📖 Tier 1: Prepare & Study Guide ✓ 100% Solved with Rationales

Mechanics & Dynamics (Physics) Solved Questions & Notes (2026) - Apex Rankers

Natural & Physical Sciences > Physics > Mechanics & Dynamics

39 Total Solved Questions
~59 mins Estimated Reading Time
1 Subject Areas / Chapters
Select Topic Area / Chapter: Click any section below to switch questions

Mechanics & Dynamics

100%
Showing 25 of 39 (64%)
🎯 Practice
Jump:
Q. 1 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is the rate of change of momentum of a body directly proportional to, according to Newton's Second Law of Motion?
A
The applied net external force
✓ Correct
B
The kinetic energy
C
The velocity of the body
D
The gravitational potential
💡 Step-by-Step Explanation & Concept Rationale
Newton's Second Law states that F = dp/dt (the rate of change of momentum is equal to the net applied force).
Q. 2 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
A ball is projected horizontally from a height h with initial velocity v. At the same instant, an identical ball is dropped vertically from the same height. Neglecting air resistance, which ball hits the ground first?
A
Both balls hit the ground at exactly the same time
✓ Correct
B
The dropped ball hits first
C
The horizontally projected ball hits first
D
It depends on the mass of the balls
💡 Step-by-Step Explanation & Concept Rationale
Vertical and horizontal motions are completely independent; both balls experience the identical vertical acceleration g and initial vertical velocity zero, taking time t = sqrt(2h/g).
Q. 3 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is the centripetal acceleration of an object of mass m moving at constant speed v in a circular path of radius r?
A
a = v² / r directed towards the center
✓ Correct
B
a = v / r² directed tangentially
C
a = v² * r directed outward
D
a = zero because speed is constant
💡 Step-by-Step Explanation & Concept Rationale
Centripetal acceleration is a = v²/r directed radially inward toward the center of rotation.
Q. 4 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In an isolated system, what quantities are conserved during a perfectly elastic collision between two particles?
A
Both total linear momentum and total kinetic energy
✓ Correct
B
Only total linear momentum, kinetic energy is lost
C
Only kinetic energy, momentum changes
D
Neither momentum nor kinetic energy
💡 Step-by-Step Explanation & Concept Rationale
In elastic collisions, both total momentum and total kinetic energy are strictly conserved.
Q. 5 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
What is the moment of inertia of a uniform solid cylinder of mass M and radius R rotating about its central longitudinal axis?
A
I = (1/2) M R²
✓ Correct
B
I = M R²
C
I = (2/5) M R²
D
I = (1/12) M R²
💡 Step-by-Step Explanation & Concept Rationale
For a solid cylinder or disk of mass M and radius R about its cylindrical axis, I = 1/2 MR².
Q. 6 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
What is the dimensional formula of the universal gravitational constant G?
A
[M^-1 L^3 T^-2]
✓ Correct
B
[M L^2 T^-2]
C
[M^-1 L^2 T^-1]
D
[M L^3 T^-2]
💡 Step-by-Step Explanation & Concept Rationale
From F = G*m1*m2/r^2, G = F*r^2/(m1*m2) = [M L T^-2][L^2]/[M^2] = [M^-1 L^3 T^-2].
Q. 7 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
What is the apparent weight of a person in an elevator falling freely under gravity?
A
2 mg
B
mg
C
Zero
✓ Correct
D
mg / 2
💡 Step-by-Step Explanation & Concept Rationale
In free fall, a = g, so apparent weight N = m(g - a) = m(g - g) = 0 (weightlessness).
Q. 8 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #13: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
The incident mass stops and target mass moves with original velocity
✓ Correct
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 9 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #19: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both bounce back with equal speeds
B
Both masses stick together and move with half velocity
C
The incident mass stops and target mass moves with original velocity
✓ Correct
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 10 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #25: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
The incident mass stops and target mass moves with original velocity
✓ Correct
B
Both masses stick together and move with half velocity
C
Both bounce back with equal speeds
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 11 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In classical mechanics problem #31: In an elastic collision in one dimension between two identical masses where one is initially at rest, what occurs after collision?
A
Both bounce back with equal speeds
B
Both masses stick together and move with half velocity
