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MATSEC IM Applied Mathematics study materials
MATSEC IM Applied Mathematics study resources

MATSEC IM Applied Mathematics practice questions

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Question 1Applied Mathematics (Intermediate Matriculation level)

A particle moves along a straight line so that its displacement s metres after t seconds is s = 3t² − 2t. What is its velocity when t = 4 s?

  1. 46 m/s
  2. 22 m/s Correct answer.
  3. 24 m/s
  4. 10 m/s

Explanation

Velocity is the derivative of displacement with respect to time: ds/dt = 6t − 2. Substituting t = 4 gives 24 − 2 = 22 m/s. The displacement value itself (40 m) is not the velocity.

Question 2Applied Mathematics (Intermediate Matriculation level)

A body moves in a straight line with velocity v = 4t − t² m/s for 0 ≤ t ≤ 4. At what time does the body momentarily stop?

  1. t = 4 s only
  2. t = 1 s only
  3. t = 2 s only Correct answer.
  4. t = 0 s and t = 4 s

Explanation

The body is momentarily at rest when v = 0, i.e. 4t − t² = t(4 − t) = 0. This gives t = 0 and t = 4, but t = 0 is the start of the interval where motion begins; within the interval the body reverses direction at t = 2? Careful: solving gives t = 0 or t = 4.

Question 3Applied Mathematics (Intermediate Matriculation level)

A constant force of 12 N acts on a mass of 4 kg on a smooth horizontal surface. What is the acceleration of the mass?

  1. 16 m/s²
  2. 48 m/s²
  3. 0.33 m/s²
  4. 3 m/s² Correct answer.

Explanation

Newton's second law gives F = ma, so a = F/m = 12/4 = 3 m/s². The surface being smooth means no friction needs to be subtracted.

Question 4Applied Mathematics (Intermediate Matriculation level)

A box of mass 5 kg rests on a rough horizontal floor with coefficient of friction μ = 0.4. Taking g = 10 m/s², what is the minimum horizontal force needed to just start it moving?

  1. 20 N Correct answer.
  2. 2 N
  3. 50 N
  4. 12.5 N

Explanation

The normal reaction equals the weight, R = mg = 5 × 10 = 50 N. The limiting friction is F = μR = 0.4 × 50 = 20 N, which is the minimum applied force to overcome it.

Question 5Applied Mathematics (Intermediate Matriculation level)

A projectile is launched with speed 20 m/s at 30° above the horizontal. Taking g = 10 m/s² and ignoring air resistance, what is the horizontal component of its initial velocity?

  1. 17.3 m/s Correct answer.
  2. 10 m/s
  3. 20 m/s
  4. 5 m/s

Explanation

The horizontal component is u cos θ = 20 × cos 30° = 20 × 0.866 ≈ 17.3 m/s. The vertical component would be u sin θ = 10 m/s.

Question 6Applied Mathematics (Intermediate Matriculation level)

For the projectile launched at 20 m/s at 30° to the horizontal, taking g = 10 m/s², what is the time of flight over level ground?

  1. 4 s
  2. 3.46 s
  3. 1 s
  4. 2 s Correct answer.

Explanation

Time of flight over level ground is T = 2u sin θ / g = 2 × 20 × 0.5 / 10 = 2 s. The vertical velocity component is 10 m/s, so ascent takes 1 s and descent another 1 s.

Question 7Applied Mathematics (Intermediate Matriculation level)

A force of 50 N acts at 60° to a displacement of 4 m along a horizontal line. What is the work done by the force?

  1. 50 J
  2. 173 J
  3. 200 J
  4. 100 J Correct answer.

Explanation

Work done is W = Fd cos θ = 50 × 4 × cos 60° = 200 × 0.5 = 100 J. Only the component of force along the displacement does work.

Question 8Applied Mathematics (Intermediate Matriculation level)

A car of mass 800 kg accelerates uniformly from rest to 20 m/s. What is its gain in kinetic energy?

  1. 8 000 J
  2. 320 000 J
  3. 160 000 J Correct answer.
  4. 16 000 J

Explanation

Kinetic energy is ½mv² = ½ × 800 × 20² = 400 × 400 = 160 000 J. Starting from rest means the gain equals the final kinetic energy.

