Mathematics

1. 1.

   A car moving at 12 m/s accelerates uniformly at 2 m/s^2 for 8 seconds. Find its final speed and the distance travelled in this time.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. List u, a, t then choose the constant-acceleration equation that contains exactly the quantities you know and want.
   2. Check: average speed (12 + 28)/2 = 20 m/s, and 20 * 8 = 160 m.

   Marking points

   • Uses v = u + at: v = 12 + 2 * 8 = 28 m/s.
   • Uses s = ut + (1/2)at^2.
   • s = 96 + 64 = 160 m.

   Examiner tip: List the SUVAT quantities first. Include units in the final answers.

2. 2.

   A 5 kg block on a rough horizontal surface is pulled by a horizontal force of 20 N. Friction is 8 N. Find the acceleration and the distance travelled in the first 5 seconds from rest.

   [3 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Apply Newton's second law along the direction of motion using the resultant force.
   2. Start from rest, so u = 0 in the constant-acceleration equation.

   Marking points

   • Resultant force = 20 - 8 = 12 N.
   • a = F/m = 12/5 = 2.4 m/s^2.
   • s = (1/2)(2.4)(5^2) = 30 m.

   Examiner tip: Use the resultant force (net of friction) in F = ma, not just the applied force.

3. 3.

   A ball is projected from level ground at 20 m/s at 30 degrees above the horizontal. Taking g = 9.8 m/s^2, find the time of flight, the horizontal range and the greatest height.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Resolve the velocity into components; gravity acts only on the vertical component.
   2. At the top the vertical velocity is zero, which gives the maximum height; the flight lasts twice the time to the top.

   Marking points

   • Vertical component 20 sin 30 = 10 m/s and horizontal component 20 cos 30 = 17.3 m/s.
   • Time of flight 2 * 10/9.8 = 2.04 s.
   • Range = 20 cos 30 * 2.04 = 35.3 m.
   • Greatest height = 10^2/(2 * 9.8) = 5.10 m.

   Examiner tip: Treat horizontal and vertical motion separately. Horizontal velocity stays constant when air resistance is ignored.

4. 4.

   Particles of mass 3 kg and 5 kg are connected by a light inextensible string passing over a smooth fixed pulley. They are released from rest. Taking g = 9.8 m/s^2, find the acceleration and the tension.

   [4 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Write Newton's second law separately for each particle, taking the direction of motion as positive.
   2. The same string tension and the same acceleration magnitude apply to both, so adding the equations eliminates T.

   Marking points

   • Equation for the 5 kg particle: 5g - T = 5a.
   • Equation for the 3 kg particle: T - 3g = 3a.
   • Adds them: 2g = 8a so a = g/4 = 2.45 m/s^2.
   • T = 3(g + a) = 36.75 N.

   Examiner tip: The tension is not equal to either weight. Use the two equations simultaneously.

5. 5.

   A uniform horizontal beam AB of length 4 m and mass 20 kg rests on supports at A and B. A load of mass 30 kg is placed 1 m from A. Taking g = 9.8 m/s^2, find the reactions at A and B.

   [5 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. The beam is in equilibrium: the forces balance and the moments about any point balance.
   2. Take moments about A to eliminate R_A, then use vertical equilibrium for the other reaction.

   Marking points

   • Takes moments about A: R_B * 4 = 20g * 2 + 30g * 1.
   • R_B = 70g/4 = 17.5g = 171.5 N.
   • Resolves vertically: R_A + R_B = 50g.
   • R_A = 32.5g = 318.5 N.
   • Checks the answer using moments about B.

   Examiner tip: A uniform beam's weight acts at its midpoint. Take moments about a point where an unknown force acts to eliminate it.

6. 6.

   A particle moves along a straight line with acceleration a = 6t - 12 m/s^2. When t = 0 the particle is at the origin with velocity 9 m/s. Find (a) the velocity and displacement as functions of t, (b) the times when the particle is at rest, (c) the total distance travelled in the first 4 seconds.

   [6 marks] · no calculator

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Integrate acceleration to get velocity, then velocity to get displacement, using the initial conditions to find each constant.
   2. Distance differs from displacement when the particle reverses direction, so split the journey at the times when v = 0.

   Marking points

   • Integrates a: v = 3t^2 - 12t + C and uses v(0) = 9 to get v = 3t^2 - 12t + 9.
   • Integrates v: x = t^3 - 6t^2 + 9t (x(0) = 0).
   • v = 3(t - 1)(t - 3) = 0 so the particle is at rest at t = 1 and t = 3.
   • Positions: x(1) = 4, x(3) = 0, x(4) = 4.
   • The direction changes at t = 1 and t = 3, so the journey is 0 to 4, 4 to 0, 0 to 4.
   • Total distance = 4 + 4 + 4 = 12 m.

   Examiner tip: Distance is not the same as displacement when the direction changes. Split at each point where v = 0.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.