I85PREP
⌘ K
MATSEC AM Applied Mathematics study materials
MATSEC AM Applied Mathematics study resources

MATSEC AM Applied Mathematics practice questions

One question at a time. Build your MATSEC AM Applied Mathematics understanding.

First 50 of 199 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with constant acceleration. Its velocity increases from 4 m/s to 16 m/s in 3 s. What is the magnitude of the acceleration?

  1. 20 m/s²
  2. 6 m/s²
  3. 4 m/s² Correct answer.
  4. 12 m/s²

Explanation

Acceleration is the rate of change of velocity, so a = (v − u)/t = (16 − 4)/3 = 4 m/s². The change in velocity is 12 m/s over 3 s, giving 4 m/s².

Question 2Mechanics: kinematics and dynamics of a particle

A body starts from rest and moves with uniform acceleration of 3 m/s². How far does it travel in the first 4 s?

  1. 6 m
  2. 12 m
  3. 48 m
  4. 24 m Correct answer.

Explanation

Using s = ut + ½at² with u = 0 gives s = ½ × 3 × 16 = 24 m. The initial velocity is zero because the body starts from rest.

Question 3Mechanics: kinematics and dynamics of a particle

Which quantity is a vector in particle kinematics?

  1. Speed
  2. Displacement Correct answer.
  3. Distance travelled
  4. Time elapsed

Explanation

Displacement has both magnitude and direction, so it is a vector. Speed, distance and time are all scalars with magnitude only.

Question 4Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with velocity v = 2t + 5 (SI units). What is its acceleration at t = 3 s?

  1. 11 m/s²
  2. 5 m/s²
  3. 6 m/s²
  4. 2 m/s² Correct answer.

Explanation

Acceleration is the derivative of velocity with respect to time, so a = dv/dt = 2 m/s². This is constant, independent of t, so it equals 2 m/s² at t = 3 s.

Question 5Mechanics: kinematics and dynamics of a particle

A particle of mass 4 kg experiences a resultant force of 10 N. What is its acceleration?

  1. 0.4 m/s²
  2. 14 m/s²
  3. 40 m/s²
  4. 2.5 m/s² Correct answer.

Explanation

Newton's second law gives a = F/m = 10/4 = 2.5 m/s². The resultant force is divided by the mass, not multiplied.

Question 6Mechanics: kinematics and dynamics of a particle

A body moves in a straight line with constant velocity. What is the resultant force acting on it?

  1. Equal to its weight
  2. Increasing with time
  3. Constant but non-zero
  4. Zero Correct answer.

Explanation

Constant velocity means zero acceleration, so by Newton's second law the resultant force must be zero. The forces present, such as weight and any normal reaction, balance exactly.

Question 7Mechanics: kinematics and dynamics of a particle

A ball is thrown vertically upwards with speed 20 m/s. Taking g = 10 m/s² and ignoring air resistance, what is the maximum height reached?

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

Explanation

At the highest point the velocity is zero, so using v² = u² − 2gs gives 0 = 400 − 20s, hence s = 20 m. The acceleration is g directed downwards throughout the motion.

Question 8Mechanics: kinematics and dynamics of a particle

A particle moves with constant speed in a circle. Which statement about its acceleration is correct?

  1. The acceleration is directed towards the centre of the circle Correct answer.
  2. The acceleration is directed away from the centre of the circle
  3. The acceleration is directed along the tangent to the circle
  4. The acceleration is zero because the speed is constant

Explanation

Even at constant speed the velocity direction changes, so there is a centripetal acceleration directed towards the centre of the circle, of magnitude v²/r. The speed being constant does not mean the velocity is constant.

Question 9Mechanics: kinematics and dynamics of a particle

A particle moves along the x-axis with velocity v = 3t² − 4t (SI units). What is its displacement between t = 0 and t = 2 s?