Question 9Applied Mathematics (Intermediate Matriculation level)

A ball of mass 0.2 kg hits a wall horizontally at 10 m/s and rebounds horizontally at 8 m/s. What is the magnitude of the change in momentum?

  1. 2.0 kg m/s
  2. 3.6 kg m/s Correct answer.
  3. 0.4 kg m/s
  4. 1.6 kg m/s

Explanation

Taking the initial direction as positive, initial momentum is 0.2 × 10 = 2.0 and final is 0.2 × (−8) = −1.6. The change is −1.6 − 2.0 = −3.6, so the magnitude is 3.6 kg m/s.

Question 10Applied Mathematics (Intermediate Matriculation level)

Two particles of masses 3 kg and 2 kg move towards each other at 4 m/s and 6 m/s respectively and stick together. What is their common velocity after collision?

  1. 2 m/s in the direction of the 3 kg mass
  2. 5 m/s
  3. 1 m/s in the direction of the 2 kg mass
  4. 0 m/s Correct answer.

Explanation

Taking the 3 kg direction as positive: total momentum = 3×4 + 2×(−6) = 12 − 12 = 0. Since total momentum is zero and they coalesce, the common velocity is 0 m/s.

Question 11Applied Mathematics (Intermediate Matriculation level)

A particle of mass 2 kg moves in a circle of radius 3 m at a constant speed of 6 m/s. What is the centripetal force acting on it?

  1. 12 N
  2. 72 N
  3. 24 N Correct answer.
  4. 6 N

Explanation

Centripetal force is F = mv²/r = 2 × 36 / 3 = 24 N. It acts towards the centre of the circular path.

Question 12Applied Mathematics (Intermediate Matriculation level)

A particle moves in a circle of radius 2 m with angular speed 3 rad/s. What is its linear speed?

  1. 9 m/s
  2. 1.5 m/s
  3. 6 m/s Correct answer.
  4. 0.67 m/s

Explanation

Linear speed is v = rω = 2 × 3 = 6 m/s. Angular speed in rad/s multiplied by radius gives the tangential speed.

Question 13Applied Mathematics (Intermediate Matriculation level)

A force of 30 N acts at a perpendicular distance of 0.5 m from a pivot. What is the moment of the force about the pivot?

  1. 30.5 N m
  2. 60 N m
  3. 0.017 N m
  4. 15 N m Correct answer.

Explanation

The moment of a force is F × d = 30 × 0.5 = 15 N m. The distance must be measured perpendicular to the line of action of the force.

Question 14Applied Mathematics (Intermediate Matriculation level)

A uniform rod of length 4 m is pivoted at its centre. A 10 N weight hangs 1 m to the left of the pivot and a force F hangs 2 m to the right. For balance, what is F?

  1. 10 N
  2. 20 N
  3. 2.5 N
  4. 5 N Correct answer.

Explanation

Taking moments about the pivot: 10 × 1 = F × 2, so F = 5 N. The rod being uniform and pivoted at its centre means its own weight produces no moment.

Question 15Applied Mathematics (Intermediate Matriculation level)

A body of weight 60 N is placed on a rough plane inclined at 30° to the horizontal. What is the component of its weight acting down the plane?

  1. 60 N
  2. 30 N Correct answer.
  3. 52 N
  4. 17.3 N

Explanation

The component along the plane is W sin θ = 60 × sin 30° = 30 N. The component perpendicular to the plane is W cos θ ≈ 52 N.

Question 16Applied Mathematics (Intermediate Matriculation level)

A force of 8 N and a force of 6 N act at right angles to each other at a point. What is the magnitude of their resultant?

  1. 2 N
  2. 14 N
  3. 10 N Correct answer.
  4. 48 N

Explanation

For perpendicular forces the resultant is √(8² + 6²) = √(64 + 36) = √100 = 10 N. This is the Pythagorean combination of the two components.

Question 17Applied Mathematics (Intermediate Matriculation level)

A particle of mass 3 kg hangs in equilibrium from two light strings, each making 45° with the vertical. Taking g = 10 m/s², what is the tension in each string?

  1. 15 N
  2. 21.2 N Correct answer.
  3. 30 N
  4. 42.4 N

Explanation

By symmetry each tension T is equal. Vertical equilibrium gives 2T cos 45° = mg = 30, so T = 30 / (2 × 0.707) ≈ 21.2 N.

