Chapter 05: Laws of Motion
Physics XI : Complete NCERT Exercise Solutions
Practice these problems based on Newton's laws, friction, circular motion, and equilibrium. Each solution is explained step-by-step.
Question 5.1
(Take g = 10 m s⁻²) Give the magnitude and direction of the net force acting on:
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, free of electric and magnetic fields.
(a) a drop of rain falling down with a constant speed,
(b) a cork of mass 10 g floating on water,
(c) a kite skillfully held stationary in the sky,
(d) a car moving with constant velocity of 30 km/h on a rough road,
(e) a high-speed electron in space far from all material objects, free of electric and magnetic fields.
Answer & Explanation:
(a) Net force = 0 (constant velocity → zero acceleration).
(b) Net force = 0 (floating → weight balanced by buoyant force).
(c) Net force = 0 (stationary → forces balanced).
(d) Net force = 0 (constant velocity → net external force zero).
(e) Net force = 0 (no external force in space).
Physics Concept: Newton's First Law of Motion - An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
(b) Net force = 0 (floating → weight balanced by buoyant force).
(c) Net force = 0 (stationary → forces balanced).
(d) Net force = 0 (constant velocity → net external force zero).
(e) Net force = 0 (no external force in space).
Physics Concept: Newton's First Law of Motion - An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Question 5.2
A pebble of mass 0.05 kg is thrown vertically upwards. Give direction and magnitude of net force on the pebble:
(a) during upward motion,
(b) during downward motion,
(c) at highest point.
Ignore air resistance.
(a) during upward motion,
(b) during downward motion,
(c) at highest point.
Ignore air resistance.
Answer & Explanation:
Force = weight = mg = 0.05 × 10 = 0.5 N downwards in all cases.
At highest point, velocity = 0 but acceleration = g, so force is still 0.5 N downward.
Key Concept: Gravity acts constantly on the pebble regardless of its direction of motion or momentary zero velocity at the highest point.
At highest point, velocity = 0 but acceleration = g, so force is still 0.5 N downward.
Key Concept: Gravity acts constantly on the pebble regardless of its direction of motion or momentary zero velocity at the highest point.
Question 5.3
Give magnitude and direction of net force on a stone of mass 0.1 kg:
(a) just after dropping from window of stationary train,
(b) just after dropping from train running at constant velocity 36 km/h,
(c) just after dropping from train accelerating at 1 m s⁻²,
(d) lying on floor of train accelerating at 1 m s⁻², stone at rest relative to train.
(a) just after dropping from window of stationary train,
(b) just after dropping from train running at constant velocity 36 km/h,
(c) just after dropping from train accelerating at 1 m s⁻²,
(d) lying on floor of train accelerating at 1 m s⁻², stone at rest relative to train.
Answer & Explanation:
Force = weight = 0.1 × 10 = 1 N downwards in (a), (b), (c) because only gravity acts once dropped.
(d) Stone accelerates with train → net force = ma = 0.1 × 1 = 0.1 N in direction of train's acceleration.
Key Concept: In cases (a), (b), (c), the stone is in free fall after being dropped, so only gravitational force acts on it.
(d) Stone accelerates with train → net force = ma = 0.1 × 1 = 0.1 N in direction of train's acceleration.
Key Concept: In cases (a), (b), (c), the stone is in free fall after being dropped, so only gravitational force acts on it.
Question 5.4
One end of string length l connected to mass m, other to peg on smooth table. If particle moves in circle speed v, net force towards centre is:
(i) T, (ii) T – mv²/l, (iii) T + mv²/l, (iv) 0
(i) T, (ii) T – mv²/l, (iii) T + mv²/l, (iv) 0
Answer & Explanation:
(i) T
On smooth horizontal table, only tension provides centripetal force: T = mv²/l.
So net force towards centre = T.
Physics Principle: In uniform circular motion on a horizontal smooth surface, tension provides the necessary centripetal force.
On smooth horizontal table, only tension provides centripetal force: T = mv²/l.
So net force towards centre = T.
Physics Principle: In uniform circular motion on a horizontal smooth surface, tension provides the necessary centripetal force.
Question 5.5
A constant retarding force 50 N applied to body mass 20 kg moving initially at 15 m/s. How long to stop?
