Dynamics Chapter 3 - Complete 100-Question Answer-Sheet Review

Class 9 Dynamics: Complete 100-Question Answer-Sheet Review

Student grading · corrected model answers · formulas · explanations · worked reasoning

How to use this page: Attempt each question first, then open the model answer. Green = correct, amber = partially correct, red = incorrect or unattempted. A partial grade means the core idea was present but a definition, direction, calculation, derivation or explanation was incomplete.
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Correct
24
Partially correct
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Incorrect
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Unattempted

Jump to: Test Your Concepts 1-50 · Concept Application 1-50 · Revision priorities

Part 1: Test Your Concepts - Questions 1-50

Question 1: State the law of conservation of momentum.

Status: Correct
Show model answer and explanation
If no net external force acts on a system, its total momentum remains constant: Σp before = Σp after. For two bodies, m1u1 + m2u2 = m1v1 + m2v2.

Question 2: Give two examples of renewable sources of energy.

Status: Correct
Show model answer and explanation
Solar, wind, tidal, hydro, geothermal and biogas are renewable. Any two, such as wind and tidal energy, are sufficient.

Question 3: If no external _____ acts, the total momentum of colliding bodies is conserved.

Status: Correct
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The blank is force. Momentum conservation applies to an isolated system whose net external force is zero.

Question 4: What is normal reaction or normal force?

Status: Partially correct
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The normal reaction is the contact force exerted by a surface perpendicular to itself. Saying only “a perpendicular force” misses which surface exerts it and on what body.

Question 5: Give two examples of non-renewable sources of energy.

Status: Correct
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Coal and petroleum are correct examples. Natural gas and nuclear fuels are also non-renewable.

Question 6: Why does a person getting down from a fast-moving bus fall forward?

Status: Correct
Show model answer and explanation
Because of inertia of motion (Newton’s first law): the feet slow on touching the ground, while the upper body tends to keep moving forward.

Question 7: Define a force field.

Status: Partially correct
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A force field is the region around a body in which another suitable body experiences a non-contact force, for example a gravitational, electric or magnetic field.

Question 8: Define viscous force.

Status: Correct
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Viscous force is the frictional resistance between adjacent layers of a fluid, or between a body and the fluid through which it moves. It opposes relative motion.

Question 9: State Hooke’s law.

Status: Correct
Show model answer and explanation
Within the elastic limit, stress is directly proportional to strain. For a spring this is F = kx in magnitude; the restoring force is F = -kx.

Question 10: Name the force responsible for circular motion.

Status: Correct
Show model answer and explanation
Centripetal force, directed toward the centre, keeps a body moving in a circular path: Fc = mv²/r.

Question 11: Define the SI unit of work.

Status: Partially correct
Show model answer and explanation
One joule is the work done when a force of 1 N displaces a body by 1 m in the force’s direction: 1 J = 1 N m. The student had the main idea but the wording was unclear.

Question 12: A change in state of rest or uniform motion is produced by _____.

Status: Correct
Show model answer and explanation
An unbalanced or net external force. A balanced force cannot change velocity.

Question 13: Name the apparent outward force in circular motion.

Status: Correct
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Centrifugal force. It is a pseudo-force observed in a rotating frame and points away from the centre.

Question 14: What are lubricants?

Status: Partially correct
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Lubricants are substances placed between moving surfaces to reduce friction, heating and wear. “Decrease friction” states the purpose but not the definition.

Question 15: Convert 1 newton to dyne.

Status: Unattempted
Show model answer and explanation
Key concept behind this hard question: The numerical value of a force changes when the base units change. Convert kilograms to grams and metres to centimetres; seconds already match.
Given: 1 N = 1 kg m s⁻²
Choose the idea: Unit conversion using 1 kg = 10³ g and 1 m = 10² cm.
  1. Replace kg: 1 kg = 10³ g.
  2. Replace m: 1 m = 10² cm.
  3. Therefore 1 N = (10³ g)(10² cm)s⁻² = 10⁵ g cm s⁻².
  4. By definition, 1 dyne = 1 g cm s⁻².
Final answer: 1 N = 10⁵ dyne.

Reasonableness check: A newton is much larger than a dyne because kilograms and metres are larger units than grams and centimetres.

Question 16: Can all bodies have equal momentum?

Status: Incorrect
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No. Momentum p = mv depends on both mass and velocity. Different bodies can have equal momentum only when their mass-velocity products are equal.

Question 17: What type of energy does a flying bird possess?

Status: Correct
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It has kinetic energy because it is moving and gravitational potential energy because it is above the ground.

Question 18: Friction always acts _____ to the surfaces in contact.

