Problem 1: Kinematics

Problem Statement

A particle moves along a straight line such that its position \( x \) at time \( t \) is given by \( x(t) = 3t^3 – 12t^2 + 9t + 6 \). Find the velocity and acceleration of the particle at \( t = 2 \) seconds.

Concepts Involved

This problem involves understanding the relationship between position, velocity, and acceleration. Velocity is the first derivative of position with respect to time, and acceleration is the second derivative.

### v(t) = \frac{dx}{dt} = 9t^2 – 24t + 9 ###

### a(t) = \frac{d^2x}{dt^2} = 18t – 24 ###

Solution

To find the velocity, differentiate the position function with respect to time:

### v(t) = \frac{dx}{dt} = 9t^2 – 24t + 9 ###

Substitute \( t = 2 \) into the velocity equation:

### v(2) = 9(2)^2 – 24(2) + 9 = 36 – 48 + 9 = -3 \, \text{m/s} ###

To find the acceleration, differentiate the velocity function with respect to time:

### a(t) = \frac{d^2x}{dt^2} = 18t – 24 ###

Substitute \( t = 2 \) into the acceleration equation:

### a(2) = 18(2) – 24 = 36 – 24 = 12 \, \text{m/s}^2 ###

Key Takeaways

Understanding differentiation is crucial for solving kinematics problems. Always verify your calculations by re-substituting values to ensure accuracy.

Problem 2: Newton’s Laws

Problem Statement

A block of mass \( m = 5 \, \text{kg} \) is placed on a frictionless horizontal surface. A horizontal force \( F = 20 \, \text{N} \) is applied to the block. Find the acceleration of the block.

Concepts Involved

This problem applies Newton’s Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

### F = ma ###

Solution

Using Newton’s Second Law:

### a = \frac{F}{m} = \frac{20 \, \text{N}}{5 \, \text{kg}} = 4 \, \text{m/s}^2 ###

Key Takeaways

Always ensure that the forces are balanced and the surface is frictionless unless stated otherwise. This simplifies the problem and allows direct application of Newton’s Laws.

Problem 3: Work and Energy

Problem Statement

A force \( F = 5x^2 \) acts on a particle, where \( x \) is the displacement in meters. Calculate the work done by the force as the particle moves from \( x = 0 \) to \( x = 2 \) meters.

Concepts Involved

Work done by a variable force is calculated by integrating the force over the displacement.

### W = \int_{x_1}^{x_2} F(x) \, dx ###

Solution

Substitute the given force into the work formula:

### W = \int_{0}^{2} 5x^2 \, dx = 5 \left[ \frac{x^3}{3} \right]_{0}^{2} = 5 \left( \frac{8}{3} – 0 \right) = \frac{40}{3} \, \text{J} ###

Key Takeaways

Integration is a powerful tool for calculating work done by variable forces. Always double-check the limits of integration and the units.

Problem 4: Thermodynamics

Problem Statement

A gas expands from a volume of \( 1 \, \text{m}^3 \) to \( 3 \, \text{m}^3 \) at a constant pressure of \( 2 \times 10^5 \, \text{Pa} \). Calculate the work done by the gas.

Concepts Involved

Work done by a gas during expansion at constant pressure is given by \( W = P \Delta V \).

Solution

Calculate the change in volume:

### \Delta V = V_f – V_i = 3 \, \text{m}^3 – 1 \, \text{m}^3 = 2 \, \text{m}^3 ###

Now, calculate the work done:

### W = P \Delta V = 2 \times 10^5 \, \text{Pa} \times 2 \, \text{m}^3 = 4 \times 10^5 \, \text{J} ###

Key Takeaways

Understanding the relationship between pressure, volume, and work is essential in thermodynamics. Always ensure units are consistent.

Problem 5: Electrostatics

Problem Statement

Two point charges \( q_1 = 2 \, \mu\text{C} \) and \( q_2 = -3 \, \mu\text{C} \) are placed 0.5 meters apart. Find the force between them.

Concepts Involved

Coulomb’s Law states that the force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.

### F = k \frac{q_1 q_2}{r^2} ###

Solution

Substitute the given values into Coulomb’s Law:

### F = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \times \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{(0.5)^2} = 216 \, \text{N} ###

Key Takeaways

Always convert units to SI before substituting into formulas. The sign of the charges determines the direction of the force.