Answer & Explanation:
F = ma → a = F/m = –50/20 = –2.5 m/s².
v = u + at → 0 = 15 – 2.5 t → t = 6 s.
Formulas used: Newton's Second Law (F = ma) and First Equation of Motion (v = u + at).
v = u + at → 0 = 15 – 2.5 t → t = 6 s.
Formulas used: Newton's Second Law (F = ma) and First Equation of Motion (v = u + at).
Question 5.6
Constant force on mass 3 kg changes speed from 2 m/s to 3.5 m/s in 25 s. Direction unchanged. Find magnitude and direction of force.
Answer & Explanation:
a = (v – u)/t = (3.5 – 2)/25 = 0.06 m/s².
F = ma = 3 × 0.06 = 0.18 N in direction of motion.
F = ma = 3 × 0.06 = 0.18 N in direction of motion.
Formula used: F = m × a, where a = (v - u) / t
Question 5.7
Mass 5 kg acted by perpendicular forces 8 N and 6 N. Find magnitude and direction of acceleration.
Answer & Explanation:
Net force = √(8² + 6²) = 10 N.
a = F/m = 10/5 = 2 m/s².
Direction at θ = tan⁻¹(6/8) = 36.9° from 8 N force.
Method: Resultant of perpendicular forces found using Pythagorean theorem.
a = F/m = 10/5 = 2 m/s².
Direction at θ = tan⁻¹(6/8) = 36.9° from 8 N force.
Method: Resultant of perpendicular forces found using Pythagorean theorem.
Question 5.8
Three-wheeler speed 36 km/h stops in 4 s to save child. Mass of vehicle = 400 kg, driver = 65 kg. Find average retarding force.
Answer & Explanation:
u = 36 × (5/18) = 10 m/s, v = 0, t = 4 s.
a = (0 – 10)/4 = –2.5 m/s².
Total mass = 400 + 65 = 465 kg.
F = ma = 465 × (–2.5) = –1162.5 N (retarding).
a = (0 – 10)/4 = –2.5 m/s².
Total mass = 400 + 65 = 465 kg.
F = ma = 465 × (–2.5) = –1162.5 N (retarding).
Conversion: km/h to m/s → multiply by 5/18
Question 5.9
Rocket lift-off mass 20,000 kg, initial acceleration 5 m/s² upward. Calculate initial thrust.
Answer & Explanation:
Net upward force = ma = 20000 × 5 = 100000 N.
Thrust F = mg + ma = 20000×10 + 100000 = 300000 N = 3 × 10⁵ N.
Physics Concept: Thrust must overcome both gravity (mg) and provide acceleration (ma).
Thrust F = mg + ma = 20000×10 + 100000 = 300000 N = 3 × 10⁵ N.
Physics Concept: Thrust must overcome both gravity (mg) and provide acceleration (ma).
Question 5.10
Body mass 0.4 kg moves north at 10 m/s. Constant force 8 N south applied for 30 s. Predict position at t = –5 s, 25 s, 100 s.
Answer & Explanation:
a = F/m = –8/0.4 = –20 m/s² (south).
At t = –5 s: no force, so uniform motion: x = 10 × (–5) = –50 m (north of origin).
At t = 25 s: force acts for 25 s: x = 10×25 + ½×(–20)×25² = –6000 m (south).
At t = 100 s: force stops at 30 s; after 30 s, velocity = 10 + (–20)×30 = –590 m/s.
Position at 100 s: x = position at 30 s + (–590)×70 = –6000 – 41300 = –47300 m.
Physics Concept: Equations of motion: v = u + at, s = ut + ½at²
At t = –5 s: no force, so uniform motion: x = 10 × (–5) = –50 m (north of origin).
At t = 25 s: force acts for 25 s: x = 10×25 + ½×(–20)×25² = –6000 m (south).
At t = 100 s: force stops at 30 s; after 30 s, velocity = 10 + (–20)×30 = –590 m/s.
Position at 100 s: x = position at 30 s + (–590)×70 = –6000 – 41300 = –47300 m.
Physics Concept: Equations of motion: v = u + at, s = ut + ½at²
Question 5.11
Truck starts from rest, accelerates uniformly at 2 m/s². At t = 10 s, stone dropped from top (height 6 m). Find at t = 11 s:
(a) velocity, (b) acceleration of stone. Neglect air resistance.
