acceleration **Answer:** B **Solution:** v = 0, a = g = 9.8 m/s² downward. **Shortcut:** At peak, v = 0 but gravity still pulls. **Tag:** Free-fall13. A 2 kg & 3 kg masses collide & stick. Initial momenta 6 kg·m/s & 4 kg·m/s same direction. Final speed is
A. 1 m/s B. 2 m/s C. 3 m/s D. 4 m/s **Answer:** B **Solution:** ptotal = 6 + 4 = 10 kg·m/s; mtotal = 5 kg; v = 10/5 = 2 m/s. **Shortcut:** v = (p1+p2)/(m1+m2). **Tag:** Inelastic collision14. Sudden removal of table-cloth without disturbing dishes demonstrates
A. Inertia of rest B. Inertia of motion C. Gravity D. Friction **Answer:** A **Solution:** Dishes tend to stay at rest. **Shortcut:** “Dishes don’t move” ⇒ rest inertia. **Tag:** Newton-I15. A 50 g bullet at 200 m/s embeds in 950 g stationary block. Velocity of combined mass is
A. 5 m/s B. 10 m/s C. 20 m/s D. 40 m/s **Answer:** B **Solution:** 0.05×200 = (0.95+0.05)v ⇒ v = 10/1 = 10 m/s. **Shortcut:** v = (mbullet/mtotal) vbullet. **Tag:** Inelastic collisionSpeed Tricks
| Situation | Shortcut | Example |
|---|---|---|
| Recoil speed | vgun = (mbullet/mgun) vbullet | 10 g bullet, 5 kg gun, 400 m/s ⇒ vgun = 0.8 m/s |
| Force from momentum change | F = Δp /Δt; if direction reverses Δp = 2mv | Ball rebounds 0.1 kg, 20 m/s, 0.01 s ⇒ F = 400 N |
| Collision velocity (stick) | vfinal = (m1v1+m2v2)/(m1+m2) | 2 kg at 3 m/s + 3 kg at 2 m/s ⇒ v = 2.4 m/s |
| Deceleration to stop | a = v²/(2s); F = ma | Car 20 m/s stops in 40 m ⇒ a = 5 m/s² |
| Inertia quick test | Heavier object ⇒ more force to change state | 10 kg vs 1 kg: need 10× force for same a |
Quick Revision
| Point | Detail |
|---|---|
| 1 | Newton-I: Inertia ∝ mass. |
| 2 | Newton-II: Vector form F⃗ = ma⃗; direction of a same as Fnet. |
| 3 | Newton-III: Forces always come in pairs on different bodies. |
| 4 | Momentum conserved only if no external force. |
| 5 | Impulse = area under F-t graph. |
| 6 | Recoil problems: initial momentum = 0. |
| 7 | In perfectly inelastic collision, bodies stick; KE not conserved. |
| 8 | Static friction ≤ μsN; kinetic friction = μkN (μs > μk). |
| 9 | For same force, lighter body gets bigger acceleration. |
| 10 | Always convert grams → kg & cm → m before substituting in formulas. |
BETA
