Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
80 original questions across 14 selected topic areas. No account or payment required to practise.
6 questions in this topic. Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
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Original Cambridge 2027 concept practice for source-backed gaps. Indicative self-marking only, not an official paper or complete syllabus coverage.
Selected subtopics: Projectile trajectories · Composite centres of mass and limiting equilibrium · Circular motion and elastic energy · Variable resistance and oblique impact
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Question 1A particle is projected from level ground with horizontal velocity 20 m s^-1 and upward vertical velocity 15 m s^-1. Ignore air resistance and use g=10 m s^-2. Derive its Cartesian trajectory, time of flight, horizontal range and greatest height. Find its speed and direction 1.0 s after launch.6 indicative marks
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A particle is projected from level ground with horizontal velocity 20 m s^-1 and upward vertical velocity 15 m s^-1. Ignore air resistance and use g=10 m s^-2. Derive its Cartesian trajectory, time of flight, horizontal range and greatest height. Find its speed and direction 1.0 s after launch.
Cambridge 2027: Direct conceptual match
Independently derived the trajectory and roots, located the vertex, and computed the speed sqrt425 and atan(1/4) angle.
The horizontal component stays constant only in the no-drag model; vertical acceleration is -g.
The root t=0 describes launch, not landing. At t=1 the vertical velocity is still positive, so the direction is above, not below, the horizontal. Real drag and nonlevel landing would change the model outputs.
Self-marking points
The component equations are x=20t and y=15t-5t^2.
Eliminating t gives y=(3/4)x-x^2/80.
The next positive zero of y occurs at t=3 s, so range is 60 m.
At the top v_y=15-10t=0, t=1.5 s; greatest height is 11.25 m.
At t=1, velocity components are (20,5) m s^-1, giving speed sqrt(425), approximately 20.6 m s^-1.
The direction is arctan(5/20), approximately 14.0 degrees above horizontal.
Examiner tip
Use velocity components, not position coordinates, to find the direction of motion.
Question 2A uniform lamina is the rectangle 0<=x<=4,0<=y<=3 with the rectangle 2<=x<=4,1<=y<=3 removed. Dimensions are metres. Its mass is 10 kg. It stands in its own vertical plane on its full bottom edge on a horizontal floor. A horizontal force P acts to the right at height 2 m. With g=10 m s^-2 and floor coefficient mu=0.80, find its centre of mass and decide whether sliding or toppling occurs first as P increases slowly from zero.6 indicative marks
Show your reasoning6 indicative marksCalculator allowed
A uniform lamina is the rectangle 0<=x<=4,0<=y<=3 with the rectangle 2<=x<=4,1<=y<=3 removed. Dimensions are metres. Its mass is 10 kg. It stands in its own vertical plane on its full bottom edge on a horizontal floor. A horizontal force P acts to the right at height 2 m. With g=10 m s^-2 and floor coefficient mu=0.80, find its centre of mass and decide whether sliding or toppling occurs first as P increases slowly from zero.
Cambridge 2027: Direct conceptual match
Recomputed negative-area moments and COM (1.5,1.25), then independently compared the 80 N sliding and 125 N toppling thresholds.
Use area weighting because density and thickness are uniform. The original and removed centres follow symmetry of rectangles.
Before either limit, the floor reaction can shift along the base to supply a balancing moment. At toppling its line of action reaches the right edge; at sliding static friction reaches its limit. Compare both thresholds rather than assume one mode.
Self-marking points
Treat the removed area as negative mass: the full rectangle has area 12, centre (2,1.5), and removed area 4 has centre (3,2).
The remaining area is 8, so x_bar=(12*2-4*3)/8=1.5 m.
Similarly y_bar=(12*1.5-4*2)/8=1.25 m.
Vertical equilibrium gives total normal reaction R=100 N; limiting sliding force is mu*R=80 N.
For impending clockwise toppling about the right edge x=4, moments give 2P=100*(4-1.5), so P=125 N.
Since 80<125, sliding starts first under the model; the toppling-only threshold cannot be reached while remaining in static equilibrium.
Examiner tip
Use the horizontal distance of the centre of mass from the pivot, not its height, for the weight moment.
Question 3(a) A 0.50 kg particle moves in a horizontal circle of radius 1.0 m on a light string at 45 degrees to the vertical. Find its speed and string tension. (b) A small particle moves without losses in contact with the inside surface of a smooth vertical circular track of radius 1.0 m. Find its minimum bottom speed for continuous contact and the top speed in that limiting case. Use g=10 m s^-2.6 indicative marks
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(a) A 0.50 kg particle moves in a horizontal circle of radius 1.0 m on a light string at 45 degrees to the vertical. Find its speed and string tension. (b) A small particle moves without losses in contact with the inside surface of a smooth vertical circular track of radius 1.0 m. Find its minimum bottom speed for continuous contact and the top speed in that limiting case. Use g=10 m s^-2.
Cambridge 2027: Direct conceptual match
Independently resolved the string forces and combined top radial contact with a 2R rise in potential energy; verified squared speeds 10 and 50.
Centripetal force is the inward resultant, not a separate additional force. The horizontal-circle vertical resultant is zero.
For the inside loop contact force cannot pull the bead towards the track. The limiting case occurs at the top; a bead merely reaching that height at zero speed would already have lost contact.
Self-marking points
For the horizontal circle, vertical balance is T cos45=mg=5 N.
