Hard A-Level Connected particles Questions

Challenging, exam-style A-Level Connected particles questions with worked solutions. Stretch yourself on the hardest pulley, acceleration, tension, table problems.

pulleyaccelerationtensiontabletow-barbraking
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
A lift of mass MM carries a person of mass mm. The lift accelerates upward at aa. Consider (P) the tension in the supporting cable and (Q) the normal reaction between the person and the floor. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which pair of expressions is correct?
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Worked solution

  1. Treat the lift and person as one system

    total mass M+m\text{total mass }M+m

    The cable supports both the lift and the person.

  2. Apply Newton's second law to the system

    P−(M+m)g=(M+m)aP-(M+m)g=(M+m)a

    The cable tension minus the total weight gives the acceleration.

  3. Solve for the cable tension

    P=(M+m)(g+a)P=(M+m)(g+a)

    Factorise the total mass.

  4. Now isolate the person

    R↑, mg↓R\uparrow,\ mg\downarrow

    Consider only the forces on the person.

  5. Apply Newton's second law to the person

    Q−mg=maQ-mg=ma

    The floor's reaction provides the person's upward acceleration.

  6. Solve for the reaction

    Q=m(g+a)Q=m(g+a)

    Factorise the person's mass.

  7. Check consistency with the cable

    P=(M+m)(g+a)P=(M+m)(g+a)

    The whole-system result includes the person's share m(g+a)m(g+a).

  8. Reject the cable-carries-lift-only option

    P≠M(g+a)P\neq M(g+a)

    The cable must also support the person's weight and acceleration.

  9. Reject the downward-acceleration option

    P≠(M+m)(g−a)P\neq (M+m)(g-a)

    That applies when the lift accelerates downward.

  10. Reject the static option

    P≠(M+m)gP\neq (M+m)g

    That applies only at constant velocity.

  11. Reject the mismatched-reaction option

    Q≠mgQ\neq mg

    The reaction on the person also increases while accelerating upward.

  12. Compare P and Q

    P=Q+M(g+a)P=Q+M(g+a)

    The cable additionally supports the lift itself.

  13. State both expressions

    P=(M+m)(g+a), Q=m(g+a)P=(M+m)(g+a),\ Q=m(g+a)

    Both increase above their static values while accelerating upward.

  14. Reject the cable-carries-lift-only option

    P≠M(g+a)P\neq M(g+a)

    The cable must also support the person's weight and acceleration.

  15. Confirm the correct pair

    P=(M+m)(g+a), Q=m(g+a)P=(M+m)(g+a),\ Q=m(g+a)

    The cable carries the whole system; the floor carries only the person.

Answer
P=(M+m)(g+a), Q=m(g+a)P=(M+m)(g+a),\ Q=m(g+a)
Question 2
8 markschallenging
Two particles are connected by a light inextensible string over a smooth pulley. One particle has mass 3 kg3\,\text{kg} and the system accelerates at 2.8 m s−22.8\,\text{m s}^{-2} when released from rest. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the mass of the other particle.
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Worked solution

  1. Set up the model and mark the forces

    heavier m1↓, 3 kg↑, tension T\text{heavier }m_1\downarrow,\ \text{3\,kg}\uparrow,\ \text{tension }T

    Light inextensible string over a smooth pulley: one tension, one acceleration magnitude aa.

  2. Write the acceleration formula for the system

    a=(m1−3)×9.8m1+3a=\dfrac{(m_1-3)\times 9.8}{m_1+3}

    Adding the two Newton's-second-law equations eliminates the tension.

  3. Newton's second law for the heavier particle

    m1g−T=m1am_1 g-T=m_1 a

    Its weight exceeds the tension.

  4. Newton's second law for the lighter particle

    T−3g=3aT-3g=3 a

    The tension exceeds its weight.

  5. Substitute the known acceleration

    2.8=(m1−3)×9.8m1+32.8=\dfrac{(m_1-3)\times 9.8}{m_1+3}

    Insert a=2.8a=2.8.

  6. Clear the fraction

    2.8(m1+3)=(m1−3)×9.82.8(m_1+3)=(m_1-3)\times 9.8

    Multiply both sides by the total mass.

  7. Expand both sides

    2.8m1+2.8×3=9.8m1−9.8×32.8 m_1+2.8\times 3=9.8 m_1-9.8\times 3

    Distribute the brackets.

  8. Collect the terms in m1m_1

    (9.8−2.8)m1=2.8×3+9.8×3(9.8-2.8)m_1=2.8\times 3+9.8\times 3

    Gather the unknown on one side.

  9. Make m1m_1 the subject

    m1=2.8×3+9.8×39.8−2.8m_1=\dfrac{2.8\times 3+9.8\times 3}{9.8-2.8}

    Divide by the coefficient of m1m_1.