Question 18Applied Mathematics (Intermediate Matriculation level)

The displacement of a particle is s = t³ − 6t² + 9t metres. What is its acceleration when t = 2 s?

  1. 0 m/s² Correct answer.
  2. 12 m/s²
  3. −6 m/s²
  4. 6 m/s²

Explanation

Velocity is ds/dt = 3t² − 12t + 9, and acceleration is dv/dt = 6t − 12. At t = 2, acceleration = 12 − 12 = 0 m/s².

Question 19Applied Mathematics (Intermediate Matriculation level)

A particle moves with velocity v = 2t + 1 m/s. If its displacement is zero at t = 0, what is its displacement at t = 3 s?

  1. 6 m
  2. 12 m Correct answer.
  3. 21 m
  4. 9 m

Explanation

Displacement is the integral of velocity: s = ∫(2t + 1)dt = t² + t + C. With s = 0 at t = 0, C = 0, so s(3) = 9 + 3 = 12 m.

Question 20Applied Mathematics (Intermediate Matriculation level)

A force F = 3i + 4j newtons moves a particle through a displacement d = 2i + j metres. What is the work done by the force?

  1. 14 J
  2. 10 J Correct answer.
  3. 4 J
  4. 6 J

Explanation

Work is the scalar product F·d = (3)(2) + (4)(1) = 6 + 4 = 10 J. The scalar product multiplies corresponding components and adds.

Question 21Applied Mathematics (Intermediate Matriculation level)

A block slides down a smooth incline of height 5 m from rest. Taking g = 10 m/s², what is its speed at the bottom?

  1. 10 m/s Correct answer.
  2. 25 m/s
  3. 50 m/s
  4. 5 m/s

Explanation

By conservation of energy, ½mv² = mgh, so v = √(2gh) = √(2 × 10 × 5) = √100 = 10 m/s. The mass cancels and the smooth surface means no energy is lost.

Question 22Applied Mathematics (Intermediate Matriculation level)

A particle of mass 0.5 kg is attached to a string of length 1 m and whirled in a vertical circle. At the lowest point its speed is 6 m/s. Taking g = 10 m/s², what is the tension in the string there?

  1. 13 N
  2. 18 N
  3. 5 N
  4. 23 N Correct answer.

Explanation

At the lowest point, T − mg = mv²/r, so T = mg + mv²/r = 5 + (0.5 × 36)/1 = 5 + 18 = 23 N. Both weight and centripetal requirement act along the string.

Question 23Applied Mathematics (Intermediate Matriculation level)

A 2 kg mass moving at 5 m/s collides with a stationary 3 kg mass. After the collision the 2 kg mass moves at 1 m/s in the same direction. What is the velocity of the 3 kg mass?

  1. 4 m/s
  2. 2.67 m/s Correct answer.
  3. 1.5 m/s
  4. 3 m/s

Explanation

Conservation of momentum: 2×5 + 3×0 = 2×1 + 3v, so 10 = 2 + 3v, giving v = 8/3 ≈ 2.67 m/s. The total momentum before and after must be equal.

Question 24Applied Mathematics (Intermediate Matriculation level)

A particle moves so that its position vector is r = (2t)i + (t²)j metres. What is the magnitude of its velocity at t = 2 s?

  1. 4 m/s
  2. 2 m/s
  3. √5 m/s
  4. 2√5 m/s Correct answer.

Explanation

Velocity is dr/dt = 2i + 2tj. At t = 2 this is 2i + 4j, whose magnitude is √(4 + 16) = √20 = 2√5 m/s.

Question 25Applied Mathematics (Intermediate Matriculation level)

A force of 20 N acts on a body and its line of action passes through the pivot point. What is the moment of this force about the pivot?

  1. 10 N m
  2. 0 N m Correct answer.
  3. 40 N m
  4. 20 N m

Explanation

The moment is F × d where d is the perpendicular distance from the pivot to the line of action. If the line of action passes through the pivot, d = 0, so the moment is zero.

Question 26Applied Mathematics (Intermediate Matriculation level)

A particle moves in a straight line with constant acceleration 3 m/s². Its initial velocity is 4 m/s. What is its velocity after 5 s?