Status: Correct
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Tangential (parallel) to the contacting surfaces and opposite the relative motion or tendency of motion.

Question 19: What is meant by unbalanced force?

Status: Correct
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Forces are unbalanced when their vector sum is non-zero. The resulting net force produces acceleration.

Question 20: When is work maximum for a given force and displacement?

Status: Incorrect
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Key concept: Only the component of force parallel to the displacement does work.

Suppose an object moves horizontally through displacement s, while a force F acts at an angle θ above the horizontal. Resolve the force into two perpendicular components:

Horizontal component: Fx = F cosθ
Vertical component: Fy = F sinθ

Why is cosine horizontal and sine vertical? In the right-angled component triangle, F is the hypotenuse. The angle θ is measured from the horizontal direction. The horizontal side lies adjacent to θ, so cosθ = adjacent/hypotenuse = Fx/F, giving Fx = F cosθ. The vertical side lies opposite θ, so sinθ = opposite/hypotenuse = Fy/F, giving Fy = F sinθ.

↑ F sinθ (vertical/opposite) │ │ ↗ F │ θ/ ──────────────●────────→ F cosθ (horizontal/adjacent) → displacement s

Because the displacement is horizontal, the vertical component is perpendicular to the motion and does no work: Wvertical = (F sinθ)s cos90° = 0. Therefore only the horizontal component contributes:

W = Fxs = (F cosθ)s = Fs cosθ

When is work maximum? For fixed F and s, cosθ has its greatest possible value, 1, at θ = 0°. Therefore Wmax = Fs. This happens when force and displacement point in exactly the same direction.

Sample: Let F = 10 N and s = 5 m. At 0°, W = 10 × 5 × cos0° = 50 J. At 60°, W = 10 × 5 × cos60° = 25 J. At 90°, W = 0 J. Thus turning the force away from the direction of motion reduces its useful horizontal component and therefore reduces the work.

Important: Cosine is not always “horizontal” and sine is not always “vertical.” That is true here only because θ is measured from the horizontal. If θ were measured from the vertical, the adjacent vertical component would use cosine and the opposite horizontal component would use sine.

Question 21: The impulse of a body is equal to _____.

Status: Correct
Show model answer and explanation
Impulse J equals change in momentum: J = Δp = FΔt for a constant force. Unit: N s, equivalent to kg m s⁻¹.

Question 22: Define inertia of motion and inertia of direction.

Status: Partially correct
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Inertia of motion is the tendency of a moving body to remain in uniform motion. Inertia of direction is its resistance to a change in the direction of motion. The second definition in the answer was incomplete.

Question 23: 1 kgf is equal to how many newtons?

Status: Correct
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Given: m = 1 kg, g = 9.8 m s⁻²
Choose the idea: Kilogram-force is the weight of a 1 kg mass.
  1. Use weight W = mg.
  2. W = 1 × 9.8 = 9.8 N.
Final answer: 1 kgf = 9.8 N (often rounded to 10 N when g = 10 m s⁻²).

Question 24: Define the SI unit of force.

Status: Correct
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One newton is the force that gives a mass of 1 kg an acceleration of 1 m s⁻²: 1 N = 1 kg m s⁻².

Question 25: What is the relation between momentum and kinetic energy?

Status: Correct
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Given: p = mv and K = ½mv²
Choose the idea: Eliminate velocity v between the two formulas.
  1. From p = mv, v = p/m.
  2. Substitute into K: K = ½m(p/m)².
  3. Simplify: K = p²/(2m).
  4. Rearranging gives p = √(2mK).
Final answer: K = p²/(2m), or p = √(2mK).

Question 26: What is the period of a seconds pendulum?

Status: Correct
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2 seconds. One complete oscillation takes 2 s, so each swing from one extreme to the other takes 1 s.

Question 27: Give dimensional formulae of force, momentum and impulse.

Status: Correct
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Force: [MLT⁻²]. Momentum: [MLT⁻¹]. Impulse: [MLT⁻¹], since impulse equals change in momentum.

Question 28: A mass m1 exerts force on m2 and produces acceleration a2. Find acceleration of m1.

Status: Partially correct
Show model answer and explanation
Given: Force on m₂ is m₂a₂
Choose the idea: Newton’s third law gives equal magnitudes and opposite directions.
  1. F₂ = m₂a₂.
  2. The reaction on m₁ has magnitude F₁ = F₂, so m₁a₁ = m₂a₂.
  3. Divide by m₁: a₁ = (m₂/m₁)a₂.
Final answer: a₁ = (m₂/m₁)a₂ in the direction opposite to a₂.

Reasonableness check: The smaller mass must have the larger acceleration for the same force.

Question 29: State the work-energy theorem.