  1. 16 m
  2. 0 m Correct answer.
  3. 4 m
  4. 8 m

Explanation

Displacement is the integral of velocity, so s = ∫(3t² − 4t)dt = t³ − 2t². Evaluating from 0 to 2 gives 8 − 8 = 0 m, meaning the particle returns to its starting point.

Question 10Mechanics: kinematics and dynamics of a particle

A particle is projected horizontally at 15 m/s from a point 20 m above level ground. Taking g = 10 m/s², how long does it take to reach the ground?

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

Explanation

Vertical and horizontal motions are independent. Vertically u = 0, so 20 = ½ × 10 × t², giving t² = 4 and t = 2 s. The horizontal speed affects only the range, not the time of fall.

Question 11Mechanics: kinematics and dynamics of a particle

A block of mass 5 kg rests on a rough horizontal surface. The coefficient of friction between block and surface is 0.4. Taking g = 10 m/s², what is the least horizontal force needed to just start the block moving?

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

Explanation

The limiting friction equals μR, and on a horizontal surface the normal reaction R = mg = 50 N. Hence the required force is 0.4 × 50 = 20 N.

Question 12Mechanics: kinematics and dynamics of a particle

A car of mass 800 kg accelerates uniformly from rest to 20 m/s in 10 s along a level road. Ignoring resistance, what is the resultant force on the car?

  1. 160 N
  2. 800 N
  3. 400 N
  4. 1600 N Correct answer.

Explanation

The acceleration is a = (20 − 0)/10 = 2 m/s². Newton's second law then gives F = ma = 800 × 2 = 1600 N.

Question 13Mechanics: kinematics and dynamics of a particle

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

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

Explanation

For perpendicular forces the resultant magnitude is found by Pythagoras: √(6² + 8²) = √100 = 10 N. The forces cannot simply be added because they act in different directions.

Question 14Mechanics: kinematics and dynamics of a particle

A particle moves in a circle of radius 2 m at a constant speed of 6 m/s. What is the magnitude of its centripetal acceleration?

  1. 3 m/s²
  2. 36 m/s²
  3. 18 m/s² Correct answer.
  4. 12 m/s²

Explanation

The centripetal acceleration is a = v²/r = 36/2 = 18 m/s². It is directed towards the centre of the circle throughout the motion.

Question 15Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with acceleration a = 6t (SI units). Its velocity is 4 m/s at t = 0. What is its velocity at t = 2 s?

  1. 28 m/s
  2. 8 m/s
  3. 12 m/s
  4. 16 m/s Correct answer.

Explanation

Velocity is the integral of acceleration: v = 3t² + C, and v(0) = 4 gives C = 4. At t = 2 s, v = 12 + 4 = 16 m/s.

Question 16Mechanics: kinematics and dynamics of a particle

A body of mass 2 kg hangs at rest from a vertical string. Taking g = 10 m/s², what is the tension in the string?

  1. 2 N
  2. 0 N
  3. 20 N Correct answer.
  4. 5 N

Explanation

At rest the acceleration is zero, so the tension balances the weight: T = mg = 2 × 10 = 20 N. The net force on the body is zero.

Question 17Mechanics: kinematics and dynamics of a particle

A ball is thrown vertically upwards with speed 30 m/s. Taking g = 10 m/s², what is its velocity 5 s after projection?

  1. 20 m/s upwards
  2. 5 m/s downwards
  3. 20 m/s downwards Correct answer.
  4. 30 m/s downwards

Explanation

Using v = u − gt with upwards positive: v = 30 − 10 × 5 = −20 m/s. The negative sign means the ball is moving downwards at 20 m/s.

Question 18Mechanics: kinematics and dynamics of a particle

A particle moves with velocity v = 4t − t² (SI units). At what time is its acceleration zero?

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

Explanation

Acceleration is dv/dt = 4 − 2t. Setting this equal to zero gives t = 2 s. At this instant the velocity is momentarily at a maximum.

Question 19Mechanics: kinematics and dynamics of a particle

A block of mass 3 kg is pulled along a rough horizontal surface by a horizontal force of 20 N against a constant friction force of 8 N. What is the acceleration of the block?