  10. Evaluate the mass

    m1=5.4 kgm_1=5.4\,\text{kg}

    This is the required mass.

  11. Check the acceleration

    (5.4−3)×9.85.4+3=2.8 m s−2\dfrac{(5.4-3)\times 9.8}{5.4+3}=2.8\,\text{m s}^{-2}

    Substituting back reproduces the given acceleration.

  12. State the assumptions

    light string, smooth pulley\text{light string, smooth pulley}

    These justify one tension and equal accelerations.

  13. Confirm the units

    m1 in kilogramsm_1\ \text{in kilograms}

    The answer is a mass.

  14. Newton's second law for the heavier particle

    m1g−T=m1am_1 g-T=m_1 a

    Its weight exceeds the tension.

  15. State the heavier mass

    m1=5.4 kgm_1=5.4\,\text{kg}

    This mass gives the required acceleration.

Answer
m1=5.4 kgm_1=5.4\,\text{kg}
Question 3
8 markschallenging
Two particles of masses 5 kg5\,\text{kg} and 2 kg2\,\text{kg} are connected by a light inextensible string over a smooth pulley. The system is released from rest with the heavier particle 1 m1\,\text{m} above the floor. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the further height the lighter particle rises after the heavier particle hits the floor. Give your answer to 3 significant figures.
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Worked solution

  1. Set up the model and mark the forces

    5 kg↓, 2 kg↑, tension T\text{5\,kg}\downarrow,\ \text{2\,kg}\uparrow,\ \text{tension }T

    While the string is taut both particles share the acceleration magnitude aa.

  2. Apply Newton's second law to each particle

    5(9.8)−T=5a,T−2(9.8)=2a5(9.8)-T=5 a,\quad T-2(9.8)=2 a

    Take the direction of motion as positive.

  3. Add the equations to eliminate TT

    (5−2)×9.8=(5+2)a(5-2)\times 9.8=(5+2)a

    Adding cancels the tension.

  4. Evaluate the acceleration

    a=4.2 m s−2a=4.2\,\text{m s}^{-2}

    The common acceleration before the heavier particle lands.

  5. Speed when the heavier particle reaches the floor

    v2=2×4.2×1v^2=2\times 4.2\times 1

    Using v2=u2+2asv^2=u^2+2as with u=0u=0 over the drop of 1 m.

  6. Evaluate that speed

    v=8.4=2.9 m s−1v=\sqrt{8.4}=2.9\,\text{m s}^{-1}

    Both particles are moving at this speed when the string goes slack.

  7. Describe the next phase

    string slack, lighter particle free\text{string slack, lighter particle free}

    Once the heavier particle lands the string becomes slack and the lighter particle moves freely.

  8. Model the free flight

    a=−9.8 m s−2 (decelerating)a=-9.8\,\text{m s}^{-2}\ (\text{decelerating})

    It continues upward with initial speed vv under gravity alone.

  9. Condition at the highest extra point

    vtop=0v_{\text{top}}=0

    It rises until its speed is momentarily zero.

  10. Apply the time-free equation

    0=v2−2×9.8×s0=v^2-2\times 9.8\times s

    Using v2=u2+2asv^2=u^2+2as for the free flight.

  11. Make the extra height the subject

    s=8.42×9.8s=\dfrac{8.4}{2\times 9.8}

    Rearrange for the further rise.

  12. Evaluate the extra height

    s=0.429 ms=0.429\,\text{m}

    This is how much further the lighter particle rises.

  13. Add the two contributions if required

    1+0.429=1.43 m1+0.429=1.43\,\text{m}

    The total rise is the taut-string rise plus the free rise.

  14. Confirm the units

    s in metress\ \text{in metres}

    The answer is a length.

  15. State the further height risen

    s=0.429 ms=0.429\,\text{m}

    After the heavier particle lands, the lighter one rises this much further.

Answer
s=0.429 ms=0.429\,\text{m}
Question 4
8 markschallenging
A person of mass 68 kg68\,\text{kg} stands on bathroom scales inside a lift. The lift accelerates downwards at 2.2 m s−22.2\,\text{m s}^{-2}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the reading of the scales (the normal reaction on the person). Give your answer to 3 significant figures.
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Worked solution

  1. Model the person and identify the forces

    R (up), mg=68×9.8=666 N (down)R\ (\text{up}),\ mg=68\times 9.8=666\,\text{N}\ (\text{down})

    The normal reaction RR acts upward and the weight acts downward.

  2. Apply Newton's second law (taking down as positive)

    mg−R=mamg-R=ma

    The net force equals mass times acceleration in the direction of the acceleration.