Status: Correct
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The net work done on a body equals its change in kinetic energy: Wnet = ΔK = ½mv² - ½mu².

Question 30: Define impulse.

Status: Partially correct
Show model answer and explanation
Impulse is the product of force and the time for which it acts, and equals change in momentum: J = FΔt = Δp. “Change in momentum” is correct but incomplete as a definition.

Question 31: Distinguish renewable and non-renewable energy sources.

Status: Partially correct
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Renewable sources are naturally replenished on a human timescale (solar, wind, hydro). Non-renewable sources are finite and form very slowly (coal, petroleum, natural gas). Examples and the timescale complete the distinction.

Question 32: Explain the cause of friction between two bodies in contact.

Status: Correct
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Microscopic surface irregularities interlock and molecular adhesion occurs at real contact points. These effects resist relative motion.

Question 33: A 0.5 kg ball changes velocity from +10 m/s to -20 m/s in 0.02 s. Find impulse and force.

Status: Correct
Show model answer and explanation
Given: m = 0.5 kg, u = +10 m/s, v = −20 m/s, Δt = 0.02 s
Choose the idea: Impulse is change in vector momentum; average force is impulse divided by time.
  1. Choose the initial direction as positive.
  2. Δv = v − u = −20 − 10 = −30 m/s.
  3. Impulse J = Δp = mΔv = 0.5(−30) = −15 kg m/s = −15 N s.
  4. Favg = J/Δt = −15/0.02 = −750 N.
Final answer: Impulse = −15 N s; average force = −750 N. Magnitudes are 15 N s and 750 N, opposite to the initial motion.

Reasonableness check: The large force is reasonable because the reversal occurs in only 0.02 s.

Question 34: Distinguish nuclear fission and fusion.

Status: Correct
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Fission splits a heavy nucleus (such as uranium) into lighter nuclei and releases energy. Fusion combines light nuclei (such as hydrogen isotopes) into a heavier nucleus and releases energy; it requires extremely high temperature.

Question 35: Explain Newton’s third law with an example.

Status: Partially correct
Show model answer and explanation
For every action force there is an equal and opposite reaction force acting simultaneously on a different body. A rocket moves forward because gases are expelled backward. The drawing shows the idea but should explicitly say the forces act on different bodies.

Question 36: Masses 5 kg and 3 kg move at 3 m/s and 0, stick together. Find common velocity.

Status: Incorrect
Show model answer and explanation
Given: m₁ = 5 kg, u₁ = 3 m/s; m₂ = 3 kg, u₂ = 0; bodies stick
Choose the idea: For a perfectly inelastic collision, conserve momentum and use one common final velocity v.
  1. Momentum before = 5×3 + 3×0 = 15 kg m/s.
  2. Momentum after = (5+3)v = 8v.
  3. Set before = after: 15 = 8v.
  4. v = 15/8 = 1.875 m/s.
Final answer: Common velocity = 1.875 m/s in the original direction of the 5 kg mass.

Reasonableness check: The final speed lies between 0 and 3 m/s, as expected.

Question 37: What is meant by energy crisis?

Status: Correct
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An energy crisis is a shortage or scarcity of usable energy resources relative to demand, especially as finite fuels are depleted or supply is disrupted.

Question 38: Derive the relation between newton and dyne.

Status: Correct
Show model answer and explanation
Given: 1 N = 1 kg m s⁻²
Choose the idea: Convert SI base units to CGS base units.
  1. 1 kg = 10³ g.
  2. 1 m = 10² cm.
  3. 1 N = (10³ g)(10² cm)s⁻² = 10⁵ g cm s⁻².
  4. Since 1 dyne = 1 g cm s⁻², compare the coefficients.
Final answer: 1 N = 10⁵ dyne.

Question 39: A 20 kg cart is dragged at 18 km/h, μ = 0.1. Find minimum force over 10 m.

Status: Partially correct
Show model answer and explanation
Given: m = 20 kg, final speed v = 18 km/h = 5 m/s, μ = 0.1, s = 10 m, take g = 10 m/s²
Choose the idea: The applied work must supply both the gain in KE and the work lost against friction.
  1. Convert speed: 18 km/h ÷ 3.6 = 5 m/s.
  2. Friction f = μmg = 0.1×20×10 = 20 N.
  3. Gain in KE = ½mv² = ½×20×25 = 250 J.
  4. Work against friction = fs = 20×10 = 200 J.
  5. Total applied work = 250 + 200 = 450 J.
  6. For constant applied force, Fs = 450, so F = 450/10 = 45 N.
Final answer: Minimum constant pulling force = 45 N.

Reasonableness check: 20 N only balances friction; an additional 25 N is needed to build kinetic energy over 10 m.