(a) velocity, (b) acceleration of stone. Neglect air resistance.
Answer & Explanation:
At t = 10 s, truck velocity = 0 + 2×10 = 20 m/s horizontally.
Stone inherits this horizontal velocity.
(a) At t = 11 s, stone in free fall for 1 s: v_y = gt = 10 m/s down.
Horizontal v_x = 20 m/s unchanged.
Velocity magnitude = √(20² + 10²) ≈ 22.36 m/s.
(b) Acceleration = g = 10 m/s² downward.
Stone inherits this horizontal velocity.
(a) At t = 11 s, stone in free fall for 1 s: v_y = gt = 10 m/s down.
Horizontal v_x = 20 m/s unchanged.
Velocity magnitude = √(20² + 10²) ≈ 22.36 m/s.
(b) Acceleration = g = 10 m/s² downward.
Resultant velocity: v = √(v_x² + v_y²)
Question 5.12
Bob mass 0.1 kg hung by string 2 m oscillates. Speed at mean position = 1 m/s. What is trajectory if string cut when bob is:
(a) at extreme position, (b) at mean position?
(a) at extreme position, (b) at mean position?
Answer & Explanation:
(a) At extreme: velocity = 0, so bob falls vertically down.
(b) At mean: velocity horizontal = 1 m/s, so projectile motion (parabolic path).
Physics Insight: When string is cut, the bob becomes a projectile with initial velocity equal to its velocity at that instant.
(b) At mean: velocity horizontal = 1 m/s, so projectile motion (parabolic path).
Physics Insight: When string is cut, the bob becomes a projectile with initial velocity equal to its velocity at that instant.
Question 5.13
Man mass 70 kg on weighing scale in lift moving:
(a) up with uniform speed 10 m/s,
(b) down with uniform acceleration 5 m/s²,
(c) up with uniform acceleration 5 m/s²,
(d) lift falls freely under gravity.
(a) up with uniform speed 10 m/s,
(b) down with uniform acceleration 5 m/s²,
(c) up with uniform acceleration 5 m/s²,
(d) lift falls freely under gravity.
Answer & Explanation:
Scale reads normal force N.
(a) a = 0 → N = mg = 700 N.
(b) downward a = 5 m/s²: mg – N = ma → N = m(g – a) = 70×5 = 350 N.
(c) upward a = 5 m/s²: N – mg = ma → N = m(g + a) = 70×15 = 1050 N.
(d) free fall a = g → N = 0 (weightlessness).
General Formula: Apparent weight = m(g ± a) depending on direction.
(a) a = 0 → N = mg = 700 N.
(b) downward a = 5 m/s²: mg – N = ma → N = m(g – a) = 70×5 = 350 N.
(c) upward a = 5 m/s²: N – mg = ma → N = m(g + a) = 70×15 = 1050 N.
(d) free fall a = g → N = 0 (weightlessness).
General Formula: Apparent weight = m(g ± a) depending on direction.
Question 5.14
Position-time graph of mass 4 kg given. Find:
(a) force for t < 0, t > 4 s, 0 < t < 4 s.
(b) impulse at t = 0 and t = 4 s.
(a) force for t < 0, t > 4 s, 0 < t < 4 s.
(b) impulse at t = 0 and t = 4 s.
Answer & Explanation:
Graph shows x constant for t<0 and t>4 → a = 0 → F = 0.
For 0<t<4, x linear with t → constant velocity → a = 0 → F = 0.
Impulse = change in momentum. At t=0 and t=4, velocity same (from graph slope same before and after), so impulse = 0.
Key Insight: Force is related to acceleration (second derivative of position), not velocity.
For 0<t<4, x linear with t → constant velocity → a = 0 → F = 0.
Impulse = change in momentum. At t=0 and t=4, velocity same (from graph slope same before and after), so impulse = 0.
Key Insight: Force is related to acceleration (second derivative of position), not velocity.
Question 5.15
Two masses 10 kg and 20 kg on smooth surface tied by string. Horizontal force F = 600 N applied to (i) A, (ii) B along string. Find tension in string.
Answer & Explanation:
Total mass = 30 kg, acceleration a = F/m = 600/30 = 20 m/s².
(i) Force on A: T = m_B × a = 20 × 20 = 400 N.
(ii) Force on B: T = m_A × a = 10 × 20 = 200 N.
