Ib Physics Hl Topic 9

1. State the independence of the vertical and the horizontal components of velocity for a projectile in a uniform field.

The vertical and horizontal components of velocity for a projectile in a uniform field are directly proportional to each other.

The vertical and horizontal components of velocity for a projectile in a uniform field are inversely proportional to each other.

Simple state objective.

The vertical and horizontal components of velocity for a projectile in a uniform field are dependent on the projectile's mass.

2. Describe and sketch the trajectory of projectile motion as parabolic in the absence of air resistance.

Define the trajectory of projectile motion as exponential in the absence of air resistance.

Refer to image.

Discuss the trajectory of projectile motion as circular in the absence of air resistance.

Explain the trajectory of projectile motion as linear in the absence of air resistance.

Projectile motion follows a parabolic trajectory due to the influence of gravity, where the path curves downwards in the shape of a parabola.

Explanation

Projectile motion follows a parabolic trajectory due to the influence of gravity, where the path curves downwards in the shape of a parabola.

3. Describe qualitatively the effect of air resistance on the trajectory of a projectile.

Effects include a shorter range and a lower maximum height attained during flight. Caused by drag force opposing projectile motion. Air resistance is this cause.

Air resistance causes the projectile to accelerate faster.

Air resistance increases the range and height reached by a projectile.

Air resistance has no impact on the trajectory of a projectile.

4. State and apply the expression for gravitational potential due to a point mass.

Gravitational Potential: V = GMe / r. Gravitational Potential Energy: PE = Mg?h

Gravitational Potential:V = W / m= -GMe / r Gravitational Potential Energy:PE = ?Ep= mg?h= m?V= -GMem / r

Gravitational Potential: V = -GMe / r. Gravitational Potential Energy: PE = mg?h

Gravitational Potential: V = GMe / r. Gravitational Potential Energy: PE = GMem / r

The gravitational potential due to a point mass is given by V = -GMe / r where Me is the mass of the Earth, r is the distance from the center of the Earth, and G is the gravitational constant. The gravitational potential energy is given by PE = -GMem / r.

Explanation

The gravitational potential due to a point mass is given by V = -GMe / r where Me is the mass of the Earth, r is the distance from the center of the Earth, and G is the gravitational constant. The gravitational potential energy is given by PE = -GMem / r.

5. State and apply the formula relating gravitational field strength to gravitational potential gradient.

Gravitational Field Strength (I) = ?V * ?x

Gravitational Field Strength (I) = - ?V * ?x

Gravitational Field Strength (I) = - ?V / ?x

Gravitational Field Strength (I) = ?V / ?x

The correct formula relating gravitational field strength to gravitational potential gradient is Gravitational Field Strength (I) = - ?V / ?x. The negative sign indicates that the field strength points in the direction of decreasing potential.

Explanation

The correct formula relating gravitational field strength to gravitational potential gradient is Gravitational Field Strength (I) = - ?V / ?x. The negative sign indicates that the field strength points in the direction of decreasing potential.

6. State the relation between equipotential surfaces and gravitational field lines.

Gravitational lines ? equipotential surface

Equipotential surfaces repel objects, while gravitational field lines attract objects.

Gravitational field lines are only present in outer space, while equipotential surfaces exist on Earth.

Gravitational field lines are always perpendicular to equipotential surfaces.

Equipotential surfaces are surfaces in a region where the gravitational potential is the same at every point. In the case of gravitational field lines, they represent the direction a free-falling test mass would take under the influence of gravity. The relation between them is that gravitational lines coincide with equipotential surfaces because a test mass experiences no change in potential energy when moving along a line of constant potential.

Explanation

Equipotential surfaces are surfaces in a region where the gravitational potential is the same at every point. In the case of gravitational field lines, they represent the direction a free-falling test mass would take under the influence of gravity. The relation between them is that gravitational lines coincide with equipotential surfaces because a test mass experiences no change in potential energy when moving along a line of constant potential.

7. Derive an expression for the escape speed of an object from the surface of a planet.

Vescape = 2√GM/R

|KE| = |PE||0.5mvescape2|= |-GMem/Re|vescape = ?2GM/Rvescape on earth = ?2g0Re
( g0 = GMe / Re2)

Vescape = 2GM/R

Vescape = √GM/R

The escape speed of an object from the surface of a planet is derived by equating the kinetic energy to the potential energy. The correct formula for escape speed is vescape = √2GM/R, where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet. The incorrect answers do not consider the square root of 2 in the formula, leading to incorrect calculations.