  1. 15 m/s
  2. 19 m/s Correct answer.
  3. 23 m/s
  4. 35 m/s

Explanation

For constant acceleration, v = u + at. Substituting gives v = 4 + 3 × 5 = 19 m/s. The acceleration adds 3 m/s of speed for each second, so over 5 s the speed rises by 15 m/s from 4 m/s.

Question 27Applied Mathematics (Intermediate Matriculation level)

A particle starts from rest and moves with uniform acceleration. It travels 18 m in the first 3 s. What is the magnitude of its acceleration?

  1. 2 m/s²
  2. 6 m/s²
  3. 4 m/s² Correct answer.
  4. 12 m/s²

Explanation

Using s = ut + ½at² with u = 0 gives 18 = ½a(3²) = 4.5a. Hence a = 18 ÷ 4.5 = 4 m/s². Checking: ½ × 4 × 9 = 18 m, which matches the given distance.

Question 28Applied Mathematics (Intermediate Matriculation level)

A particle moves with simple harmonic motion of amplitude 0.05 m and angular frequency 4 rad/s. What is the magnitude of its maximum acceleration?

  1. 0.20 m/s²
  2. 1.60 m/s²
  3. 0.05 m/s²
  4. 0.80 m/s² Correct answer.

Explanation

In simple harmonic motion the maximum acceleration is ω²A. Here ω² = 16 and A = 0.05, giving 16 × 0.05 = 0.80 m/s². This maximum occurs at the extremes of the motion where displacement is greatest.

Question 29Applied Mathematics (Intermediate Matriculation level)

A body of mass 5 kg rests on a rough horizontal surface with coefficient of friction 0.4. Taking g = 10 m/s², what horizontal force is needed to keep it moving at constant speed?

  1. 50 N
  2. 20 N Correct answer.
  3. 2 N
  4. 10 N

Explanation

At constant speed the applied force balances the limiting friction. The normal reaction equals the weight, 5 × 10 = 50 N, so friction is μN = 0.4 × 50 = 20 N. Hence a horizontal force of 20 N keeps the body moving uniformly.

Question 30Applied Mathematics (Intermediate Matriculation level)

A particle is projected horizontally at 8 m/s from a cliff 45 m high. Taking g = 10 m/s², how far from the base of the cliff does it land?

  1. 12 m
  2. 24 m Correct answer.
  3. 18 m
  4. 36 m

Explanation

Vertical motion from rest: 45 = ½ × 10 × t², so t² = 9 and t = 3 s. Horizontal velocity stays constant at 8 m/s, so the range is 8 × 3 = 24 m. The two components are treated independently.

Question 31Applied Mathematics (Intermediate Matriculation level)

A particle moves along a straight line so that its velocity is v = 6t − 3 m/s. At what time does the particle reverse its direction of motion?

  1. t = 2 s
  2. t = 0.5 s Correct answer.
  3. t = 3 s
  4. t = 6 s

Explanation

Reversal occurs when the velocity passes through zero. Setting 6t − 3 = 0 gives t = 0.5 s. Before this time v is negative and afterwards positive, so the direction changes at that instant.

Question 32Applied Mathematics (Intermediate Matriculation level)

A light inextensible string passes over a smooth pulley with masses 4 kg and 2 kg hanging on either side. Taking g = 10 m/s², what is the acceleration of the system?

  1. 6.67 m/s²
  2. 3.33 m/s² Correct answer.
  3. 10 m/s²
  4. 1.67 m/s²

Explanation

For an Atwood machine, a = (m₁ − m₂)g/(m₁ + m₂). Substituting gives (4 − 2) × 10 ÷ 6 = 20/6 ≈ 3.33 m/s². The heavier mass descends and the lighter rises with this common acceleration.

Question 33Applied Mathematics (Intermediate Matriculation level)

A force of 12 N acts on a body of mass 3 kg for 4 s. What is the impulse imparted to the body?

  1. 16 N s
  2. 3 N s
  3. 48 N s Correct answer.
  4. 36 N s

Explanation

Impulse equals force multiplied by the time for which it acts, so J = 12 × 4 = 48 N s. By the impulse–momentum theorem this also equals the change in momentum of the body.

Question 34Applied Mathematics (Intermediate Matriculation level)

Two forces of magnitudes 5 N and 12 N act at a point with an angle of 90° between them. What is the magnitude of their resultant?

