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IGCSE Physics Recap – Forces, Moments and Momentum | Adrian (2026)

Effects of Forces, Moments and Momentum

IGCSE Physics

Adrian (2026)

For use on Monday, 7 September 2026

01

Effects of Forces

Motion of a Lift

A lift of total mass 650 kg is initially at rest. Its motor produces an upward tension of 7800 N in the supporting cable. A constant frictional force of 650 N opposes the motion of the lift.

Take the gravitational field strength, g, to be 10 N/kg. Assume that the forces remain constant during each stage of the motion.
Tension = 7800 N Weight Friction = 650 N LIFT Mass = 650 kg Direction of motion

The diagram is not drawn to scale. Friction acts in the direction opposite to the motion.

  1. State two possible effects that a resultant force can have on an object.
  2. Calculate the weight of the lift.
  3. Calculate the magnitude and direction of the resultant force on the lift as it begins to move upwards.
  4. Calculate the initial acceleration of the lift.
  5. The lift accelerates from rest for 3.0 s. Calculate:
    1. its speed after 3.0 s;
    2. the distance it travels during this time.
  6. After 3.0 s, the tension in the cable is reduced to 5850 N. The lift is still moving upwards, so friction continues to act downwards. Calculate the new resultant force, including its direction.
  7. Calculate the magnitude and direction of the lift’s new acceleration.
  8. Determine the time taken for the lift to come momentarily to rest after the tension is reduced.
  9. Calculate the additional distance travelled by the lift before it comes momentarily to rest.
18 marks
02

Turning Effect

Equilibrium of a Loaded Beam

A uniform horizontal beam AB is 5.0 m long and has a weight of 180 N. It is supported on a narrow pivot located 2.0 m from end A. A crate of weight 240 N is placed at end A, and an object of unknown weight W is placed at end B. The beam is initially in equilibrium.

A B Crate 240 N Object W Pivot Beam’s weight 180 N 2.0 m 3.0 m

The beam is uniform, so its weight acts at the geometric centre of the beam.

  1. State the principle of moments for an object in equilibrium.
  2. State the distance between the pivot and the point at which the beam’s weight acts.
  3. Calculate the anticlockwise moment produced by the crate about the pivot.
  4. Calculate the clockwise moment produced by the weight of the beam about the pivot.
  5. Use the principle of moments to calculate the unknown weight W.
  6. Calculate the upward force exerted by the pivot on the beam.
  7. The object of weight W is now moved 0.40 m from B towards the pivot. The crate remains at A.
    1. Calculate the total clockwise moment about the pivot.
    2. Calculate the resultant moment about the pivot.
    3. State which end of the beam begins to move downwards.
  8. The object remains in its new position. Calculate how far the crate must be moved from A towards the pivot to restore equilibrium.
18 marks
03

Momentum

Collision, Impulse and Energy

Two trolleys move along the same straight, horizontal track. Trolley A has a mass of 0.80 kg and travels to the right at 6.0 m/s. Trolley B has a mass of 1.20 kg and travels to the left at 1.0 m/s. The trolleys collide and lock together. The collision lasts for 0.060 s.

Take motion to the right as positive. Resistance from the track is negligible during the collision.
Trolley A Mass = 0.80 kg 6.0 m/s Trolley B Mass = 1.20 kg 1.0 m/s Straight horizontal track

The velocities shown are the velocities immediately before the collision.