C
The incident mass stops and target mass moves with original velocity
✓ Correct
D
Both stop immediately
💡 Step-by-Step Explanation & Concept Rationale
For elastic collision of identical masses in 1D, velocities are completely exchanged upon impact (v1' = 0, v2' = v1).
Q. 12 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
Terminal velocity of a spherical body falling through a viscous medium is directly proportional to:
A
Square of its radius (r^2)
✓ Correct
B
Radius of sphere (r)
C
Cube of its radius (r^3)
D
Inverse of its radius (1/r)
💡 Step-by-Step Explanation & Concept Rationale
According to Stokes' law, v_t = (2/9) * r^2 * g * (rho - sigma) / eta, which is proportional to r^2.
Q. 13 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #8: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Froude Number
B
Mach Number
C
Reynolds Number (Re)
✓ Correct
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 14 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #14: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Reynolds Number (Re)
✓ Correct
B
Mach Number
C
Froude Number
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 15 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #20: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Froude Number
B
Mach Number
C
Reynolds Number (Re)
✓ Correct
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 16 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In fluid mechanics concept #26: What is the ratio of inertial forces to viscous forces in fluid flow analysis?
A
Reynolds Number (Re)
✓ Correct
B
Mach Number
C
Froude Number
D
Prandtl Number
💡 Step-by-Step Explanation & Concept Rationale
The Reynolds number (Re = rho*v*L/mu) is the dimensionless parameter quantifying the ratio of inertial to viscous forces.
Q. 17 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In an adiabatic process involving an ideal gas, which quantity remains zero throughout the process?
A
Temperature change (delta T = 0)
B
Work done (W = 0)
C
Heat exchanged (delta Q = 0)
✓ Correct
D
Pressure change (delta P = 0)
💡 Step-by-Step Explanation & Concept Rationale
An adiabatic process is one in which no heat enters or leaves the system (Q = 0).
Q. 18 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #8: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.67 (5/3)
✓ Correct
B
1.40 (7/5)
C
1.33 (4/3)
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 19 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #14: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.33 (4/3)
B
1.40 (7/5)
C
1.67 (5/3)
✓ Correct
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 20 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #20: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.67 (5/3)
✓ Correct
B
1.40 (7/5)
C
1.33 (4/3)
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 21 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
In thermodynamic analysis #26: What is the molar specific heat ratio (gamma = Cp/Cv) for a standard monoatomic ideal gas?
A
1.33 (4/3)
B
1.40 (7/5)
C
1.67 (5/3)
✓ Correct
D
2.00
💡 Step-by-Step Explanation & Concept Rationale
For a monoatomic gas with 3 degrees of freedom, Cp = 5/2 R, Cv = 3/2 R, so gamma = 5/3 ≈ 1.67.
Q. 22 Physics (Secondary & College Level)
Difficulty: MEDIUM (1 Mark)
The phenomenon of polarization in light waves confirms conclusively that light waves are:
A
Transverse waves
✓ Correct
B
Longitudinal waves
C
Stationary waves
D
Scalar waves
💡 Step-by-Step Explanation & Concept Rationale
Only transverse waves can exhibit polarization because their oscillations are perpendicular to the direction of propagation.
Q. 23 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #8: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
pi * sqrt(L * g)
B
2 * pi * sqrt(g / L)
C
2 * pi * sqrt(L / g)
✓ Correct
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 24 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #14: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
2 * pi * sqrt(L / g)
✓ Correct
B
2 * pi * sqrt(g / L)
C
pi * sqrt(L * g)
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Q. 25 Physics (Secondary & College Level)
Difficulty: EASY (1 Mark)
In wave theory #20: What is the time period T of a simple pendulum of length L in a gravitational field g?
A
pi * sqrt(L * g)
B
2 * pi * sqrt(g / L)
C
2 * pi * sqrt(L / g)
✓ Correct
D
2 * pi * (L / g)^2
💡 Step-by-Step Explanation & Concept Rationale
The period of oscillation for a simple pendulum for small angles is T = 2*pi*sqrt(L/g).
Study Stream Progress: Showing 25 of 39 Questions (64%)
Jump to:

Ready to Test Your Retention & Speed?

Now that you have reviewed the study questions and rationales, test yourself in our interactive 1-by-1 practice engine or take the full official timed mock exam.