Physics Concept: When force is applied to mass A, tension must accelerate mass B. When applied to B, tension accelerates mass A.
(i) Force on A: T = m_B × a = 20 × 20 = 400 N.
(ii) Force on B: T = m_A × a = 10 × 20 = 200 N.
Physics Concept: When force is applied to mass A, tension must accelerate mass B. When applied to B, tension accelerates mass A.
Question 5.16
Two masses 8 kg and 12 kg connected over frictionless pulley. Find acceleration and tension.
Answer & Explanation:
Let heavier (12 kg) descend.
Net force = (12 – 8)g = 4g = 40 N.
Total mass = 20 kg → a = 40/20 = 2 m/s².
For 8 kg moving up: T – 80 = 8×2 → T = 96 N.
Net force = (12 – 8)g = 4g = 40 N.
Total mass = 20 kg → a = 40/20 = 2 m/s².
For 8 kg moving up: T – 80 = 8×2 → T = 96 N.
Atwood machine formula: a = (m₁ - m₂)g / (m₁ + m₂)
Question 5.17
Nucleus at rest disintegrates into two smaller nuclei. Show products move in opposite directions.
Answer & Explanation:
Initially momentum = 0.
After disintegration: m₁v₁ + m₂v₂ = 0 → v₂ = –(m₁/m₂) v₁.
So velocities opposite.
Physics Law: Conservation of Linear Momentum - Total momentum of an isolated system remains constant.
After disintegration: m₁v₁ + m₂v₂ = 0 → v₂ = –(m₁/m₂) v₁.
So velocities opposite.
Physics Law: Conservation of Linear Momentum - Total momentum of an isolated system remains constant.
Question 5.18
Two billiard balls mass 0.05 kg each, speed 6 m/s opposite directions collide and rebound same speed. Impulse on each due to other?
Answer & Explanation:
Change in momentum for one ball = m(v_f – v_i) = 0.05(6 – (–6)) = 0.05×12 = 0.6 N s.
Impulse = 0.6 N s.
Formula: Impulse = Change in momentum = m(v_f - v_i)
Impulse = 0.6 N s.
Formula: Impulse = Change in momentum = m(v_f - v_i)
Question 5.19
Shell mass 0.02 kg fired from gun mass 100 kg with muzzle speed 80 m/s. Find recoil speed of gun.
Answer & Explanation:
By conservation of momentum: 0 = 0.02×80 + 100×V → V = –1.6/100 = –0.016 m/s.
Recoil speed = 0.016 m/s opposite to shell.
Recoil speed = 0.016 m/s opposite to shell.
Conservation of momentum: m₁v₁ + m₂v₂ = 0
Question 5.20
Batsman deflects ball by 45° without changing speed 54 km/h. Mass ball = 0.15 kg. Find impulse imparted.
Answer & Explanation:
u = 54×5/18 = 15 m/s.
Change in momentum vector magnitude = m√(u² + u² – 2u²cos45°)
Impulse magnitude = 0.15×15×√(2 – √2) ≈ 3.15 N s.
Change in momentum vector magnitude = m√(u² + u² – 2u²cos45°)
Impulse magnitude = 0.15×15×√(2 – √2) ≈ 3.15 N s.
Vector subtraction: |Δp| = m√(u² + v² - 2uv cosθ)
Question 5.21
Stone mass 0.25 kg whirled radius 1.5 m at 40 rev/min horizontal plane. Find tension. Max tension 200 N, find max speed.
Answer & Explanation:
ω = 40 rev/min = (40×2π)/60 = 4.19 rad/s.
v = ωr = 4.19×1.5 = 6.285 m/s.
T = mv²/r = 0.25×(6.285²)/1.5 ≈ 6.57 N.
Max v when T_max = mv²/r → v_max = √(T_max r/m) = √(200×1.5/0.25) = √1200 ≈ 34.64 m/s.
Physics Principle: Centripetal force = Tension = mv²/r in horizontal circular motion.
v = ωr = 4.19×1.5 = 6.285 m/s.
T = mv²/r = 0.25×(6.285²)/1.5 ≈ 6.57 N.
Max v when T_max = mv²/r → v_max = √(T_max r/m) = √(200×1.5/0.25) = √1200 ≈ 34.64 m/s.