  1. Define momentum and state whether it is a scalar or vector quantity.
  2. Calculate the momentum of each trolley immediately before the collision. Include the correct signs.
  3. Calculate the total momentum of the two-trolley system before the collision.
  4. Use conservation of momentum to calculate the velocity of the joined trolleys immediately after the collision. State both the magnitude and direction.
  5. Calculate the impulse acting on trolley A during the collision. Include the correct sign and state the direction of the impulse.
  6. Calculate the average force acting on trolley A during the collision.
  7. State the magnitude and direction of the average force exerted on trolley B. Explain your answer using Newton’s third law.
  8. Calculate:
    1. the total kinetic energy of the trolleys before the collision;
    2. the kinetic energy of the joined trolleys after the collision;
    3. the kinetic energy transferred to other forms during the collision.
  9. Explain why momentum is conserved in this collision even though kinetic energy is not conserved.
22 marks

Question 1

  1. A resultant force can:
    • change the speed of an object;
    • change the direction of motion;
    • cause acceleration or deceleration;
    • change the shape or size of an object.
    Any two suitable effects are acceptable.
  2. Weight = Mass × Gravitational field strength
    W = mg
    W = 650 × 10 = 6500 N
    Weight of the lift = 6500 N downwards.
  3. The upward force is 7800 N. The total downward force is:
    6500 + 650 = 7150 N
    Therefore:
    Resultant force = 7800 − 7150 = 650 N
    Resultant force = 650 N upwards.
  4. Using F = ma:
    a = F / m
    a = 650 / 650 = 1.0 m/s2
    Initial acceleration = 1.0 m/s2 upwards.
    1. Using v = u + at:
      v = 0 + (1.0 × 3.0)
      v = 3.0 m/s
      Speed after 3.0 s = 3.0 m/s.
    2. Using s = ut + ½at2:
      s = (0 × 3.0) + ½(1.0)(3.0)2
      s = 4.5 m
      Distance travelled = 4.5 m.
  6. The lift is still moving upwards, so both its weight and friction act downwards.
    Total downward force = 6500 + 650 = 7150 N
    Resultant force = 7150 − 5850 = 1300 N
    New resultant force = 1300 N downwards.
  7. a = F / m
    a = 1300 / 650 = 2.0 m/s2
    Acceleration = 2.0 m/s2 downwards. If upwards is taken as positive, the acceleration is −2.0 m/s2.
  8. At the start of this stage, the upward velocity is 3.0 m/s. At the instant the lift stops, v = 0.
    v = u + at
    0 = 3.0 + (−2.0)t
    t = 1.5 s
    Time taken to come momentarily to rest = 1.5 s.
  9. Using s = ut + ½at2:
    s = (3.0 × 1.5) + ½(−2.0)(1.5)2
    s = 4.5 − 2.25
    s = 2.25 m
    Additional distance travelled = 2.25 m upwards.
Important: A downward resultant force does not mean that the lift immediately moves downwards. While the lift is still moving upwards, the downward resultant force causes it to slow down.

Question 2

  1. For an object in equilibrium: the total clockwise moment about a pivot equals the total anticlockwise moment about the same pivot.
  2. The beam is uniform, so its weight acts at its geometric centre, 2.5 m from A. The pivot is 2.0 m from A.
    Distance = 2.5 − 2.0 = 0.50 m
    The beam’s weight acts 0.50 m to the right of the pivot.
  3. Moment = Force × Perpendicular distance
    Moment of crate = 240 × 2.0
    Moment of crate = 480 N m
    Anticlockwise moment of the crate = 480 N m.
  4. Moment of beam’s weight = 180 × 0.50
    Moment of beam’s weight = 90 N m
    Clockwise moment of the beam’s weight = 90 N m.
  5. The object at B is 3.0 m from the pivot. For equilibrium:
    Total clockwise moment = Total anticlockwise moment
    90 + 3.0W = 480
    3.0W = 390
    W = 130 N
    Weight of the object = 130 N.
  6. The upward force from the pivot must balance the total downward force.
    Upward force = 240 + 180 + 130
    Upward force = 550 N
    Upward force exerted by the pivot = 550 N.
  7. The object is moved 0.40 m towards the pivot. Its new distance from the pivot is:
    3.0 − 0.40 = 2.60 m
    1. The object’s new clockwise moment is:
      130 × 2.60 = 338 N m
      The beam also produces a clockwise moment of 90 N m.
      Total clockwise moment = 338 + 90
      Total clockwise moment = 428 N m
      Total clockwise moment = 428 N m.
    2. The anticlockwise moment remains 480 N m.
      Resultant moment = 480 − 428
      Resultant moment = 52 N m
      Resultant moment = 52 N m anticlockwise.
    3. An anticlockwise turning effect causes the left-hand side of the beam to move down. End A begins to move downwards.
  8. For equilibrium, the crate must provide an anticlockwise moment of 428 N m. Let the new distance of the crate from the pivot be d.
    240d = 428
    d = 428 / 240
    d = 1.783 m
    The crate was originally 2.0 m from the pivot.
    Distance moved = 2.0 − 1.783
    Distance moved = 0.217 m
    The crate must be moved approximately 0.22 m from A towards the pivot.
Key distinction: Translational equilibrium requires the resultant force to be zero. Rotational equilibrium requires the resultant moment to be zero. A stationary beam must satisfy both conditions.