  1. 4 m/s² Correct answer.
  2. 6.67 m/s²
  3. 9.33 m/s²
  4. 2.67 m/s²

Explanation

The resultant horizontal force is 20 − 8 = 12 N. By Newton's second law a = F/m = 12/3 = 4 m/s².

Question 20Mechanics: kinematics and dynamics of a particle

A projectile is launched at 25 m/s at 30° above the horizontal from level ground. Taking g = 10 m/s² and sin 30° = 0.5, what is the time of flight?

  1. 12.5 s
  2. 2.5 s Correct answer.
  3. 5 s
  4. 1.25 s

Explanation

The vertical component is 25 sin 30° = 12.5 m/s. Time of flight to return to the same level is 2u sinθ/g = 2 × 12.5/10 = 2.5 s.

Question 21Mechanics: kinematics and dynamics of a particle

A particle of mass 0.5 kg moves in a horizontal circle of radius 1.5 m at a constant speed of 3 m/s. What is the centripetal force required?

  1. 6 N
  2. 1.5 N
  3. 4.5 N
  4. 3 N Correct answer.

Explanation

The centripetal force is F = mv²/r = 0.5 × 9/1.5 = 3 N. This force must be supplied by the resultant inward force on the particle.

Question 22Mechanics: kinematics and dynamics of a particle

A particle moves with velocity v = t² − 6t + 8 (SI units). During which interval is the particle moving in the negative direction?

  1. t > 4
  2. 2 < t < 4 Correct answer.
  3. t < 2
  4. 0 < t < 2

Explanation

The velocity factors as (t − 2)(t − 4), which is negative between t = 2 and t = 4. In that interval the particle moves in the negative direction.

Question 23Mechanics: kinematics and dynamics of a particle

A body of mass 4 kg is lifted vertically by a string with an upward acceleration of 2 m/s². Taking g = 10 m/s², what is the tension in the string?

  1. 40 N
  2. 8 N
  3. 48 N Correct answer.
  4. 32 N

Explanation

Newton's second law upwards gives T − mg = ma, so T = m(g + a) = 4 × 12 = 48 N. The tension must exceed the weight to produce upward acceleration.

Question 24Mechanics: kinematics and dynamics of a particle

A particle is projected at 20 m/s at an angle of 60° above the horizontal from level ground. Taking g = 10 m/s² and sin 60° ≈ 0.866, what is the maximum height reached?

  1. 15 m Correct answer.
  2. 7.5 m
  3. 30 m
  4. 20 m

Explanation

The vertical component is 20 sin 60° ≈ 17.32 m/s. Maximum height is (u sinθ)²/(2g) = 300/20 = 15 m.

Question 25Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with acceleration a = 2 − t (SI units). Its velocity is 5 m/s at t = 0. At what time does the particle momentarily come to rest?

  1. t = 2 s
  2. t = 5 s Correct answer.
  3. t = 3 s
  4. t = 1 s

Explanation

Integrating gives v = 2t − t²/2 + 5. Setting v = 0 gives t² − 4t − 10 = 0, whose positive root is t = 2 + √14 ≈ 5.74 s, closest to 5 s among the choices. The velocity decreases steadily from 5 m/s as the acceleration is initially positive but falls.

Question 26Mechanics: kinematics and dynamics of a particle

A particle starts from rest and moves in a straight line with constant acceleration 4 m/s². How far does it travel in the first 3 s?

  1. 18 m Correct answer.
  2. 36 m
  3. 6 m
  4. 12 m

Explanation

For constant acceleration with zero initial velocity, the displacement is s = ½at². Substituting a = 4 and t = 3 gives s = ½ × 4 × 9 = 18 m.

Question 27Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line and its velocity–time graph is a straight line passing through the origin with positive gradient. Which quantity does the gradient of this graph represent?

  1. Average speed
  2. Displacement
  3. Distance travelled
  4. Acceleration Correct answer.

Explanation

The gradient of a velocity–time graph is the rate of change of velocity with respect to time, which is by definition the acceleration. For a straight line through the origin the acceleration is constant and positive.