Question 40: Define periodic motion and oscillatory motion.

Status: Correct
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Periodic motion repeats after equal time intervals. Oscillatory motion is repeated to-and-fro motion about a mean position.

Question 41: Show experimentally that inertia depends on mass.

Status: Partially correct
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Push or strike a light ball and a heavier ball with the same force. The heavier ball changes its state less, showing greater inertia. The answer states the conclusion but needs the equal-force comparison and observation.

Question 42: A 10 g body reaches 20 m/s in 10 s. Find work and power.

Status: Correct
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Given: m = 10 g = 0.010 kg, u = 0, v = 20 m/s, t = 10 s
Choose the idea: Use the work–energy theorem, then power = work/time.
  1. Convert mass to kilograms: 10 g = 0.010 kg.
  2. W = ΔK = ½m(v²−u²).
  3. W = ½×0.010×20² = 0.005×400 = 2 J.
  4. P = W/t = 2/10 = 0.2 W.
Final answer: Work done = 2 J; average power = 0.2 W.

Question 43: Define simple harmonic motion.

Status: Unattempted
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Key concept behind this hard question: SHM is recognized by a restoring influence: the farther the object moves from equilibrium, the stronger the force pulling it back. The minus sign records the opposite direction.

Definition: Simple harmonic motion is oscillatory motion in which restoring acceleration is directly proportional to displacement from the mean position and is oppositely directed.

a = −ω²x   or   F = −kx

Example: A small mass on an ideal spring. At x = 0, acceleration is zero and speed is greatest. At x = ±A, speed is zero and acceleration toward the centre is greatest.

Question 44: State Newton’s laws of motion.

Status: Correct
Show model answer and explanation
First: a body remains at rest or uniform straight-line motion unless acted on by net external force. Second: net force equals rate of change of momentum; for constant mass F = ma. Third: interaction forces are equal, opposite and act on different bodies.

Question 45: A 5 kg body falls from 20 m. Find PE and KE after falling 15 m, and KE at ground.

Status: Partially correct
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Given: m = 5 kg, starting height H = 20 m, distance fallen = 15 m, take g = 10 m/s²
Choose the idea: With air resistance ignored, PE + KE is constant. Height remaining—not distance fallen—is used in PE = mgh.
  1. Initial total energy = mgH = 5×10×20 = 1000 J.
  2. After falling 15 m, remaining height h = 20−15 = 5 m.
  3. PE there = mgh = 5×10×5 = 250 J.
  4. KE there = total − PE = 1000−250 = 750 J.
  5. At ground h = 0, PE = 0, so all 1000 J is kinetic.
Final answer: After 15 m: PE = 250 J and KE = 750 J. At ground: KE = 1000 J.

Reasonableness check: After falling three quarters of the height, three quarters of the initial PE has become KE.

Question 46: Describe an experiment to find g using a simple pendulum.

Status: Partially correct
Show model answer and explanation
Measure length L from pivot to bob centre. Time 10-20 oscillations repeatedly, find T = t/n, and use g = 4π²L/T². Repeat for several lengths and average or plot L against T². The answer gave only the formula.

Question 47: State and verify the law of conservation of momentum.

Status: Partially correct
Show model answer and explanation
For an isolated system, total momentum remains constant. For two colliding bodies, m1u1+m2u2=m1v1+m2v2 because internal action-reaction impulses are equal and opposite. The answer states the equation but does not complete the verification.

Question 48: Define and derive gravitational potential energy.

Status: Partially correct
Show model answer and explanation
Near Earth, raising mass m slowly through height h requires force mg through displacement h. Work done W = Fh = mgh; this work is stored as gravitational PE. The response named mgh but did not derive it clearly.

Question 49: State conservation of energy and verify it for a freely falling body.

Status: Partially correct
Show model answer and explanation
Energy cannot be created or destroyed, only transformed. From height H, total energy is mgH. At height h, PE=mgh and KE=mg(H-h), so PE+KE=mgH. At ground KE=mgH. The statement was correct but the falling-body verification was missing.

Question 50: Derive F = ma from Newton’s second law.

Status: Partially correct
Show model answer and explanation
F ∝ Δp/Δt. For constant mass, Δp/Δt = mΔv/Δt = ma. Choosing SI units makes the proportionality constant 1, so F = ma. The core derivation was present but needed the constant-mass and SI steps.

Part 2: Concept Application - Questions 1-50

Application 1: True/False: A stretched spring has potential energy.

Status: Correct
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True. Work done in stretching is stored as elastic potential energy, U = ½kx².

Application 2: True/False: Work and energy have the same SI unit.

