9702/Oct Nov/42/2015/Q1

Solution:

Reference: Past Exam Paper – Oct Nov 2015 Paper 42 Q1

9702/May June/4/2003/Q1

(a) Define gravitational potential.

(b) Explain why values of gravitational potential near to an isolated mass are all negative.

(c) Earth may be assumed to be an isolated sphere of radius 6.4 × 10³ km with its mass of 6.0 × 10²⁴ kg concentrated at its centre. An object is projected vertically from the surface of the Earth so that it reaches an altitude of 1.3 × 10⁴ km.

Calculate, for this object,

(i) change in gravitational potential,

(ii) speed of projection from the Earth’s surface, assuming air resistance is negligible.

(d) Suggest why the equation

v² = u² + 2as

is not appropriate for calculation in (c)(ii).

Solution:

(a) Gravitational potential (at a point) is defined as the work done in bringing/moving unit mass from infinity to the point.

(b) The potential at infinity is defined as being zero. The forces are always attractive, so work got out in moving to point (work is done on the mass when moving it to infinity).

(c)(i)

Gravitational potential, φ = - GM / R = - GM × (1/R)

{The distances should be converted in metre. Initial position is at the surface of the Earth, which is a distance of 6.4×10⁶m from the centre of the Earth. The final distance is 1.3×10⁷m (altitude = distance above surface) + 6.4×10⁶m (distance of surface of the centre) = 1.94×10⁷m}

Change in potential = (6.67×10^-11) (6.0×10²⁴) × ({6.4×10⁶}^-1 – {1.94×10⁷}^-1)

Change in potential = 4.19 × 10⁷ J kg^-1 (ignore sign)

(ii)

The kinetic energy is converted to gravitational potential energy as the height of the object from the surface of the Earth increases.

½ mv² = mΔφ

v² = 2 × 4.19×10⁷ = 8.38 × 10⁷

Speed v = 9150 m s^-1

(d) The acceleration is not constant.

Reference: Past Exam Paper – June 2003 Paper 4 Q1

9702/Oct Nov/11/2010/Q13

A rigid L-shaped lever arm is pivoted at point P.

Three forces act on the lever arm, as shown in the diagram.

What is the magnitude of the resultant moment of these forces about point P?

A 15 N m 

B 20 N m 

C 35 N m 

D 75 N m

Answer: A.

We need to consider which forces will cause a clockwise moment and which cause an anticlockwise moment, that is consider how the lever would rotate about P due to each force.

Clockwise moment is caused by the 5N upward and 10N horizontal forces.

Anticlockwise moment is caused by the 15N downward force.

As for the distances, we need to consider the perpendicular distances from the line of action of the forces from point P.

Perpendicular distance of 5N force: (3m + 1m) – 2m = 2m

Perpendicular distance of 10N force: 2m

Perpendicular distance of 15N force: 3m

Magnitude of resultant moment = Anticlockwise moment – Clockwise moment

Magnitude of resultant moment = (15 × 3) – [(10 × 2) + (5 × 2)] = 15 N m

Reference: PYQ - Oct/Nov 2010 Paper 11 Q13