  1. 17 N
  2. 13 N Correct answer.
  3. 8.5 N
  4. 7 N

Explanation

For perpendicular forces the resultant is found from Pythagoras: R = √(5² + 12²) = √169 = 13 N. The 5-12-13 triple confirms the arithmetic.

Question 35Applied Mathematics (Intermediate Matriculation level)

A particle is in equilibrium under three coplanar forces. Two of them are 6 N acting due east and 8 N acting due north. What is the magnitude of the third force?

  1. 10 N Correct answer.
  2. 14 N
  3. 7 N
  4. 2 N

Explanation

For equilibrium the third force must exactly cancel the resultant of the other two. The resultant of perpendicular 6 N and 8 N forces is √(36 + 64) = 10 N, so the third force has magnitude 10 N and points in the opposite direction.

Question 36Applied Mathematics (Intermediate Matriculation level)

A uniform beam of weight 200 N is supported at two points 3 m apart, with its centre of gravity midway between them. What is the reaction at each support?

  1. 200 N
  2. 50 N
  3. 400 N
  4. 100 N Correct answer.

Explanation

By symmetry the centre of gravity lies midway between the supports, so each carries half the weight. Taking moments about one support gives the other reaction as 100 N, and vertical equilibrium gives 100 N for the first.

Question 37Applied Mathematics (Intermediate Matriculation level)

A block of mass 2 kg is pulled 5 m along a rough horizontal floor by a horizontal force of 15 N against a constant friction force of 5 N. What is the net work done on the block?

  1. 75 J
  2. 25 J
  3. 100 J
  4. 50 J Correct answer.

Explanation

The net horizontal force is 15 − 5 = 10 N. Net work is this force times the displacement, 10 × 5 = 50 J. Equivalently, the applied force does 75 J and friction removes 25 J.

Question 38Applied Mathematics (Intermediate Matriculation level)

A pump raises 300 kg of water through a height of 10 m in 20 s. Taking g = 10 m/s², what useful power does the pump deliver?

  1. 3000 W
  2. 150 W
  3. 600 W
  4. 1500 W Correct answer.

Explanation

The work done is mgh = 300 × 10 × 10 = 30 000 J. Power is work divided by time, 30 000 ÷ 20 = 1500 W. This is the useful output, before any losses in the pump.

Question 39Applied Mathematics (Intermediate Matriculation level)

A body of mass 4 kg moving at 6 m/s collides head-on with a stationary body of mass 2 kg. They stick together. What is their common velocity after the collision?

  1. 3 m/s
  2. 4 m/s Correct answer.
  3. 2 m/s
  4. 6 m/s

Explanation

Momentum is conserved: 4 × 6 = (4 + 2)v, so 24 = 6v and v = 4 m/s. The combined mass moves in the original direction of the 4 kg body.

Question 40Applied Mathematics (Intermediate Matriculation level)

A particle of mass 0.5 kg moves in a horizontal circle of radius 2 m at a constant speed of 4 m/s. What is the centripetal acceleration?

  1. 16 m/s²
  2. 8 m/s² Correct answer.
  3. 2 m/s²
  4. 4 m/s²

Explanation

Centripetal acceleration is v²/r = 4² ÷ 2 = 8 m/s². It is directed towards the centre of the circle, even though the speed is constant.

Question 41Applied Mathematics (Intermediate Matriculation level)

A particle moves in a circle of radius 0.5 m with a period of 2 s. What is its angular speed?

  1. 4π rad/s
  2. 2π rad/s
  3. π rad/s Correct answer.
  4. 0.5π rad/s

Explanation

Angular speed is ω = 2π/T. With T = 2 s this gives ω = 2π ÷ 2 = π rad/s. The radius does not affect the angular speed, only the linear speed.

Question 42Applied Mathematics (Intermediate Matriculation level)

A particle moves in a straight line so that its displacement is s = 2t³ − 9t² + 12t metres. At what time is the particle momentarily at rest?

  1. t = 0 s and t = 3 s
  2. t = 1.5 s only
  3. t = 2 s only
  4. t = 1 s and t = 2 s Correct answer.