Question 3

  1. Momentum is the product of mass and velocity.
    Momentum, p = mv
    Momentum is a vector quantity because it has both magnitude and direction.
  2. Taking motion to the right as positive:
    Momentum of A = 0.80 × 6.0
    Momentum of A = +4.8 kg m/s
    Momentum of B = 1.20 × (−1.0)
    Momentum of B = −1.2 kg m/s
    Trolley A: +4.8 kg m/s
    Trolley B: −1.2 kg m/s
  3. Total momentum = +4.8 + (−1.2)
    Total momentum = +3.6 kg m/s
    Total momentum before collision = 3.6 kg m/s to the right.
  4. The trolleys lock together, so their combined mass is:
    Combined mass = 0.80 + 1.20 = 2.00 kg
    By conservation of momentum:
    Total momentum before = Total momentum after
    3.6 = 2.00v
    v = 1.8 m/s
    Velocity after collision = 1.8 m/s to the right.
  5. Impulse is equal to the change in momentum. For trolley A:
    Impulse = m(v − u)
    Impulse = 0.80(1.8 − 6.0)
    Impulse = 0.80(−4.2)
    Impulse = −3.36 N s
    Impulse on trolley A = −3.36 N s, or 3.36 N s to the left.
  6. Average force = Impulse / Time
    F = −3.36 / 0.060
    F = −56 N
    Average force on trolley A = 56 N to the left.
  7. Newton’s third law states that when two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction. Therefore: The average force on trolley B is 56 N to the right. The force on A and the force on B act on different objects, so they do not cancel each other on either individual trolley.
    1. Kinetic energy of trolley A:
      KEA = ½mv2
      KEA = ½(0.80)(6.0)2
      KEA = 14.4 J
      Kinetic energy of trolley B:
      KEB = ½(1.20)(1.0)2
      KEB = 0.60 J
      Total KE before = 14.4 + 0.60
      Total KE before = 15.0 J
      Total kinetic energy before collision = 15.0 J.
    2. The combined mass is 2.00 kg and the final speed is 1.8 m/s.
      KE after = ½(2.00)(1.8)2
      KE after = 3.24 J
      Kinetic energy after collision = 3.24 J.
    3. Energy transferred = 15.0 − 3.24
      Energy transferred = 11.76 J
      Kinetic energy transferred to other forms = 11.76 J, approximately 11.8 J.
  9. The total momentum is conserved because the two trolleys form an approximately isolated system during the short collision. The resultant external force and external impulse are negligible. However, the collision is inelastic because the trolleys lock together. Some kinetic energy is transferred into:
    • thermal energy;
    • sound;
    • deformation of the trolleys and coupling mechanism.
    Therefore: total momentum is conserved, but total kinetic energy is not conserved.
Direction rule: Momentum, impulse, velocity and force are vector quantities. A negative value indicates a direction opposite to the direction chosen as positive; it does not indicate a negative magnitude.
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