Physics Principle: Centripetal force = Tension = mv²/r in horizontal circular motion.
Question 5.22
If string breaks when speed exceeds max, stone moves:
(a) radially outwards,
(b) tangentially,
(c) at angle to tangent depending on speed.
(a) radially outwards,
(b) tangentially,
(c) at angle to tangent depending on speed.
Answer & Explanation:
(b) the stone flies off tangentially
When string breaks, centripetal force vanishes, stone continues with velocity at that instant (tangential).
Physics Insight: In circular motion, velocity is always tangential. When centripetal force is removed, the object continues in a straight line (tangent) due to inertia.
When string breaks, centripetal force vanishes, stone continues with velocity at that instant (tangential).
Physics Insight: In circular motion, velocity is always tangential. When centripetal force is removed, the object continues in a straight line (tangent) due to inertia.
Question 5.23
Explain:
(a) horse cannot pull cart in empty space,
(b) passengers thrown forward when bus stops suddenly,
(c) easier to pull lawn mower than push,
(d) cricketer moves hands back while catching.
(a) horse cannot pull cart in empty space,
(b) passengers thrown forward when bus stops suddenly,
(c) easier to pull lawn mower than push,
(d) cricketer moves hands back while catching.
Answer & Explanation:
(a) In space, no reaction force from ground (Newton's 3rd law).
(b) Inertia: body tends to continue motion (Newton's 1st law).
(c) Pulling reduces normal force, hence friction.
(d) Increases time of catch → reduces force (impulse-momentum).
Physics Principles Applied:
• Newton's Third Law (action-reaction pairs)
• Newton's First Law (inertia)
• Friction dependence on normal force
• Impulse = Force × Time
(b) Inertia: body tends to continue motion (Newton's 1st law).
(c) Pulling reduces normal force, hence friction.
(d) Increases time of catch → reduces force (impulse-momentum).
Physics Principles Applied:
• Newton's Third Law (action-reaction pairs)
• Newton's First Law (inertia)
• Friction dependence on normal force
• Impulse = Force × Time
Question 5.24
Position-time graph of mass 0.04 kg given. Suggest physical context. Find time between consecutive impulses and magnitude of each impulse.
Answer & Explanation:
Graph shows periodic impacts (like ball bouncing between walls).
Time between impulses = period from graph ≈ 4 s.
Impulse = change in momentum = mΔv. From graph, Δv = slope change → calculate.
Physics Context: Ball bouncing elastically between two walls, receiving impulses at each collision.
Time between impulses = period from graph ≈ 4 s.
Impulse = change in momentum = mΔv. From graph, Δv = slope change → calculate.
Physics Context: Ball bouncing elastically between two walls, receiving impulses at each collision.
Question 5.25
Man stationary on conveyor belt accelerating at 1 m/s². Net force on man? If μ_s = 0.2, max belt acceleration for man to remain stationary?
Answer & Explanation:
Net force on man = ma = 65×1 = 65 N forward (friction provides).
Max static friction = μ_s mg = 0.2×65×10 = 130 N.
Max a = μ_s g = 2 m/s².
Physics Principle: Static friction provides the force for acceleration, up to its maximum value μ_sN.
Max static friction = μ_s mg = 0.2×65×10 = 130 N.
Max a = μ_s g = 2 m/s².
Physics Principle: Static friction provides the force for acceleration, up to its maximum value μ_sN.
Question 5.26
Stone in vertical circle radius R. Choose correct net force at lowest and highest points directed vertically downward.
Answer & Explanation:
At lowest: mg – T₁ = mv₁²/R → net downward = mg – T₁.
At highest: mg + T₂ = mv₂²/R → net downward = mg + T₂.
So option showing these expressions is correct.
Key Concept: At lowest point, tension and weight oppose; at highest point, both act downward.
At highest: mg + T₂ = mv₂²/R → net downward = mg + T₂.
So option showing these expressions is correct.
Key Concept: At lowest point, tension and weight oppose; at highest point, both act downward.
Question 5.27
Helicopter mass 1000 kg rises with vertical acceleration 15 m/s². Crew+passengers = 300 kg. Find force on floor by crew, action of rotor on air, force on helicopter by air.
Answer & Explanation:
Force on floor by crew = m_person(g + a) = 300×(10+15) = 7500 N down.