  3. State the weight of the person

    mg=68×9.8=666 Nmg=68\times 9.8=666\,\text{N}

    Weight is mass times gg.

  4. Rearrange for the reaction

    R=m(9.8−a)R=m(9.8-a)

    Make RR the subject.

  5. Substitute the mass and acceleration

    R=68(9.8−2.2)R=68(9.8-2.2)

    Insert the numbers.

  6. Evaluate the bracket

    9.8−2.2=7.69.8-2.2=7.6

    Combine the gravitational and acceleration terms.

  7. Multiply out

    R=68×7.6R=68\times 7.6

    Multiply by the mass.

  8. State the numerical value

    R=517 NR=517\,\text{N}

    This is the normal reaction.

  9. Interpret the apparent weight

    R<mgR<mg

    The apparent weight is less than the true weight.

  10. Compare with the true weight

    mg=666 Nmg=666\,\text{N}

    The true weight does not change.

  11. Note the net force

    R−mg=68×(−2.2)R-mg=68\times(-2.2)

    The net force provides the acceleration.

  12. Check the limiting case

    a=0⇒R=mga=0\Rightarrow R=mg

    At constant velocity the reaction equals the weight.

  13. State the modelling assumption

    person modelled as a particle\text{person modelled as a particle}

    The reaction acts through a single point.

  14. State the reaction with units

    R=517 NR=517\,\text{N}

    The floor pushes up on the person with this force.

  15. State the normal reaction

    R=517 N (apparent weight)R=517\,\text{N}\ \text{(apparent weight)}

    This is the force the floor exerts on the person.

Answer
R=517 NR=517\,\text{N}
Question 5
8 markschallenging
A car of mass 1000 kg1000\,\text{kg} tows a trailer of mass 1000 kg1000\,\text{kg} by a light rigid tow-bar along a straight horizontal road. The car brakes with a braking force of 3000 N3000\,\text{N}. The resistances to motion are 500 N500\,\text{N} on the car and 500 N500\,\text{N} on the trailer. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the magnitude of the force in the tow-bar, stating whether it is a tension or a thrust.
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Worked solution

  1. Model the car and trailer and mark the forces

    braking 3000 N, resistances 500 N, 500 N\text{braking }3000\,\text{N},\ \text{resistances }500\,\text{N},\ 500\,\text{N}

    Braking force acts on the car; both vehicles decelerate together through the light rigid tow-bar.

  2. Apply Newton's second law to the whole system

    −3000−500−500=(1000+1000)a-3000-500-500=(1000+1000)a

    Taking the direction of motion as positive, all forces oppose the motion.

  3. Solve for the acceleration

    a=−3000−500−5001000+1000a=\dfrac{-3000-500-500}{1000+1000}

    Divide the net force by the total mass.

  4. Evaluate the (negative) acceleration

    a=−2 m s−2a=-2\,\text{m s}^{-2}

    The negative sign shows the system is decelerating.

  5. Consider the trailer alone

    C−500=1000aC-500=1000 a

    CC is the force in the tow-bar on the trailer (positive if a tension).

  6. Substitute the acceleration

    C=1000(−2)+500C=1000(-2)+500

    Insert the deceleration and the trailer's resistance.

  7. Evaluate the tow-bar force

    C=−1500 NC=-1500\,\text{N}

    A negative value means the tow-bar pushes the trailer, i.e. a thrust.

  8. Interpret the sign

    tow-bar in thrust\text{tow-bar in thrust}

    During braking the tow-bar is typically in thrust (compression).

  9. State the magnitude

    ∣C∣=1500 N|C|=1500\,\text{N}

    This is the magnitude of the force in the tow-bar.

  10. Check with the car alone

    500+1500−3000 balances 1000a500+1500-3000\ \text{balances }1000 a

    The car's equation is consistent.

  11. State the assumptions

    light rigid tow-bar\text{light rigid tow-bar}

    The tow-bar has no mass and can push or pull.

  12. Confirm the units

    a in m s−2, ∣C∣ in Na\ \text{in m s}^{-2},\ |C|\ \text{in N}

    Deceleration and force respectively.

  13. Summarise

    decel 2 m s−2, thrust 1500 N\text{decel }2\,\text{m s}^{-2},\ thrust\ 1500\,\text{N}

    Deceleration and tow-bar force found.

  14. Solve for the acceleration

    a=−3000−500−5001000+1000a=\dfrac{-3000-500-500}{1000+1000}

    Divide the net force by the total mass.

  15. State the force in the tow-bar

    thrust =1500 Nthrust\ =1500\,\text{N}

    This is the magnitude and nature of the tow-bar force.

Answer
thrust 1500 N\text{thrust }1500\,\text{N}

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