A-Level Connected particles Practice Questions

Free A-Level Connected particles practice questions with full step-by-step worked solutions. Covers lift, normal-reaction, pulley, acceleration. Practise exam-style problems and check your method.

liftnormal-reactionpulleyaccelerationtensiontable
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A person of mass 70 kg70\,\text{kg} stands in a lift that is accelerating upwards at 2 m s−22\,\text{m s}^{-2}. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the normal reaction between the person and the floor of the lift.
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Worked solution

  1. Model the person and identify the forces

    R (up), mg=70×9.8=686 N (down)R\ (\text{up}),\ mg=70\times 9.8=686\,\text{N}\ (\text{down})

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

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

    R−mg=maR-mg=ma

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

  3. State the normal reaction

    R=826 N (apparent weight)R=826\,\text{N}\ \text{(apparent weight)}

    This is the force the floor exerts on the person.

Answer
R=826 NR=826\,\text{N}
Question 2
2 markseasy
A particle moves along a straight line at constant velocity. What can be said about the resultant force acting on it?
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Worked solution

  1. Apply Newton's second law

    F=maF=ma

    The resultant force equals mass times acceleration.

  2. Note constant velocity means no acceleration

    a=0a=0

    Velocity is not changing.

  3. State the resultant force

    F=0F=0

    A particle at constant velocity is in equilibrium.

Answer
F=0F=0
Question 3
3 marksintermediate
Two particles connected by a light inextensible string hang over a smooth pulley (an Atwood machine), with masses MM and mm where M>mM>m. Which statement about the string tension TT is correct while the system moves?
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Worked solution

  1. Consider the heavier particle

    Mg−T=MaMg-T=Ma

    It accelerates downward, so its weight exceeds the tension: T<MgT<Mg.

  2. Deduce the upper bound

    T<MgT<Mg

    Rearranging the heavier particle's equation gives T<MgT<Mg.

  3. Consider the lighter particle

    T−mg=maT-mg=ma

    It accelerates upward, so the tension exceeds its weight: T>mgT>mg.

  4. Deduce the lower bound

    T>mgT>mg

    Rearranging the lighter particle's equation gives T>mgT>mg.

  5. Combine the two inequalities

    mg<T<Mgmg<T<Mg

    The tension lies strictly between the two weights.

  6. State the range of the tension

    ∴ mg<T<Mg\therefore\ mg<T<Mg

    This is why the tension is never equal to either weight while moving.

Answer
mg<T<Mgmg<T<Mg
Question 4
5 markshard
A person of mass mm stands on scales in a lift. When the lift accelerates upward at aa, the scales read a value RR. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, which statement is correct?
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Worked solution

  1. Identify the forces on the person

    R↑, mg↓R\uparrow,\ mg\downarrow

    The scales push up with RR; gravity pulls down with mgmg.

  2. Apply Newton's second law (up positive)

    R−mg=maR-mg=ma

    The net upward force gives the upward acceleration.

  3. Rearrange for the reading

    R=mg+maR=mg+ma

    Add mgmg to both sides.

  4. Factorise

    R=m(g+a)R=m(g+a)

    Take out the common factor mm.

  5. Compare with the true weight

    R>mgR>mg

    Since a>0a>0, the reading exceeds the true weight.

  6. Interpret physically

    feels heavier\text{feels heavier}

    The person feels heavier while accelerating upward.

  7. Reject the downward-acceleration option

    R≠m(g−a)R\neq m(g-a)

    That expression applies when the lift accelerates downward.

  8. Reject the constant-velocity option

    R≠mgR\neq mg

    That applies only when the acceleration is zero.

  9. Reject the gravity-ignored option

    R≠maR\neq ma

    Gravity cannot be ignored; the weight still acts.

  10. State the scale reading

    R=m(g+a)R=m(g+a)

    The apparent weight is greater than the true weight.

Answer
R=m(g+a)R=m(g+a)
Question 5
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)

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