Problem 6: Optics

Problem Statement

A convex lens has a focal length of 20 cm. An object is placed 30 cm in front of the lens. Find the position and nature of the image formed.

Concepts Involved

The lens formula is used to determine the image position and nature. The formula is \( \frac{1}{f} = \frac{1}{v} – \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.

Solution

Substitute the given values into the lens formula:

### \frac{1}{20} = \frac{1}{v} – \frac{1}{-30} ###

### \frac{1}{v} = \frac{1}{20} + \frac{1}{30} = \frac{5}{60} = \frac{1}{12} ###

### v = 12 \, \text{cm} ###

The positive value of \( v \) indicates that the image is real and formed on the opposite side of the lens.

Key Takeaways

Understanding the sign convention is crucial in optics. A positive image distance indicates a real image, while a negative value indicates a virtual image.

Problem 7: Waves

Problem Statement

A wave has a frequency of 50 Hz and a wavelength of 2 meters. Find the speed of the wave.

Concepts Involved

The speed of a wave is given by the product of its frequency and wavelength.

### v = f \lambda ###

Solution

Substitute the given values into the wave speed formula:

### v = 50 \, \text{Hz} \times 2 \, \text{m} = 100 \, \text{m/s} ###

Key Takeaways

Always ensure that the units for frequency and wavelength are compatible. The speed of a wave depends on the medium it travels through.

Problem 8: Magnetism

Problem Statement

A current-carrying wire of length 0.5 meters is placed perpendicular to a magnetic field of 0.2 Tesla. If the current in the wire is 5 Amperes, find the force acting on the wire.

Concepts Involved

The force on a current-carrying wire in a magnetic field is given by \( F = ILB \), where \( I \) is the current, \( L \) is the length of the wire, and \( B \) is the magnetic field strength.

Solution

Substitute the given values into the force formula:

### F = 5 \, \text{A} \times 0.5 \, \text{m} \times 0.2 \, \text{T} = 0.5 \, \text{N} ###

Key Takeaways

Ensure that the wire is perpendicular to the magnetic field for the maximum force. The direction of the force can be determined using Fleming’s Left-Hand Rule.

Problem 9: Modern Physics

Problem Statement

Calculate the de Broglie wavelength of an electron moving with a velocity of \( 10^6 \, \text{m/s} \). The mass of the electron is \( 9.11 \times 10^{-31} \, \text{kg} \).

Concepts Involved

The de Broglie wavelength is given by \( \lambda = \frac{h}{p} \), where \( h \) is Planck’s constant and \( p \) is the momentum of the particle.

Solution

Calculate the momentum of the electron:

### p = mv = 9.11 \times 10^{-31} \, \text{kg} \times 10^6 \, \text{m/s} = 9.11 \times 10^{-25} \, \text{kg m/s} ###

Now, calculate the de Broglie wavelength:

### \lambda = \frac{6.626 \times 10^{-34} \, \text{J s}}{9.11 \times 10^{-25} \, \text{kg m/s}} = 7.27 \times 10^{-10} \, \text{m} ###

Key Takeaways

The de Broglie wavelength is significant in quantum mechanics and helps explain the wave-particle duality of matter. Always use consistent units in calculations.

Problem 10: Rotational Motion

Problem Statement

A solid cylinder of mass \( m = 2 \, \text{kg} \) and radius \( r = 0.1 \, \text{m} \) rolls down an inclined plane without slipping. Find the acceleration of the cylinder.

Concepts Involved

For rolling motion without slipping, the acceleration is given by \( a = \frac{g \sin \theta}{1 + \frac{I}{mr^2}} \), where \( I \) is the moment of inertia of the cylinder.

Solution

The moment of inertia for a solid cylinder is \( I = \frac{1}{2} mr^2 \). Substitute this into the acceleration formula:

### a = \frac{g \sin \theta}{1 + \frac{\frac{1}{2} mr^2}{mr^2}} = \frac{g \sin \theta}{1.5} ###

Assuming the plane is inclined at \( \theta = 30^\circ \):

### a = \frac{9.8 \, \text{m/s}^2 \times \sin 30^\circ}{1.5} = 3.27 \, \text{m/s}^2 ###

Key Takeaways

Understanding the moment of inertia and its role in rotational motion is crucial. Always consider the angle of inclination in problems involving inclined planes.