Explanation

The escape speed of an object from the surface of a planet is derived by equating the kinetic energy to the potential energy. The correct formula for escape speed is vescape = √2GM/R, where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet. The incorrect answers do not consider the square root of 2 in the formula, leading to incorrect calculations.

8. Define electric potential and electric potential energy.

3. Electric Potential (V) - the rate of flow of charges in an electric circuit

1. Electric Potential (V) - the heat generated when an object with electric charge is exposed to sunlight

1. Electric Potential (V) -work done per unit charge moving a small positive test charge in frominfinity to a point in an electric field. (V = W/q) (V = kq/r) (NOTE: the work done is pathindependent) 2. *Electric Potential Energy (Ee)- energy that a charge has due to its position in anelectric field

2. Electric Potential Energy (Ee) - the energy stored in a chemical battery

Electric potential and electric potential energy are fundamental concepts in the field of electromagnetism. Electric potential refers to the work done per unit charge in moving a charge in an electric field. On the other hand, electric potential energy is the energy associated with the position of a charge in an electric field. The incorrect answers provided do not accurately define these concepts.

Explanation

Electric potential and electric potential energy are fundamental concepts in the field of electromagnetism. Electric potential refers to the work done per unit charge in moving a charge in an electric field. On the other hand, electric potential energy is the energy associated with the position of a charge in an electric field. The incorrect answers provided do not accurately define these concepts.

9. State and apply the formula relating electric field strength to electric potential gradient.

Potential gradient: The rate of change of resistance with respect to electric field strength

Potential gradient: Electric field strength in the opposite direction of the potential gradient

Potential gradient: The ratio of current to resistance in an electric circuit

Potential gradient: Workdone to move a charge from one potential to another
= q?VW = F?xF?x = q?V

The correct formula for potential gradient is the work done to move a charge from one potential to another is equal to the charge multiplied by the change in electric potential.

Explanation

The correct formula for potential gradient is the work done to move a charge from one potential to another is equal to the charge multiplied by the change in electric potential.

10. 9.3.4 Determine the potential due to one or more point charges.

See image.

The potential due to point charges is always zero.

The potential due to point charges cannot be determined.

The potential due to point charges depends on the shape of the charges.

The potential due to point charges can be determined using appropriate mathematical formulas and principles.

Explanation

The potential due to point charges can be determined using appropriate mathematical formulas and principles.

11. 9.3.5 Describe and sketch the pattern of equipotential surfaces due to one and two point charges.

See image.

The equipotential surfaces are hyperbolic for one point charge and ellipses for two point charges.

The equipotential surfaces are irregular for one point charge and spirals for two point charges.

The equipotential surfaces are linear for one point charge and concentric circles for two point charges.

Equipotential surfaces are imaginary surfaces where every point on the surface has the same electrical potential. The correct answer describes the pattern of equipotential surfaces due to one and two point charges accurately.

Explanation

Equipotential surfaces are imaginary surfaces where every point on the surface has the same electrical potential. The correct answer describes the pattern of equipotential surfaces due to one and two point charges accurately.

12. 9.3.6 State the relation between equipotential surfaces and electric field lines.

Parallel

Coincident

Tangential

Perpendicular.

Equipotential surfaces are surfaces along which the electric potential is constant while electric field lines represent the direction of the force that a positive test charge would experience at any given point. The relation between equipotential surfaces and electric field lines is that they are perpendicular to each other, meaning that the electric field lines at any point on an equipotential surface are always at right angles to the surface.

Explanation

Equipotential surfaces are surfaces along which the electric potential is constant while electric field lines represent the direction of the force that a positive test charge would experience at any given point. The relation between equipotential surfaces and electric field lines is that they are perpendicular to each other, meaning that the electric field lines at any point on an equipotential surface are always at right angles to the surface.

13. State the role of gravitation in circular orbital motion.

Causes the object to move in a straight line.

Has no effect on the direction of motion.

State.

Creates the tangential velocity of the object.

14. 9.4.4 Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite.

See image.

Diagrams showing the variation with orbital radius of the color, texture, and shape of a satellite.

Graphs showing the variation with orbital radius of the temperature, atmospheric pressure, and wind speed of a satellite.

Tables showing the variation with orbital radius of the mass, volume, and density of a satellite.

The question is specifically asking about the variation of kinetic energy, gravitational potential energy, and total energy with orbital radius of a satellite. The correct answer should involve graphs depicting these variations.

Explanation

The question is specifically asking about the variation of kinetic energy, gravitational potential energy, and total energy with orbital radius of a satellite. The correct answer should involve graphs depicting these variations.