Rotor action on air = total thrust = M_total(g + a) = 1300×25 = 32500 N down.
Force on helicopter by air = 32500 N up (reaction).
Physics Concepts: Apparent weight in accelerating frame, Newton's Third Law action-reaction pairs.
Rotor action on air = total thrust = M_total(g + a) = 1300×25 = 32500 N down.
Force on helicopter by air = 32500 N up (reaction).
Physics Concepts: Apparent weight in accelerating frame, Newton's Third Law action-reaction pairs.
Question 5.28
Water stream speed 15 m/s, area 10⁻² m² hits vertical wall. Force on wall assuming no rebound.
Answer & Explanation:
Mass flow rate = ρAv = 1000×0.01×15 = 150 kg/s.
Force = rate of change of momentum = (dm/dt) × v = 150×15 = 2250 N.
Force = rate of change of momentum = (dm/dt) × v = 150×15 = 2250 N.
Formula: F = (dm/dt) × v = ρAv²
Question 5.29
Ten coins mass m each stacked. Find force on 7th coin due to top coins, force by 8th coin, reaction of 6th on 7th.
Answer & Explanation:
(a) Force on 7th due to top 3 coins = 3mg down.
(b) Force by 8th on 7th = weight of coins above 8th = 3mg down.
(c) Reaction of 6th on 7th = force by 7th on 6th = weight above 7th = 3mg up.
Note: Count coins from top: coins 1-3 are above 7th, coin 8 is just above 7th.
(b) Force by 8th on 7th = weight of coins above 8th = 3mg down.
(c) Reaction of 6th on 7th = force by 7th on 6th = weight above 7th = 3mg up.
Note: Count coins from top: coins 1-3 are above 7th, coin 8 is just above 7th.
Question 5.30
Block mass 15 kg on rough incline (θ = 30°, μ = 0.1). Find acceleration. If μ = 0.8, what happens?
Answer & Explanation:
For θ = 30°, sinθ = 0.5, cosθ ≈ 0.866.
Net force down incline = mg sinθ – μ mg cosθ.
For μ = 0.1: a = g(sinθ – μ cosθ) = 10(0.5 – 0.1×0.866) ≈ 4.13 m/s².
For μ = 0.8: sinθ – μ cosθ = 0.5 – 0.8×0.866 = –0.193 → negative, so block won't slide (friction sufficient).
Net force down incline = mg sinθ – μ mg cosθ.
For μ = 0.1: a = g(sinθ – μ cosθ) = 10(0.5 – 0.1×0.866) ≈ 4.13 m/s².
For μ = 0.8: sinθ – μ cosθ = 0.5 – 0.8×0.866 = –0.193 → negative, so block won't slide (friction sufficient).
Condition for sliding: μ < tanθ
Question 5.31
Masses 5 kg, 3 kg, 2 kg on smooth surface. Force 20 N on 5 kg. Find tensions T₁ (between 5 & 3 kg) and T₂ (between 3 & 2 kg).
Answer & Explanation:
Total mass = 10 kg, a = F/m = 20/10 = 2 m/s².
For 2 kg mass: T₂ = m×a = 2×2 = 4 N.
For 3 kg mass: T₁ – T₂ = 3×2 → T₁ = 6 + 4 = 10 N.
Method: Consider each mass separately using F = ma.
For 2 kg mass: T₂ = m×a = 2×2 = 4 N.
For 3 kg mass: T₁ – T₂ = 3×2 → T₁ = 6 + 4 = 10 N.
Method: Consider each mass separately using F = ma.
Question 5.32
Block mass 2 kg on smooth incline (θ = 30°). Find acceleration and normal force.
Answer & Explanation:
Acceleration down incline = g sinθ = 10×0.5 = 5 m/s².
Normal force = mg cosθ = 2×10×0.866 = 17.32 N.
Normal force = mg cosθ = 2×10×0.866 = 17.32 N.
On smooth incline: a = g sinθ, N = mg cosθ
Question 5.33
Particle mass m moves in circle radius r with angular speed ω. Find centripetal force.
Answer & Explanation:
Centripetal force F = mω²r.
Also F = mv²/r where v = ωr.
Physics Formula: Centripetal force required for circular motion.
Also F = mv²/r where v = ωr.
Physics Formula: Centripetal force required for circular motion.
