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
-
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-I -
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 momentum -
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: Impulse -
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-I -
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-energy -
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-III -
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-II -
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: Impulse -
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 momentum -
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-III -
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-II -
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-fall -
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 collision -
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-I -
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 collision
Speed 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. |