01 · Initial state
Find the release height.
Hint
Evaluate at t=0.
Worked solution
h(0)=8 m above ground.
Understand · explore · practise
Modelling in mechanics: mixed practice
Twenty-four independent questions combining model domains, idealisations, SI units, force diagrams, motion signs and vector journeys.
Before you startThe preceding modelling-in-mechanics lessons.
01 / Try the questions before opening the solutions
State assumptions and units as part of each answer.
Correct arithmetic is necessary but not sufficient.
Check the model’s scope, the chosen body, the coordinate convention and the requested quantity. These questions use original constructed situations and values.
Audit the claim before calculatingExplore
Reveal the audit
02 / A motion formula needs a physical interval
Use h=8+6t−2t² metres above level ground after release at t=0.
Watch: select the root in the experiment
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 · Contact
Find the first ground-contact time after release and state the free-flight domain.
Hint
Solve 8+6t−2t²=0.
Worked solution
t²−3t−4=0 gives t=−1 or 4. Use first contact t=4 s and domain 0≤t≤4; the negative root is outside the experiment.
03 · Maximum
Find the maximum modelled height and the time at which it occurs.
Hint
Complete the square.
Worked solution
h=12.5−2(t−1.5)². The maximum is 12.5 m at t=1.5 s, within the physical domain.
04 · Later value
Does h(5)=−12 m describe the object in free flight after it lands on level ground?
Hint
Check the contact time first.
Worked solution
No. The equation was valid only until t=4 s. Its later polynomial values do not describe post-contact motion under the original assumptions.
03 / Separate geometric and force assumptions
State what each modelling word actually removes.
05 · Particle
Does treating a falling ball as a particle justify neglecting drag automatically?
Hint
Size representation and force choice are separate.
Worked solution
No. A particle model can include effective drag. Negligible drag needs a separate justification.
06 · Rigid
What does a rigid-rod assumption neglect, and does it imply uniform mass distribution?
Hint
Consider shape and distribution separately.
Worked solution
It neglects deformation. It does not imply uniformity, which must be stated separately.
07 · String constraint
One end of a taut inextensible string over a fixed pulley moves down 0.2 m. What happens to the other end in the standard two-vertical-part arrangement?
Hint
The two changing lengths must sum to a constant.
Worked solution
The other end moves up 0.2 m, provided the geometry and tautness assumptions remain valid.
08 · Light
Why is “light string means no tension” wrong?
Hint
Which property is negligible?
Worked solution
Light means negligible mass; the string can still transmit tension.
04 / Write the factors before simplifying
Pay attention to squared and cubed units.
09 · Speed
Convert 90 km/h to m/s.
Hint
Multiply by 1000/3600.
Worked solution
25 m/s.
10 · Small length unit
Convert 2.4 cm/s to m/s.
Hint
One centimetre is 0.01 m.
Worked solution
0.024 m/s.
11 · Area
Convert 0.3 m² to cm².
Hint
Square the factor 100.
Worked solution
0.3×100²=3000 cm².
12 · Density
Convert 1.2 g/cm³ to kg/m³.
Hint
Convert grams and cubic centimetres separately.
Worked solution
0.0012 kg / 10⁻⁶ m³ = 1200 kg/m³.
05 / Use a diagram of one chosen body
Do not assume a particular reaction before considering all forces.
13 · Weight
Find the weight of a 4 kg object using g=9.8 m/s².
Hint
Weight is mg, in newtons.
Worked solution
39.2 N downward.
14 · Extra push
That object rests on a horizontal table with an additional downward push of 8 N. Find the normal reaction magnitude, assuming no other vertical forces.
Hint
Vertical acceleration is zero.
Worked solution
N=39.2+8=47.2 N upward.
15 · Contact direction
For an object on a smooth inclined plane, describe the normal reaction direction and the weight direction.
Hint
They are not generally opposite.
Worked solution
Normal reaction is perpendicular to the plane, away from it. Weight is vertically downward. Smooth contact omits friction, not the normal force.
16 · Friction
Must a motionless object on a rough horizontal surface have nonzero friction if no horizontal interaction tends to move it?
Hint
Availability of friction is not a requirement for nonzero friction.
Worked solution
No. Static friction can be zero in this situation.
06 / Distinguish direction, magnitude and interval behaviour
Take right as positive.
17 · Same negative signs
For v=−6 m/s and a=−1 m/s², state direction and local speed change.
Hint
Compare the two directions.
Worked solution
The object moves left and speeds up locally.
18 · Reversal time
For v=2−4t m/s, when is the object instantaneously at rest?
Hint
Set the velocity to zero.
Worked solution
t=0.5 s. Acceleration remains −4 m/s² in this model.
19 · Endpoints
Compare the velocities and speeds at t=0 and t=1 s in that model.
Hint
Use the magnitude for speed.
Worked solution
Velocities are +2 and −2 m/s. Both speeds are 2 m/s, but the directions are opposite; between them the object stops and reverses.
20 · Position sign
At x=−5 m and v=+2 m/s, is the object moving left simply because its position is negative?
Hint
Interpret position and velocity separately.
Worked solution
No. It is 5 m left of the origin and moving right.
07 / Use the appropriate vector total
Direction angles below are anticlockwise from positive i.
21 · Magnitude
Find the speed corresponding to v=−8i+6j m/s.
Hint
Square the perpendicular components.
Worked solution
√(64+36)=10 m/s.
22 · Direction
Find that velocity direction as an angle between 0° and 360°.
Hint
The vector lies in quadrant II.
Worked solution
180°−tan⁻¹(6/8)≈143.13° anticlockwise from positive i.
23 · Two-leg route
Two straight displacements are 8i+6j m then −8i+6j m. Find distance and net displacement.
Hint
Each leg has length 10 m.
Worked solution
Distance=20 m. Net displacement=12j m, with magnitude 12 m.
24 · Average quantities
If the two-leg route takes 20 s, find average speed and average velocity.
Hint
Use the two different totals.
Worked solution
Average speed=20/20=1 m/s. Average velocity=12j/20=0.6j m/s, whose magnitude is 0.6 m/s.
08 / Review the reason behind an error
Return to the relevant lesson before repeating a similar question.
Check whether the error came from the physical domain, an unstated assumption, a conversion power, the chosen body, a direction convention or confusion between a path and a net vector. A complete answer explains the model and preserves units.
Section 1 of 8 · Try the questions before opening the solutions