Status: Correct
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True. Both are measured in joules (J).

Application 3: True/False: Friction depends on area of contact between two surfaces.

Status: Correct
Show model answer and explanation
False in the ideal dry-friction model: friction is approximately μN and is independent of apparent contact area.

Application 4: True/False: Friction can be reduced by polishing surfaces.

Status: Correct
Show model answer and explanation
True, when polishing reduces interlocking irregularities. Extremely smooth clean surfaces can show adhesion, but the school-level statement is true.

Application 5: True/False: All forces exist in pairs.

Status: Correct
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True. Interaction forces occur as Newton’s third-law pairs on different bodies.

Application 6: True/False: Impulse and momentum have similar units.

Status: Correct
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True. Both have dimensions [MLT⁻¹]; N s = kg m s⁻¹.

Application 7: True/False: Mass is a measure of inertia.

Status: Correct
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True. Greater mass means greater resistance to acceleration for the same net force.

Application 8: A cricket ball during flight possesses _____ and _____ energy.

Status: Correct
Show model answer and explanation
Kinetic energy and gravitational potential energy.

Application 9: The change in momentum has the same _____ as the applied force.

Status: Unattempted
Show model answer and explanation
Key concept behind this hard question: Force and change in momentum are vectors. From F = Δp/Δt, division by a positive time cannot reverse a direction.

Answer: The change in momentum has the same direction as the net applied force.

Example: If a force acts east for 2 s, Δp = FΔt also points east—even if the object was initially moving west.

Application 10: A moving car can be stopped by applying _____.

Status: Correct
Show model answer and explanation
An external force (for example braking friction). A net force opposite the velocity reduces momentum.

Application 11: Total momentum before collision equals _____ after collision.

Status: Correct
Show model answer and explanation
Total momentum, provided net external impulse is zero.

Application 12: To do the same work in less time, power should be _____.

Status: Correct
Show model answer and explanation
Increased, because P = W/t.

Application 13: When lifted, a body’s _____ energy changes to _____ energy.

Status: Correct
Show model answer and explanation
Kinetic energy changes to potential energy if it is thrown upward; more generally work done against gravity increases potential energy.

Application 14: _____ force opposes relative motion between bodies.

Status: Correct
Show model answer and explanation
Frictional force.

Application 15: Match inertia, action-reaction, friction, lubricants, unbalanced force, pendulum, PE, μ and work.

Status: Partially correct
Show model answer and explanation
Correct matching: Inertia-f; Action and reaction-i; Unit of friction-j; Lubricants-e; Unbalanced force-d; Effect of friction-h; Simple pendulum-b; Elastic PE-d (stretched rubber band); Coefficient of friction-c; Work-g. Several entries were correct, but at least the unbalanced-force match was wrong.

Application 16: MCQ: “Friction is a self-adjusting force” is true for which friction?

Status: Correct
Show model answer and explanation
Static friction (choice 2). It adjusts from zero up to limiting friction as needed to prevent slipping.

Application 17: MCQ: Period of a simple pendulum is independent of what?

Status: Correct
Show model answer and explanation
All of the listed bob properties - shape, material and mass (choice 4), for small oscillations. T = 2π√(L/g).

Application 18: MCQ: Masses m and 4m have equal KE. Find speed ratio.

Status: Correct
Show model answer and explanation
Given: Masses m and 4m have equal kinetic energy
Choose the idea: For equal KE, speed varies inversely as the square root of mass.
  1. ½m v₁² = ½(4m)v₂².
  2. Cancel ½m: v₁² = 4v₂².
  3. Take the positive square root for speeds: v₁ = 2v₂.
Final answer: Speed ratio v₁:v₂ = 2:1.

Application 19: MCQ: Weight of a 20 kg body.

Status: Partially correct
Show model answer and explanation
Given: m = 20 kg, g = 9.8 m/s²
Choose the idea: Weight is gravitational force: W = mg.
  1. W = 20×9.8.
  2. W = 196 N.
Final answer: Weight = 196 N.

Reasonableness check: Using g ≈ 10 gives 200 N, but 196 N matches the supplied choices when g = 9.8.

Application 20: MCQ: Spring k=2 N/m stretched 5 cm. Find stored energy.

Status: Correct
Show model answer and explanation
Given: k = 2 N/m, extension x = 5 cm = 0.05 m
Choose the idea: Elastic PE is the triangular area under the F–x graph: U = ½kx².
  1. Convert centimetres: 5 cm = 0.05 m.
  2. U = ½×2×(0.05)².
  3. (0.05)² = 0.0025.
  4. U = 0.0025 J.
Final answer: Stored elastic energy = 2.5×10⁻³ J.