The inward resultant gives T sin45=mv^2/r; division gives v^2=gr tan45=10.
Thus v=sqrt(10) m s^-1 and T=5sqrt(2) N; angular speed is v/r=sqrt(10) rad s^-1.
For marginal top contact in the vertical circle, N=0 and v_top^2=gR=10.
Energy conservation from bottom to top gives v_bottom^2=v_top^2+4gR=50.
Minimum bottom speed is sqrt(50)=5sqrt(2) m s^-1, and limiting top speed is sqrt(10) m s^-1.
Examiner tip
Separate vertical balance in the horizontal circle from energy transfer in the vertical circle.
Question 4A light elastic string of natural length 2.0 m and modulus 40 N has one end fixed at O and the other attached to a 1.0 kg particle on a smooth horizontal line through O. The particle is released from rest with the string stretched by 0.50 m. Find its initial tension, elastic energy and speed when it first reaches natural length. Describe the motion immediately after that point.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
A light elastic string of natural length 2.0 m and modulus 40 N has one end fixed at O and the other attached to a 1.0 kg particle on a smooth horizontal line through O. The particle is released from rest with the string stretched by 0.50 m. Find its initial tension, elastic energy and speed when it first reaches natural length. Describe the motion immediately after that point.
Cambridge 2027: Direct conceptual match
Checked effective stiffness 20, initial force 10 N and energy 2.5 J; independent work integral gives final speed sqrt5.
The modulus has units of force, while the equivalent spring stiffness here is lambda/l=20 N m^-1.
Use the positive extension in Hooke's law only while the string is taut. The ideal model neglects string mass, friction and losses during the specified segment.
Self-marking points
Hooke's law gives tension T=lambda*x/l=40*0.50/2=10 N.
Elastic energy is lambda*x^2/(2l)=40*0.50^2/4=2.5 J.
At natural length all stored elastic energy has become kinetic energy, so v=sqrt(2*2.5/1)=sqrt(5) m s^-1.
The string then becomes slack; unlike a spring it cannot push when shorter than natural length.
With no horizontal force immediately after slackening, the particle continues at speed sqrt(5) until another constraint or later retensioning acts.
Examiner tip
A string's force law stops at zero extension; do not model it as a compressible spring.
Question 5A 2.0 kg particle moves horizontally under a constant forward force 6.0 N and an opposing resistance 2v N, where v>=0 is its speed in m s^-1. It starts at rest at x=0. Form and solve a differential equation for v(t), find x(t), the time to reach 1.5 m s^-1, and the limiting speed.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
A 2.0 kg particle moves horizontally under a constant forward force 6.0 N and an opposing resistance 2v N, where v>=0 is its speed in m s^-1. It starts at rest at x=0. Form and solve a differential equation for v(t), find x(t), the time to reach 1.5 m s^-1, and the limiting speed.
Cambridge 2027: Direct conceptual match
Independently differentiated v and x, checked both initial conditions and verified 3(1-exp(-ln2))=1.5 and the force-balance limit.
The sign of the resistance is opposite the motion. The resulting solution is nonnegative for t>=0, so it respects the assumed speed regime.
Differentiate x to recover v, then differentiate v to check the original force equation. The limiting speed is approached asymptotically, not attained at a finite time in the model.
Self-marking points
Newton's law gives 2dv/dt=6-2v, or dv/dt=3-v.
Separation and v(0)=0 give v(t)=3(1-e^-t) m s^-1 for t>=0.
Integrating with x(0)=0 gives x(t)=3(t-1+e^-t) m.
v=1.5 gives e^-t=1/2, so t=ln2 s.
As t increases, v tends to 3 m s^-1 from below; at that speed the driving and resistive forces balance.
Examiner tip
Do not use a constant-acceleration formula when resistance depends on speed.
Question 6A smooth sphere of mass 0.50 kg strikes a smooth fixed horizontal wall. In coordinates with x parallel to the wall and y pointing away from it, its incoming velocity is (6,-8) m s^-1. The coefficient of restitution is 0.75. Find outgoing velocity, impulse on the sphere and kinetic energy lost; explain which component is unchanged.5 indicative marks
Show your reasoning5 indicative marksCalculator allowed
A smooth sphere of mass 0.50 kg strikes a smooth fixed horizontal wall. In coordinates with x parallel to the wall and y pointing away from it, its incoming velocity is (6,-8) m s^-1. The coefficient of restitution is 0.75. Find outgoing velocity, impulse on the sphere and kinetic energy lost; explain which component is unchanged.
Cambridge 2027: Direct conceptual match
Recomputed normal rebound 6, momentum change (0,7) and kinetic energies 25/18 J from both velocity components.
Restitution applies to the relative normal speed of separation and approach, not to the complete speed vector. The fixed wall has zero velocity.
The coefficient lies between zero and one. Only a perfectly elastic normal impact with e=1 would preserve the sphere's kinetic energy in this smooth fixed-wall model.
Self-marking points
The smooth wall exerts no tangential impulse, so outgoing x velocity remains 6 m s^-1.
The normal component reverses with factor e: outgoing y velocity is -e*(-8)=+6 m s^-1.
Outgoing velocity is (6,6) m s^-1.
Impulse is m(v_after-v_before)=(0,7) N s, directed away from the wall.
Initial KE is 25 J and final KE is 18 J, so loss is 7 J.
Examiner tip
Do not multiply the tangential velocity by the restitution coefficient.
