Physics Topic 9: Motion In Fields

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

A straight line

A circle

A spiral

Refer to image

In the absence of air resistance, the trajectory of projectile motion follows a parabolic path due to the effects of gravity. The correct answer is a parabolic shape, not a straight line, circle, or spiral.

Explanation

In the absence of air resistance, the trajectory of projectile motion follows a parabolic path due to the effects of gravity. The correct answer is a parabolic shape, not a straight line, circle, or spiral.

2. 9.2.2 State and apply the expression for gravitational potential due to a point mass

It is inversely proportional to the square of the distance between objects

It is dependent on the weight of the object

The expression does not exist

Refer to image

The correct answer is 'Refer to image.' as the specific expression for gravitational potential due to a point mass may vary and is generally provided in a visual form for better understanding.

Explanation

The correct answer is 'Refer to image.' as the specific expression for gravitational potential due to a point mass may vary and is generally provided in a visual form for better understanding.

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

Gravitational FieldStrength (I)= - ?V / ?x

Gravitational Field Strength (I) = ∂V / ∂y

Gravitational Field Strength (I) = - ∂V / ∂y

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

The correct formula relating gravitational field strength to gravitational potential gradient is given by I = - ∂V / ∂x. This equation shows that gravitational field strength is equal to the negative of the rate of change of gravitational potential with respect to distance in the x-direction.

Explanation

The correct formula relating gravitational field strength to gravitational potential gradient is given by I = - ∂V / ∂x. This equation shows that gravitational field strength is equal to the negative of the rate of change of gravitational potential with respect to distance in the x-direction.

4. Determine the potential due to one or more point masses

V = G {(M1 / x) - (M2 / r-x)}

V = -G {(M1 / x) - (M2 / r-x)}

V = -G {(M1/ x) + (M2 / r-x)}

V = G {(M1 / x) + (M2 / r+x)}

The correct formula for determining the potential due to one or more point masses is V = -G {(M1 / x) + (M2 / r-x)}. This formula takes into account the gravitational constant (G) and the masses of the point masses (M1, M2) at distances x and r-x respectively.

Explanation

The correct formula for determining the potential due to one or more point masses is V = -G {(M1 / x) + (M2 / r-x)}. This formula takes into account the gravitational constant (G) and the masses of the point masses (M1, M2) at distances x and r-x respectively.

5. Describe and sketch the pattern of equipotential surfaces due to one and two point masses

A) There are no equipotential surfaces due to point masses

C) Equipotential surfaces are concentric circles around each point mass

Refer to image

B) The equipotential surfaces are linear in shape

Equipotential surfaces due to one or two point masses form concentric spheres around each mass, with the potential decreasing as the distance from the mass increases.

Explanation

Equipotential surfaces due to one or two point masses form concentric spheres around each mass, with the potential decreasing as the distance from the mass increases.

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

Gravitational lines and equipotential surfaces are perpendicular to each other

Gravitational lines ? equipotential surface

Equipotential surface → gravitational lines

There is no relationship between equipotential surfaces and gravitational field lines

Equipotential surfaces are surfaces where the gravitational potential remains constant. Gravitational field lines always point from high potential to low potential, so they are perpendicular to equipotential surfaces.

Explanation

Equipotential surfaces are surfaces where the gravitational potential remains constant. Gravitational field lines always point from high potential to low potential, so they are perpendicular to equipotential surfaces.

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

Vescape = sqrt(GM/R)

Vescape = sqrt(2GM/R)

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

Vescape = 2GM/R

To derive the expression for escape speed, we equate the kinetic energy (KE) to the potential energy (PE) of the object. By applying the conservation of energy principle, we arrive at the correct expression vescape = sqrt(2GM/R). The incorrect answers lack the square root operation necessary for the correct solution.

Explanation

To derive the expression for escape speed, we equate the kinetic energy (KE) to the potential energy (PE) of the object. By applying the conservation of energy principle, we arrive at the correct expression vescape = sqrt(2GM/R). The incorrect answers lack the square root operation necessary for the correct solution.

8. 9.3.2 State and apply the expression for electric potential due to a point charge

C) V = kq * r

D) V = kq * r^2

B) V = kq / r

A) V = kq / r^2

Refer to image

The correct formula for the electric potential due to a point charge is V = kq / r, where V is the electric potential, k is the Coulomb's constant, q is the charge of the point charge, and r is the distance from the point charge. The incorrect options provided do not correctly represent the expression for electric potential due to a point charge.

Explanation

The correct formula for the electric potential due to a point charge is V = kq / r, where V is the electric potential, k is the Coulomb's constant, q is the charge of the point charge, and r is the distance from the point charge. The incorrect options provided do not correctly represent the expression for electric potential due to a point charge.

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

E = ?V/?x

E = qV

Potential gradient: Work done to move a charge from one potential to another
= q?VW = F?xF?x = q?V= ?,?V-?x.

F = q?V

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

Derive the formula for electric potential energy

Analyze the electric field at a point due to a charge distribution

Calculate the force between two point charges

Refer to image

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

Refer to graph.

Refer to diagram.

Refer to table.

Refer to image

This question tests understanding of equipotential surfaces due to point charges, requiring visual representation for explanation.

Explanation

This question tests understanding of equipotential surfaces due to point charges, requiring visual representation for explanation.

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

Electric field lines run parallel to equipotential surfaces.

Electric field lines ? equipotential surface

Electric field lines are always tangent to equipotential surfaces.

Equipotential surfaces are always aligned with electric field lines.

The correct answer is that electric field lines intersect equipotential surfaces at right angles. This relationship helps in visualizing the electric field and potential distribution in a given space.

Explanation

The correct answer is that electric field lines intersect equipotential surfaces at right angles. This relationship helps in visualizing the electric field and potential distribution in a given space.

13. 9.4.4 Sketch graphs showing the variation with orbital radius of the kinetic energy, gravitational potential energy and total energy of a satellite.

Total energy of a satellite remains constant regardless of orbital radius.

Gravitational potential energy increases as the orbital radius decreases.

Refer to image

Orbital radius does not affect the kinetic energy of a satellite.