Application 21: MCQ: Relation between momentum p and kinetic energy E.

Status: Correct
Show model answer and explanation
p = √(2mE) (choice 1), obtained from E=p²/(2m).

Application 22: MCQ: Momentum changes by 10 kg m/s in 5 s. Find force.

Status: Correct
Show model answer and explanation
Given: Δp = 10 kg m/s, Δt = 5 s
Choose the idea: Newton’s second law in momentum form: F = Δp/Δt.
  1. F = 10/5.
  2. F = 2 kg m s⁻² = 2 N.
Final answer: Average force = 2 N.

Application 23: MCQ: Motors of 1.75 hp and 3.5 hp pump for same time. Water ratio?

Status: Correct
Show model answer and explanation
Given: P₁ = 1.75 hp, P₂ = 3.5 hp; equal time and equal lifting height
Choose the idea: P = mgh/t, so when g, h and t are the same, pumped mass is proportional to power.
  1. m₁/m₂ = P₁/P₂.
  2. m₁:m₂ = 1.75:3.5.
  3. Divide both terms by 1.75.
Final answer: Water pumped ratio = 1:2.

Application 24: MCQ: Change in momentum is equal to what?

Status: Correct
Show model answer and explanation
Impulse, equal to force × time (choice 2).

Application 25: MCQ: A car coasts to rest on a rough road due to what?

Status: Correct
Show model answer and explanation
Sliding/rolling resistive friction; the intended answer is sliding friction (choice 2).

Application 26: MCQ: Rate of change of momentum is _____.

Status: Correct
Show model answer and explanation
Equal to and in the direction of net applied force, so all statements are true (choice 4).

Application 27: MCQ: Action and reaction forces are _____.

Status: Correct
Show model answer and explanation
They exist in pairs, are equal in magnitude, and opposite in direction; all are true (choice 4). They act on different bodies.

Application 28: MCQ: Identify the false statement about friction.

Status: Correct
Show model answer and explanation
“Coefficient of friction increases with area of contact” is false (choice 4).

Application 29: MCQ: A ripe fruit detaches when a branch is shaken. Which law?

Status: Correct
Show model answer and explanation
Newton’s first law (choice 1): the fruit tends to remain at rest while the branch moves.

Application 30: MCQ: Stone whirled in a vertical circle - what is true?

Status: Correct
Show model answer and explanation
At the top, PE is maximum and KE is minimum, so both statements 1 and 2 are true (choice 4).

Application 31: Bullet embeds in a block and falls from a wall. Find horizontal range.

Status: Unattempted
Show model answer and explanation
Key concept behind this hard question: This is a two-stage problem: momentum controls the very short collision; projectile motion controls the later fall. Do not use energy conservation through the embedding collision because it is inelastic.
Given: Bullet mass = 0.05 kg, bullet speed = 100 m/s; combined mass after embedding = 0.50 kg; wall height h = 19.6 m; g = 9.8 m/s²
Choose the idea: Stage 1: conserve horizontal momentum. Stage 2: horizontal speed stays constant while vertical free fall sets the time.
  1. Collision: 0.05×100 = 0.50×V, so V = 10 m/s.
  2. Vertical motion starts with zero vertical speed: h = ½gt².
  3. t = √(2h/g) = √(2×19.6/9.8) = √4 = 2 s.
  4. Horizontal range R = Vt = 10×2 = 20 m.
Final answer: Horizontal range = 20 m.

Reasonableness check: Horizontal and vertical motions share the same flight time, but only gravity changes vertical velocity.

Application 32: Discuss work while a porter carries luggage in five situations.

Status: Incorrect
Show model answer and explanation
Horizontal uniform motion: zero work by porter on luggage because support force is perpendicular to displacement. Sloping upward uniformly: positive work. Horizontal acceleration: positive work. Lifting vertically: positive work. Lowering: negative work. Several signs in the response were wrong.

Application 33: Car and trolley have equal KE and same braking force. Compare stopping distance.

Status: Incorrect
Show model answer and explanation
Given: Initial kinetic energies are equal; braking-force magnitudes are equal
Choose the idea: Stopping work removes all initial KE: Fs = KE.
  1. For the car: Fs₁ = K.
  2. For the trolley: Fs₂ = K.
  3. Because F and K are the same, s₁ = K/F = s₂.
Final answer: The stopping distances are equal.

Reasonableness check: Mass and initial speed may differ, but their combined effect is already contained in the equal kinetic energies.

Application 34: Can a person inside a car move the car by pushing its door?

Status: Incorrect
Show model answer and explanation
No sustained motion of the car-person system results from internal forces alone. The push and reaction are internal; the centre of mass cannot accelerate without net external force.