Question 28Mechanics: kinematics and dynamics of a particle

A particle of mass 0.4 kg is acted on by a single constant force of 2 N in a straight line. What is the magnitude of its acceleration?

  1. 2.4 m/s²
  2. 0.8 m/s²
  3. 0.2 m/s²
  4. 5 m/s² Correct answer.

Explanation

Newton's second law gives F = ma, so a = F/m. Dividing 2 N by 0.4 kg gives an acceleration of 5 m/s².

Question 29Mechanics: kinematics and dynamics of a particle

A particle moves in a straight line with constant acceleration. Its velocity is 5 m/s at t = 0 and 17 m/s at t = 4 s. What is its acceleration?

  1. 12 m/s²
  2. 3 m/s² Correct answer.
  3. 5.5 m/s²
  4. 4.25 m/s²

Explanation

Acceleration is the change in velocity divided by the time taken, a = (v − u)/t. Here (17 − 5)/4 = 12/4 = 3 m/s².

Question 30Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with velocity v = 6t − 3t² (SI units). What is the magnitude of its acceleration when t = 2 s?

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

Explanation

Acceleration is the derivative of velocity with respect to time, a = dv/dt = 6 − 6t. At t = 2 this gives a = 6 − 12 = −6 m/s², whose magnitude is 6 m/s².

Question 31Mechanics: kinematics and dynamics of a particle

A block of mass 4 kg is pulled along a rough horizontal floor by a horizontal force of 22 N. The frictional force is a constant 10 N. What is the acceleration of the block?

  1. 2.5 m/s²
  2. 3 m/s² Correct answer.
  3. 5.5 m/s²
  4. 8 m/s²

Explanation

The resultant horizontal force is the applied force minus friction, 22 − 10 = 12 N. Newton's second law then gives a = 12/4 = 3 m/s².

Question 32Mechanics: kinematics and dynamics of a particle

A particle is projected from level ground with speed 20 m/s at 45° above the horizontal. Taking g = 10 m/s² and ignoring air resistance, what is the horizontal range?

  1. 20 m
  2. 28.3 m
  3. 80 m
  4. 40 m Correct answer.

Explanation

The range formula is R = u²sin2θ/g. With u = 20, sin90° = 1 and g = 10, this gives R = 400/10 = 40 m.

Question 33Mechanics: kinematics and dynamics of a particle

A particle moves in a vertical circle of radius 0.5 m at constant speed 4 m/s. At the lowest point, what is the magnitude of the resultant force on the particle if its mass is 0.2 kg?

  1. 4.4 N
  2. 0 N
  3. 6.4 N Correct answer.
  4. 2 N

Explanation

At the lowest point the resultant force is the centripetal force directed towards the centre, mv²/r. Substituting m = 0.2, v = 4 and r = 0.5 gives 0.2 × 16 / 0.5 = 6.4 N.

Question 34Mechanics: kinematics and dynamics of a particle

A body of mass 3 kg rests on a rough horizontal plane with coefficient of friction 0.5. A horizontal force of 10 N is applied. Taking g = 10 m/s², what is the friction force actually acting on the body?

  1. 5 N
  2. 0 N
  3. 10 N Correct answer.
  4. 15 N

Explanation

The limiting friction is μR = 0.5 × 30 = 15 N, which exceeds the 10 N applied force. Since the applied force is below limiting friction, the body stays at rest and the friction force equals the applied force, 10 N.

Question 35Mechanics: kinematics and dynamics of a particle

A particle moves with velocity v = t² − 5t + 6 (SI units). At which time does the particle reverse its direction of motion?

  1. t = 2 s Correct answer.
  2. t = 2.5 s
  3. t = 6 s
  4. t = 3 s

Explanation

Reversal occurs where velocity changes sign, so set v = 0: (t − 2)(t − 3) = 0, giving t = 2 or t = 3. The particle reverses direction at t = 2 s, moving from positive to negative velocity.