Question 5.34
Car mass 1500 kg moving 30 m/s stops in 10 s. Find average braking force.
Answer & Explanation:
a = (0 – 30)/10 = –3 m/s².
F = ma = 1500×(–3) = –4500 N.
Negative sign: Indicates braking/retarding force.
F = ma = 1500×(–3) = –4500 N.
Negative sign: Indicates braking/retarding force.
Question 5.35
Bullet mass 0.01 kg fired into stationary block mass 0.99 kg. If block moves 0.1 m in 0.1 s, find bullet speed.
Answer & Explanation:
Combined mass = 1 kg.
Block velocity v = s/t = 0.1/0.1 = 1 m/s.
By momentum conservation: 0.01×u = 1×1 → u = 100 m/s.
Physics Principle: Perfectly inelastic collision - momentum conserved but not kinetic energy.
Block velocity v = s/t = 0.1/0.1 = 1 m/s.
By momentum conservation: 0.01×u = 1×1 → u = 100 m/s.
Physics Principle: Perfectly inelastic collision - momentum conserved but not kinetic energy.
Question 5.36
Two skaters A (60 kg) and B (40 kg) push each other. If A moves 2 m/s, find B's velocity.
Answer & Explanation:
Initial momentum = 0.
Final: 60×2 + 40×v = 0 → v = –120/40 = –3 m/s.
B moves 3 m/s opposite to A.
Final: 60×2 + 40×v = 0 → v = –120/40 = –3 m/s.
B moves 3 m/s opposite to A.
Conservation of momentum in isolated system.
Question 5.37
Ball mass 0.2 kg falls 5 m, rebounds to 3.2 m. Find impulse on ball by floor.
Answer & Explanation:
Velocity before impact: v₁ = √(2gh₁) = √(2×10×5) = 10 m/s (down).
Velocity after impact: v₂ = √(2gh₂) = √(2×10×3.2) = 8 m/s (up).
Impulse = m(v₂ – v₁) = 0.2(8 – (–10)) = 0.2×18 = 3.6 N s upward.
Note: Take upward positive, so v₁ = –10 m/s, v₂ = +8 m/s.
Velocity after impact: v₂ = √(2gh₂) = √(2×10×3.2) = 8 m/s (up).
Impulse = m(v₂ – v₁) = 0.2(8 – (–10)) = 0.2×18 = 3.6 N s upward.
Note: Take upward positive, so v₁ = –10 m/s, v₂ = +8 m/s.
Question 5.38
Sand drops on conveyor belt at rate 0.1 kg/s. Belt moves 2 m/s. Find force to keep belt moving.
Answer & Explanation:
Force = rate of change of momentum = (dm/dt) × v = 0.1 × 2 = 0.2 N.
Physics Principle: Force needed to accelerate sand from rest to belt speed.
Physics Principle: Force needed to accelerate sand from rest to belt speed.
Question 5.39
Body mass 2 kg moves with velocity (3i + 4j) m/s. Find magnitude of momentum.
Answer & Explanation:
Velocity magnitude = √(3² + 4²) = 5 m/s.
Momentum magnitude = mv = 2×5 = 10 kg m/s.
Momentum magnitude = mv = 2×5 = 10 kg m/s.
Momentum p = mv, where v = √(v_x² + v_y²)
Question 5.40
Circular loop radius R rotates angular speed ω about vertical diameter. Bead on wire remains at lowermost for ω ≤ √(g/R). For ω = √(2g/R), find angle with vertical.
Answer & Explanation:
For bead at angle θ: N cosθ = mg, N sinθ = mω²R sinθ.
Solving gives cosθ = g/(ω²R).
For ω = √(2g/R), cosθ = g/[(2g/R)×R] = 1/2 → θ = 60°.
Solving gives cosθ = g/(ω²R).
For ω = √(2g/R), cosθ = g/[(2g/R)×R] = 1/2 → θ = 60°.
Balance equations: Vertical: N cosθ = mg, Radial: N sinθ = mω²(R sinθ)
📘 Exam Preparation Tip:
These exercise questions will help you understand the fundamental laws governing motion and forces. You'll learn to apply Newton's three laws to various situations, calculate friction forces, solve momentum problems, analyze systems of connected bodies, and determine forces in inclined planes and pulley systems. Essential for mechanics and engineering applications.