Application 35: Can a machine reduce the total work needed to lift a weight?

Status: Correct
Show model answer and explanation
No ideal machine reduces work; it trades smaller force for larger distance. Real machines require more input work because efficiency is less than 100%.

Application 36: Discuss conditions under which no work is done.

Status: Correct
Show model answer and explanation
W = Fs cosθ is zero if displacement is zero, force is zero, or force is perpendicular to displacement (θ=90°).

Application 37: A compressed spring is clamped, then dissolved in acid. What happens to PE?

Status: Partially correct
Show model answer and explanation
Its elastic PE is ultimately transformed into other forms, mainly internal/chemical energy and heat, possibly sound or kinetic energy during release. Energy is conserved; it does not disappear.

Application 38: Block on an incline: how does friction vary as inclination increases?

Status: Correct
Show model answer and explanation
Before slipping, static friction increases as mg sinθ. At impending motion, tanθ=μs. Once sliding, fk=μkmg cosθ, which decreases as θ increases. The answer captured the normal-force dependence but should separate static and kinetic cases.

Application 39: A log is dragged up an incline. How does friction vary with angle?

Status: Partially correct
Show model answer and explanation
If the pull is parallel to the plane, N=mg cosθ and fk=μmg cosθ, so friction decreases as θ increases. If the pull is angled, its normal component must also be included.

Application 40: Smooth valley: ball starts at height h and reaches height H. Find required speed.

Status: Incorrect
Show model answer and explanation
Given: Initial position at height h with speed u; final highest point at height H with speed 0; smooth valley
Choose the idea: No friction means mechanical energy is conserved.
  1. Initial energy = ½mu² + mgh.
  2. At the highest final point, energy = mgH.
  3. Set equal: ½mu² + mgh = mgH.
  4. ½u² = g(H−h).
  5. u² = 2g(H−h).
Final answer: Required speed u = √[2g(H−h)].

Reasonableness check: If H = h, the required speed becomes zero; if H increases, more starting speed is needed.

Application 41: A 0.1 kg tennis ball reverses from 45 m/s to 45 m/s in 0.1 s. Find force.

Status: Correct
Show model answer and explanation
Given: m = 0.1 kg, u = −45 m/s, v = +45 m/s, Δt = 0.1 s
Choose the idea: A reversal makes the velocity change 90 m/s, not zero.
  1. Choose away from the racket as positive.
  2. Δv = v−u = 45−(−45) = 90 m/s.
  3. Δp = mΔv = 0.1×90 = 9 N s.
  4. Favg = Δp/Δt = 9/0.1 = 90 N.
Final answer: Average force = 90 N away from the racket.

Application 42: Two 2 kg balls are pushed apart by a spring and reach 10 m/s. Find acceleration if contact lasts 4 s.

Status: Correct
Show model answer and explanation
Given: Each ball: u = 0, v = 10 m/s, contact time = 4 s
Choose the idea: Average acceleration is change in velocity divided by time.
  1. a = (v−u)/t.
  2. a = (10−0)/4 = 2.5 m/s².
  3. The two accelerations have equal magnitudes but opposite directions.
Final answer: Each ball has average acceleration 2.5 m/s², in opposite directions.

Reasonableness check: For each 2 kg ball, the corresponding average force would be F = ma = 5 N.

Application 43: Is it easier to pull or push a cylindrical roller? Explain.

Status: Correct
Show model answer and explanation
Pulling upward at an angle reduces normal reaction and friction; pushing downward increases them. Therefore pulling is easier.

Application 44: A body moves in a vertical circle. Find speed at the lowest point when highest-point speed is 3 m/s and r=1 m.

Status: Partially correct
Show model answer and explanation
Given: Top speed vT = 3 m/s, radius r = 1 m, take g = 10 m/s²
Choose the idea: The bottom is 2r lower than the top, so lost PE becomes KE.
  1. At bottom: energy = ½mvL².
  2. At top: energy = ½mvT² + mg(2r).
  3. Conservation: ½mvL² = ½m(3²) + mg(2×1).
  4. Cancel ½m: vL² = 3² + 4gr = 9+40 = 49.
  5. vL = √49 = 7 m/s.
Final answer: Speed at the lowest point = 7 m/s.

Reasonableness check: It must be faster at the bottom because gravitational PE is lower there.

Application 45: Compare forces when stopping after jumping 1 m in 0.1 s and 9 m in 1 s.

