9709/Oct Nov/43/2011/Q4

ABC is a vertical cross-section of a surface. The part of the surface containing AB is smooth and A is 4m higher than B. The part of the surface containing BC is horizontal and the distance BC is 5m (see diagram). A particle of mass 0.8 kg is released from rest at A and slides along ABC. Find the speed of the particle at C in each of the following cases.

(i) The horizontal part of the surface is smooth.

(ii) The coefficient of friction between the particle and the horizontal part of the surface is 0.3.

Solution:

(i) 

mgh = 0.8 ×  10 × 4   = 32

For using ½ mv^2 = PE

[½ 0.8v^2 = 32]

Speed at C, v = 8.94 ms^–1

(ii) 

R = mg

= 0.8 x 10

8 N

Friction force, 

F = μ mg

   = 0.3 (8)

   = 2.4 N

Force from B to C = 2.4N (opposite direction of friction force)

Work done from B to C 

= 2.4 x 5

= 12 J

Conservation of Energy

½ mv^2 + 12 = PE

½ mv^2 = 32 - 12

Speed at C, v = 7.07 ms^-1

Reference: PYQ - Oct/Nov 2011 Paper 43 Q4

9709/Oct Nov/13/2016/Q8

(i) Express 4x² + 12x + 10 in the form (ax + b)² + c, where a, b and c are constants.

(ii) Functions f and g are both defined for x > 0. It is given that f(x) = x² + 1 and fg(x) = 4x² + 12x + 10. Find g(x).

(iii) Find (fg)⁻¹ (x) and give the domain of (fg)⁻¹.

Solution:

Reference: PYQ - Oct/Nov 2016 Paper 13 Q8

9709/May June/2011/42/Q6

A small smooth ring R, of mass 0.6 kg, is threaded on a light inextensible string of length 100 cm. One end of the string is attached to a fixed point A. A small bead B of mass 0.4 kg is attached to the other end of the string, and is threaded on a fixed rough horizontal rod which passes through A. The system is in equilibrium with B at a distance of 80 cm from A (see diagram).

(i) Find the tension in the string. 

(ii) Find the frictional and normal components of the contact force acting on B. 

(iii) Given that the equilibrium is limiting, find the coefficient of friction between the bead and the rod.

Solution:

Reference: PYQ - May/Jun 2011 Paper 42 Q6

9709/May June/2011/42/Q3

The velocity-time graph shown models the motion of a parachutist falling vertically. There are four
stages in the motion:
• falling freely with the parachute closed,
• decelerating at a constant rate with the parachute open,
• falling with constant speed with the parachute open,
• coming to rest instantaneously on hitting the ground.

(i) Show that the total distance fallen is 1048m.
The weight of the parachutist is 850N.
(ii) Find the upward force on the parachutist due to the parachute, during the second stage.

Solution:

Reference: PYQ - May/Jun 2011 Paper 42 Q3

9709/May June/2011/42/Q2

An object of mass 8 kg slides down a line of greatest slope of an inclined plane. Its initial speed at the top of the plane is 3ms−1 and its speed at the bottom of the plane is 8ms−1. The work done against the resistance to motion of the object is 120 J. Find the height of the top of the plane above the level of the bottom.

Solution:

Reference: PYQ - May/Jun 2011 Paper 42 Q2

9709/Oct Nov/2003/3/Q9

Compressed air is escaping from a container. The pressure of the air in the container at time t is P, and the constant atmospheric pressure of the air outside the container is A. The rate of decrease of P is proportional to the square root of the pressure difference (P − A). Thus the differential equation connecting P and t is

where k is a positive constant.
(i) Find, in any form, the general solution of this differential equation.
(ii) Given that P = 5A when t = 0, and that P = 2A when t = 2, show that k =√A.
(iii) Find the value of t when P = A.
(iv) Obtain an expression for P in terms of A and t.

Solution:

Reference: PYQ - Oct/Nov 2003 Paper 3 Q9

9709/May Jun/2008/3/Q10

The points A and B have position vectors, relative to the origin O, given by

OA = i + 2j + 3k  and  OB = 2i + j + 3k.

The line l has vector equation

r = (1 − 2t)i + (5 + t)j + (2 − t)k.

(i) Show that l does not intersect the line passing through A and B.

(ii) The point P lies on l and is such that angle PAB is equal to 60^o. Given that the position vector

of P is (1 − 2t)i + (5 + t)j + (2 − t)k, show that 3t² + 7t + 2 = 0. Hence find the only possible

position vector of P.

Solution:

Reference: PYQ - May/Jun 2008 Paper 3 Q10