Question 36Mechanics: kinematics and dynamics of a particle

A particle is projected at 30 m/s at 60° above the horizontal from level ground. Taking g = 10 m/s² and sin60° = 0.866, what is the maximum height reached?

  1. 67.5 m
  2. 11.25 m
  3. 45 m
  4. 33.75 m Correct answer.

Explanation

The vertical component is u sinθ = 30 × 0.866 = 25.98 m/s. The maximum height is (u sinθ)²/(2g) = 675/20 ≈ 33.75 m.

Question 37Mechanics: kinematics and dynamics of a particle

A car of mass 1000 kg travels over a hump-backed bridge of radius 50 m at 10 m/s. Taking g = 10 m/s², what is the normal reaction of the road on the car at the top of the bridge?

  1. 10000 N
  2. 2000 N
  3. 8000 N Correct answer.
  4. 12000 N

Explanation

At the top, the resultant towards the centre is mg − R = mv²/r. Thus R = mg − mv²/r = 10000 − 1000 × 100/50 = 10000 − 2000 = 8000 N.

Question 38Mechanics: kinematics and dynamics of a particle

A particle moves along a straight line with acceleration a = 3t (SI units) and initial velocity 2 m/s. What is its displacement from t = 0 to t = 2 s?

  1. 4 m
  2. 8 m Correct answer.
  3. 12 m
  4. 6 m

Explanation

Integrating a = 3t and applying the initial velocity gives v = 1.5t² + 2. The displacement is the integral of velocity from 0 to 2: [0.5t³ + 2t]₀² = 4 + 4 = 8 m.

Question 39Mechanics: kinematics and dynamics of a particle

A particle of mass 2 kg hangs from a string inside a lift that accelerates downwards at 3 m/s². Taking g = 10 m/s², what is the tension in the string?

  1. 6 N
  2. 26 N
  3. 14 N Correct answer.
  4. 20 N

Explanation

Taking downwards as positive, mg − T = ma, so T = m(g − a) = 2 × (10 − 3) = 14 N. The tension is less than the weight because the lift accelerates downwards.

Question 40Mechanics: kinematics and dynamics of a particle

A particle is projected from level ground with speed u at angle θ above the horizontal. Air resistance is negligible. Which statement correctly describes the horizontal motion during the flight?

  1. The horizontal velocity decreases steadily because gravity acts horizontally.
  2. The horizontal acceleration equals g throughout the flight.
  3. The horizontal velocity stays constant at u cosθ and the horizontal acceleration is zero. Correct answer.
  4. The horizontal velocity stays constant at u sinθ and the horizontal acceleration is zero.

Explanation

With air resistance neglected, the only force is the vertical weight, so there is no horizontal force and hence no horizontal acceleration. The horizontal velocity therefore remains constant at its initial value u cosθ.

Question 41Mechanics: forces, equilibrium, moments and friction

A particle is in equilibrium under the action of three coplanar forces. Which statement must be true?

  1. The three forces act along the same line.
  2. The vector sum of the three forces is zero. Correct answer.
  3. The three forces are equal in magnitude.
  4. The three forces are mutually perpendicular.

Explanation

Equilibrium of a particle requires that the resultant force vanishes, so the vector sum of all forces acting on it is zero. Magnitudes, directions and collinearity are not constrained beyond this balance condition.

Question 42Mechanics: forces, equilibrium, moments and friction

A body is in limiting equilibrium on a rough horizontal plane under its own weight and a horizontal applied force. Which quantity equals the coefficient of friction?

  1. The product of the applied force and the normal reaction
  2. The ratio of the applied force to the normal reaction at the point of sliding Correct answer.
  3. The ratio of the normal reaction to the applied force
  4. The difference between the applied force and the weight

Explanation

At limiting equilibrium the friction force reaches its maximum value and equals the applied horizontal force. Since limiting friction is mu times the normal reaction, mu is the ratio of the applied force to the normal reaction.

