Connected particles Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Connected particles questions. See exactly how to solve problems on lift, normal-reaction, pulley, acceleration.

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.

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 person of mass 60 kg60\,\text{kg} stands in a lift that is accelerating downwards at 1.5 m s−21.5\,\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.

Worked solution

  1. Model the person and identify the forces

    R (up), mg=60×9.8=588 N (down)R\ (\text{up}),\ mg=60\times 9.8=588\,\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 normal reaction

    R=498 N (apparent weight)R=498\,\text{N}\ \text{(apparent weight)}

    This is the force the floor exerts on the person.

Answer
R=498 NR=498\,\text{N}
Question 3
2 markseasy
Two particles of masses 6 kg6\,\text{kg} and 4 kg4\,\text{kg} are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the acceleration of the system.

Worked solution

  1. Set up the model and mark the forces

    6 kg↓, 4 kg↑, tension T\text{6\,kg}\downarrow,\ \text{4\,kg}\uparrow,\ \text{tension }T

    The string is light and inextensible over a smooth pulley, so the tension TT is the same throughout and both particles share the acceleration magnitude aa.

  2. Apply Newton's second law to each particle

    6(9.8)−T=6a,T−4(9.8)=4a6(9.8)-T=6 a,\quad T-4(9.8)=4 a

    Take the direction of motion as positive for each particle.

  3. State the common acceleration

    a=1.96 m s−2a=1.96\,\text{m s}^{-2}

    This is the acceleration of the system.

Answer
a=1.96 m s−2a=1.96\,\text{m s}^{-2}
Question 4
2 markseasy
Two particles of masses 3 kg3\,\text{kg} and 2 kg2\,\text{kg} are connected by a light inextensible string passing over a smooth fixed pulley. The system is released from rest. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the tension in the string. Give your answer to 3 significant figures.

Worked solution

  1. Set up the model and mark the forces

    3 kg↓, 2 kg↑, tension T\text{3\,kg}\downarrow,\ \text{2\,kg}\uparrow,\ \text{tension }T

    The string is light and inextensible over a smooth pulley, so the tension TT is the same throughout and both particles share the acceleration magnitude aa.

  2. Apply Newton's second law to each particle

    3(9.8)−T=3a,T−2(9.8)=2a3(9.8)-T=3 a,\quad T-2(9.8)=2 a

    Take the direction of motion as positive for each particle.

  3. State the tension

    T=23.5 NT=23.5\,\text{N}

    This is the tension in the string.

Answer
T=23.5 NT=23.5\,\text{N}
Question 5
2 markseasy
A particle of mass 8 kg8\,\text{kg} rests on a smooth horizontal table. It is connected by a light inextensible string, passing over a smooth pulley at the edge of the table, to a particle of mass 2 kg2\,\text{kg} hanging freely. The system is released from rest. Taking g=9.8 m s−2g=9.8\,\text{m s}^{-2}, find the acceleration of the system.

Worked solution

  1. Model the system and mark the forces

    table 8 kg, hanging 2 kg, tension T\text{table }8\,\text{kg},\ \text{hanging }2\,\text{kg},\ \text{tension }T

    The table is smooth and the string is light and inextensible over a smooth pulley, so both particles share the acceleration magnitude aa and the tension is the same throughout.

  2. Apply Newton's second law to each particle

    T=8a,2(9.8)−T=2aT=8 a,\quad 2(9.8)-T=2 a

    Horizontally for the particle on the table; vertically for the hanging particle.

  3. State the acceleration

    a=1.96 m s−2a=1.96\,\text{m s}^{-2}

    This is the acceleration of the system.

Answer
a=1.96 m s−2a=1.96\,\text{m s}^{-2}

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