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Cambridge IGCSE0625

Momentum

Physics 0625 Chapter Notes

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Momentum
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In this chapter

  1. 1Momentum
  2. 2Impulse
  3. 3Force as rate of change of momentum
  4. 4Making the force smaller
  5. 5Conservation of momentum
  6. 6Direction and rebounds

1. Momentum

Momentum is mass × velocity. Velocity and momentum are both vectors, so they have a direction. Mass must be in kg. Convert grams first.

p = mv

Examiner insight

“A noticeable proportion of candidates gave newton as the unit of momentum.” “Some candidates forgot or did not realise that if their answer was going to be in kg m / s that they needed to convert the mass in g to mass in kg.”

2. Impulse

Impulse is force × time for which the force acts. It is also equal to the change of momentum.

Both N s and kg m / s are accepted as the unit of impulse and of momentum.

impulse = FΔ t = Δ(mv)

Examiner insight

“Impulse equals change in momentum or impulse equals force times the time for which the force acts both provided clear definitions of impulse.” “Common misconceptions were that impulse is force over time, a force acting over a period of time or the change in momentum per unit time.”

Worked example 11 mark

A mass of 4.0 kg is travelling at a velocity of 6.0 m / s. A force acts on the mass and its velocity increases to 20 m / s.

Which impulse produces this change?

A 3.5 N s · B 6.5 N s · C 56 N s · D 80 N s

  1. 1

    impulse = Δ(mv) = 4.0 × (20 - 6.0) = 56 N s, so C [1]

3. Force as rate of change of momentum

The resultant force is the change in momentum per unit time.

So a larger change of momentum, or a shorter time, means a larger force.

F = (Δ p)/(Δ t)

Examiner insight

“Many candidates did not recall the definition of resultant force. Instead, many candidates had the misconception that a resultant force is required to keep an object moving at a constant speed.”

Worked example 12 marks

A student jumps from the top of a low wall to the ground. The mass of the student is 55 kg.

As the student reaches the ground, her speed is 4.6 m / s.

The student stops moving 0.95 s after her feet first touch the ground.

Calculate the resultant force exerted on the student as she stops moving.

  1. 1

    F = (Δ p)/t = (55 × 4.6)/0.95 [1]

  2. 2

    = 270 N [1]

4. Making the force smaller

For a given change of momentum, the impulse is fixed. A longer time of impact means a smaller force.

This is why bending the knees on landing reduces injury.

Examiner insight

“Stronger candidates stated that force = rate of change of momentum and that the same impulse acted over a longer period of time, so the force exerted on the child was less.” “A common misconception was that the impulse was reduced when the child hit the rubber floor.”

Worked example 13 marks

As the student lands, she bends her knees.

Explain how this prevents the student from being injured.

Use ideas about force, momentum and time in your answer.

  1. 1

    time for impact is longer [1]

  2. 2

    same impulse or same change of momentum [1]

  3. 3

    force is smaller [1]

5. Conservation of momentum

In a collision, the total momentum before equals the total momentum after. Write it as an equation first.

Use masses, never weights. If two objects join, use their combined mass and one final velocity.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Examiner insight

“The most common mistake was to substitute the wrong mass to the blocks after the collision, even substituting weight in a momentum equation” “Some candidates did not realise that the masses were combined and/or stated two separate velocities after the collision.”

6. Direction and rebounds

Take one direction as positive and the opposite direction as negative. Show the minus sign in your working.

If a ball rebounds, the velocity changes direction. For speeds v₁ and v₂ the change in velocity is v₁ + v₂.

Δ p = m(v₁ + v₂)

Examiner insight

“They needed to write down the equation they were using and to show the correct change of momentum by identifying one of the velocities as negative with respect to the other (positive) velocity.” “Therefore the change in velocity of the ball was v1 – (–v2) = (v1 + v2) and the change in momentum = m(v1 + v2).”

Worked example 11 mark

When a particle rebounds from a wall, a force is exerted on the wall.

What causes this force?

A the kinetic energy gained by the particle · B the kinetic energy lost by the particle · C the change of momentum of the particle · D the change of speed of the particle

  1. 1

    C: the change of momentum of the particle. [1]

Recap

  • Momentum = mass × velocity. It is a vector. Unit: kg m / s.
  • Impulse = force × time for which the force acts = change of momentum. Unit: N s.
  • Resultant force = change in momentum per unit time.
  • A longer time of impact gives a smaller force for the same impulse.
  • Total momentum before a collision = total momentum after.
  • Opposite directions have opposite signs. In a rebound, the change in velocity is the sum of the two speeds.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. State the equation for momentum and give a unit for it.2 marks
  2. A trolley of mass 1200 g moves at 0.50 m / s. Calculate its momentum.2 marks
  3. A force of 12 N acts on a trolley for 0.50 s. Calculate the impulse. State the change of momentum of the trolley.3 marks
  4. A boot kicks a ball of mass 40 g. The ball leaves the boot at 20 m / s from rest. The boot is in contact with the ball for 6.0 ms. Calculate the average force on the ball. Give your answer to 2 significant figures.3 marks
  5. A car of mass 800 kg slows from 15 m / s to rest in 6.0 s. Calculate the average braking force.3 marks
  6. Trolley A has mass 2.0 kg and moves at 3.0 m / s towards trolley B, which has mass 1.0 kg and is at rest. They stick together. Calculate their velocity after the collision.3 marks
  7. A truck of mass 4000 kg moves at 5.0 m / s towards a truck of mass 6000 kg moving at 2.0 m / s in the opposite direction. They couple together. Calculate their velocity after the collision.4 marks
  8. A ball of mass 0.50 kg hits a wall at 8.0 m / s and rebounds along the same line at 6.0 m / s. Calculate the change of momentum of the ball.3 marks
  9. A tennis ball of mass 60 g moves towards a racket at 15 m / s. The racket exerts an average force of 600 N on the ball for 4.0 ms. The ball leaves the racket in the opposite direction. (a) Calculate the impulse on the ball. (b) Calculate the speed of the ball as it leaves the racket. (c) The strings are loosened, so the ball is in contact with the racket for a longer time. The ball arrives and leaves at the same speeds as before. State and explain how the average force on the ball changes.7 marks
  10. On an ice rink, a puck of mass 200 g slides at 6.0 m / s and hits a stationary curling stone head-on. The puck bounces straight back at 1.5 m / s. The stone moves forwards at 0.075 m / s. Friction is negligible. (a) Calculate the mass of the stone. (b) The puck and the stone are in contact for 0.030 s. Calculate the average force on the puck. (c) A student says: "The momentum of the puck fell from 1.2 kg m / s to 0.30 kg m / s, so momentum was not conserved." Explain the two mistakes in this statement.8 marks

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