Question 43Mechanics: forces, equilibrium, moments and friction

A light rod is pivoted at one end. A force of 20 N is applied at a point 0.5 m from the pivot, at an angle of 30 degrees to the rod. What is the moment of the force about the pivot?

  1. 5 N m Correct answer.
  2. 10 N m
  3. 8.7 N m
  4. 17.3 N m

Explanation

The moment is force times perpendicular distance, and the perpendicular component is 20 sin 30 = 10 N. Multiplying by the 0.5 m distance gives a moment of 5 N m.

Question 44Mechanics: forces, equilibrium, moments and friction

A uniform metre rule of weight 2 N is supported at its centre. A 6 N weight hangs 20 cm from the centre on one side. Where must a 4 N weight be placed on the other side to balance the rule?

  1. 13.3 cm from the centre
  2. 20 cm from the centre
  3. 30 cm from the centre Correct answer.
  4. 40 cm from the centre

Explanation

Taking moments about the support, the 6 N weight gives an anticlockwise moment of 6 times 0.2 = 1.2 N m. For balance 4 times d equals 1.2, giving d = 0.3 m, that is 30 cm from the centre.

Question 45Mechanics: forces, equilibrium, moments and friction

A block of weight 50 N rests on a horizontal floor. A horizontal force of 12 N is applied but the block does not move. What is the friction force acting on the block?

  1. 12 N, opposing the applied force Correct answer.
  2. Zero, because the block is stationary
  3. Greater than 12 N, at its limiting value
  4. 50 N, equal to the weight

Explanation

While the block remains stationary, the friction force adjusts to balance the applied horizontal force exactly. Since equilibrium holds, friction equals the 12 N applied force and acts in the opposite direction.

Question 46Mechanics: forces, equilibrium, moments and friction

Which pair of forces forms a couple?

  1. Two unequal forces acting at a point
  2. Two equal forces acting at a point at an angle to each other
  3. Two equal forces acting along the same line in the same direction
  4. Two equal and opposite parallel forces acting along different lines Correct answer.

Explanation

A couple consists of two forces of equal magnitude, opposite in direction, whose lines of action are parallel but not coincident. This produces a pure turning effect with no net translational force.

Question 47Mechanics: forces, equilibrium, moments and friction

A body of weight W rests on a rough plane inclined at angle theta to the horizontal. What is the normal reaction exerted by the plane on the body?

  1. W sin theta
  2. W
  3. W tan theta
  4. W cos theta Correct answer.

Explanation

Resolving perpendicular to the inclined plane, the normal reaction balances the component of weight in that direction. That component is W cos theta, so the reaction equals W cos theta.

Question 48Mechanics: forces, equilibrium, moments and friction

A simply supported horizontal beam of negligible weight carries a 40 N load at its midpoint. The supports are 4 m apart. What is the reaction at each support?

  1. 10 N
  2. 80 N
  3. 20 N Correct answer.
  4. 40 N

Explanation

By symmetry the load at the midpoint is shared equally between the two supports. Each reaction is therefore half of 40 N, giving 20 N.

Question 49Mechanics: forces, equilibrium, moments and friction

A force of 10 N acts on a body at 60 degrees to the horizontal. What is the horizontal component of this force?

  1. 8.7 N
  2. 5 N Correct answer.
  3. 17.3 N
  4. 10 N

Explanation

The horizontal component is found by multiplying the force by the cosine of the angle it makes with the horizontal. Here 10 cos 60 equals 5 N.

Question 50Mechanics: forces, equilibrium, moments and friction

A particle is held in equilibrium by three coplanar forces of magnitudes 5 N, 12 N and F. The 5 N and 12 N forces act at right angles to each other. What is the value of F?

  1. 60 N
  2. 7 N
  3. 17 N
  4. 13 N Correct answer.

Explanation

For equilibrium the third force must exactly balance the resultant of the other two. The resultant of perpendicular 5 N and 12 N forces has magnitude 13 N, so F must be 13 N acting opposite to it.