Explanation

Velocity is ds/dt = 6t² − 18t + 12. Setting this to zero and dividing by 6 gives t² − 3t + 2 = 0, so (t − 1)(t − 2) = 0 and the particle is at rest at t = 1 s and t = 2 s.

Question 43Applied Mathematics (Intermediate Matriculation level)

A particle of mass 2 kg is attached to a light string of length 1.5 m and moves in a horizontal circle at 3 m/s. What is the tension in the string?

  1. 4 N
  2. 12 N Correct answer.
  3. 18 N
  4. 6 N

Explanation

The tension provides the centripetal force, T = mv²/r = 2 × 3² ÷ 1.5 = 18 ÷ 1.5 = 12 N. For horizontal circular motion on a smooth table the weight is balanced by the normal reaction.

Question 44Applied Mathematics (Intermediate Matriculation level)

A body is projected at 25 m/s at an angle of 37° above the horizontal. Taking sin 37° = 0.6 and cos 37° = 0.8, what is the maximum height reached?

  1. 15 m
  2. 7.5 m
  3. 20 m
  4. 11.25 m Correct answer.

Explanation

The vertical component of velocity is 25 × 0.6 = 15 m/s. Using v² = u² − 2gh with v = 0 gives h = 15² ÷ (2 × 10) = 225 ÷ 20 = 11.25 m.

Question 45Applied Mathematics (Intermediate Matriculation level)

A particle moves under the action of a single constant force. Which statement about its path is necessarily true?

  1. Its displacement is proportional to time.
  2. Its velocity is always parallel to the force.
  3. Its acceleration is constant in both magnitude and direction. Correct answer.
  4. Its speed remains constant throughout the motion.

Explanation

Newton's second law gives a = F/m. A constant force on a body of fixed mass therefore produces a constant acceleration vector, so both magnitude and direction are fixed. The velocity need not be parallel to the force, as in projectile motion.

Question 46Applied Mathematics (Intermediate Matriculation level)

Two forces act at a point: P = 4i − 3j newtons and Q = −2i + 5j newtons. What is the magnitude of their resultant?

  1. √10 N
  2. √20 N
  3. √8 N Correct answer.
  4. √26 N

Explanation

Adding the components gives R = (4 − 2)i + (−3 + 5)j = 2i + 2j. Its magnitude is √(2² + 2²) = √8 N. Components in the same direction are added and opposite ones subtracted.

Question 47Applied Mathematics (Intermediate Matriculation level)

A uniform rod of length 2 m and weight 60 N is hinged at one end and held horizontally by a vertical string attached to the other end. What is the tension in the string?

  1. 60 N
  2. 30 N Correct answer.
  3. 120 N
  4. 15 N

Explanation

The weight of a uniform rod acts at its midpoint, 1 m from the hinge. Taking moments about the hinge: T × 2 = 60 × 1, so T = 30 N. The hinge then supplies the remaining 30 N vertically.

Question 48Applied Mathematics (Intermediate Matriculation level)

A particle of mass 3 kg is lifted vertically at a constant speed of 2 m/s. Taking g = 10 m/s², what is the power delivered by the lifting force?

  1. 6 W
  2. 120 W
  3. 30 W
  4. 60 W Correct answer.

Explanation

At constant speed the lifting force equals the weight, 3 × 10 = 30 N. Power is force times speed, 30 × 2 = 60 W. No net force is needed because the speed does not change.

Question 49Applied Mathematics (Intermediate Matriculation level)

A particle moves with velocity v = 3t² − 4t m/s and its displacement is zero at t = 0. What is its displacement at t = 2 s?

  1. 0 m Correct answer.
  2. 8 m
  3. −2 m
  4. 4 m

Explanation

Integrating gives s = t³ − 2t² + C, and s(0) = 0 means C = 0. At t = 2, s = 8 − 8 = 0 m, so the particle returns to its starting point even though it has been moving.

Question 50Applied Mathematics (Intermediate Matriculation level)

A body slides down a smooth plane inclined at 30° to the horizontal. Taking g = 10 m/s², what is its acceleration along the plane?

  1. 8.66 m/s²
  2. 2.5 m/s²
  3. 10 m/s²
  4. 5 m/s² Correct answer.

Explanation

The component of gravity along the plane is g sin θ. With θ = 30° this is 10 × 0.5 = 5 m/s². The normal reaction cancels the perpendicular component of the weight.