Status: Correct
Show model answer and explanation
Given: Case A: fall 1 m and stop in 0.1 s; Case B: fall 9 m and stop in 1 s
Choose the idea: Impact speed v = √(2gh); for the same mass, average stopping force (ignoring weight during the brief stop) is proportional to v/t.
  1. F₁/F₂ = (v₁/t₁)/(v₂/t₂).
  2. v₁/v₂ = √(h₁/h₂) = √(1/9) = 1/3.
  3. F₁/F₂ = (1/0.1)/(3/1) = 10/3.
Final answer: Stopping-force ratio F₁:F₂ = 10:3.

Reasonableness check: A shorter stopping time can produce a larger force even from a smaller height.

Application 46: Are applied force and acceleration always in the same direction?

Status: Partially correct
Show model answer and explanation
Acceleration is in the direction of net force, not necessarily one applied force. An angled pull can have vertical components balanced by weight/normal while the net force and acceleration are horizontal.

Application 47: Can a man in a pan lift himself by pulling down on a rope over a fixed pulley?

Status: Partially correct
Show model answer and explanation
Treat man plus pan as one system. Two upward tensions act (one on the man’s hands and one on the pan). It rises if 2T exceeds total weight. Newton’s third law alone is not a complete explanation.

Application 48: Block pulled up an incline by force F at angle φ; derive acceleration with friction μ.

Status: Unattempted
Show model answer and explanation
Key concept behind this hard question: Draw axes parallel and perpendicular to the plane. Friction needs the normal reaction first. Because the pull is angled away from the plane, its perpendicular component reduces the normal reaction and therefore reduces friction.
Given: Mass m on a plane of angle θ; pull F at angle φ above the plane; coefficient of friction μ
Choose the idea: Resolve forces separately along and perpendicular to the plane.
  1. Perpendicular balance: N + F sinφ − mg cosθ = 0.
  2. Therefore N = mg cosθ − F sinφ.
  3. Since motion is up the plane, friction acts down the plane: f = μN.
  4. Along the plane: ma = F cosφ − mg sinθ − μN.
  5. Substitute N: ma = F cosφ − mg sinθ − μ(mg cosθ − F sinφ).
  6. Divide by m.
Final answer: a = [F cosφ − mg sinθ − μ(mg cosθ − F sinφ)]/m.

Reasonableness check: If μ = 0, the friction term disappears; if φ increases, the pull lifts more strongly and reduces N.

Application 49: Nine 0.5 kg books stacked; μ=0.2. Maximum force to pull fifth book?

Status: Incorrect
Show model answer and explanation
Given: Nine books, each m = 0.5 kg; fifth book from top is pulled; μ = 0.2; g = 10 m/s²
Choose the idea: The fifth book rubs against books on both faces. Each interface has a different normal load.
  1. Four books lie above it, so upper-interface normal Ntop = 4mg = 4×0.5×10 = 20 N.
  2. Upper friction = μNtop = 0.2×20 = 4 N.
  3. At the lower face, the lower stack supports the pulled book plus four above: Nbottom = 5mg = 25 N.
  4. Lower friction = μNbottom = 0.2×25 = 5 N.
  5. Both friction forces oppose pulling, so add them: 4+5 = 9 N.
Final answer: Maximum resisting force = 9 N; the applied force must exceed 9 N to pull the fifth book out.

Reasonableness check: The common 8 N mistake counts four books at both interfaces and misses the pulled book’s own load on the lower face.

Application 50: Bouncing ball: at h2/2 the KE ratio downward:upward is 2:1. Find h1:h2.

Status: Correct
Show model answer and explanation
Given: At height h₂/2, KEdown:KEup = 2:1; initial drop height = h₁; rebound height = h₂
Choose the idea: At the same reference height, KE equals the loss of gravitational PE from that motion’s highest point.
  1. On the way down from h₁: KEdown = mg(h₁−h₂/2).
  2. On the way up toward h₂: KEup = mg(h₂−h₂/2) = mgh₂/2.
  3. Use the ratio: [h₁−h₂/2]/(h₂/2) = 2.
  4. h₁−h₂/2 = h₂.
  5. h₁ = 3h₂/2.
Final answer: h₁:h₂ = 3:2.

Reasonableness check: Because energy is lost during the bounce, the original height h₁ is greater than rebound height h₂.

Highest-priority revision topics

  1. Work-energy theorem and recognizing positive, negative and zero work.
  2. Conservation of momentum in collisions and Newton’s third-law pairs.
  3. Conservation of mechanical energy in falling bodies and vertical circles.
  4. Friction on horizontal and inclined surfaces, including normal reaction.
  5. Complete derivations: F=ma, PE=mgh, pendulum g and inclined-plane acceleration.
  6. Always carry units, define directions/signs, and finish numerical answers.

Grading is based on the legible content of the photographed handwriting. When an answer showed a correct idea but lacked a required step, unit, direction or explanation, it was marked partially correct.

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