Momentum

1. Overview

Momentum is a fundamental concept in physics that describes the "quantity of motion" an object possesses. It is a vector quantity that accounts for both the mass and the velocity of an object, helping us predict the outcomes of collisions and explosions in everyday life, from car crashes to sports.

Key Definitions

• Momentum: The product of an object's mass and its velocity.
• Impulse: The product of the force acting on an object and the time for which it acts, resulting in a change in momentum.
• Conservation of Momentum: The principle stating that the total momentum of a closed system remains constant, provided no external forces act on it.
• Resultant Force: In terms of momentum, it is defined as the rate of change of momentum per unit time.

Core Content

Note: The IGCSE syllabus classifies the specific learning objectives for Momentum under the Supplement (Extended) curriculum. Refer to the section below for full details.

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Extended Content (Extended curriculum only)

A. Calculating Momentum

Momentum ($p$) depends on mass and velocity. Because velocity is a vector, momentum is also a vector. This means the direction is just as important as the magnitude.

• Equation: $p = mv$
• If an object is at rest (velocity = 0), its momentum is 0.

B. Impulse and Change in Momentum

Impulse describes the effect of a force acting over a period of time. When a resultant force acts on an object, it causes the object's velocity to change, which means its momentum changes.

• Equation: $Impulse = F\Delta t = \Delta(mv)$
• $\Delta(mv)$ represents the change in momentum (Final Momentum – Initial Momentum).

C. The Conservation of Momentum

In any collision or explosion involving two or more objects, the total momentum before the event is equal to the total momentum after the event, provided no external forces (like friction) act.

• Total momentum before = Total momentum after
• $(m_1 \times u_1) + (m_2 \times u_2) = (m_1 \times v_1) + (m_2 \times v_2)$

Worked Example: A 2 kg trolley moving at 3 m/s hits a stationary 1 kg trolley. They stick together and move off. Calculate their common velocity.

1. Momentum before = $(2 \text{ kg} \times 3 \text{ m/s}) + (1 \text{ kg} \times 0 \text{ m/s}) = 6 \text{ kg m/s}$.
2. Momentum after = $(2 \text{ kg} + 1 \text{ kg}) \times v = 3v$.
3. Apply conservation: $6 = 3v$.
4. Solve: $v = 2 \text{ m/s}$.

D. Force as Rate of Change of Momentum

Newton's Second Law can be expressed more precisely using momentum. The resultant force acting on an object is equal to the change in momentum divided by the time taken for that change.

• Equation: $F = \frac{\Delta p}{t}$ or $F = \frac{mv - mu}{t}$

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Key Equations

Equation	Symbols	Units
$p = mv$	$p$ = momentum, $m$ = mass, $v$ = velocity	kg m/s, kg, m/s
$Impulse = F\Delta t$	$F$ = Force, $\Delta t$ = time interval	N s (Newton-seconds)
$\Delta p = mv - mu$	$mu$ = initial momentum, $mv$ = final momentum	kg m/s
$F = \frac{\Delta p}{t}$	$F$ = Resultant Force, $t$ = time	N, kg m/s, s

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Common Mistakes to Avoid

• ❌ Wrong: Calculating change in momentum for a rebounding ball by subtracting speeds (e.g., $10 \text{ m/s} - 7 \text{ m/s} = 3 \text{ m/s}$).
  • ✅ Right: Treat velocity as a vector. If a ball hits a wall at $+10 \text{ m/s}$ and bounces back at $7 \text{ m/s}$, its new velocity is $-7 \text{ m/s}$. The change in velocity is $10 - (-7) = 17 \text{ m/s}$.
• ❌ Wrong: Dividing mass by velocity ($m/v$) to find momentum.
  • ✅ Right: Always multiply mass and velocity ($mv$).
• ❌ Wrong: Forgetting to convert time into seconds (e.g., using milliseconds or minutes directly in the $F = \Delta p/t$ formula).
  • ✅ Right: Always ensure time is in seconds (s) and mass is in kilograms (kg).
• ❌ Wrong: Assigning a positive sign to momentum in both directions in a collision.
  • ✅ Right: Assign one direction as positive (e.g., Right = $+$) and the opposite as negative (e.g., Left = $-$).

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Exam Tips

1. Check for "Sticking Together": In collision questions, check if the objects move off separately or stick together. If they stick together, add their masses for the "momentum after" calculation.
2. The Unit of Impulse: Remember that Impulse can be measured in either $N \cdot s$ or $kg \cdot m/s$. They are equivalent!
3. Show Your Workings: In momentum conservation problems, clearly state "Total momentum before = Total momentum after" to pick up easy method marks, even if you make a calculation error later.

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A remote-controlled toy car of mass 0.50 kg is travelling in a straight line at a constant speed of 2.0 m/s. It then collides head-on with a stationary toy truck of mass 0.80 kg. After the collision, the car rebounds with a speed of 0.50 m/s in the opposite direction.

(a) Define momentum. [1]

(b) Calculate the momentum of the car before the collision. [2]

(c) Calculate the speed of the truck immediately after the collision. Assume the principle of conservation of momentum applies. [2]

Worked Solution:

(a)

1. Momentum is the product of mass and velocity. [Definition of momentum]

How to earn full marks:

• State that momentum is the product of mass and velocity (or equivalent wording).

(b)

1. Calculate the momentum of the car: $p = mv = 0.50 \text{ kg} \times 2.0 \text{ m/s} = 1.0 \text{ kg m/s}$ [Applying the formula for momentum]

How to earn full marks:

• Correct substitution of mass and velocity into the momentum formula.
• Correct final answer with correct units.

(c)

1. Apply the principle of conservation of momentum: Total momentum before = Total momentum after $m_c v_{c,i} + m_t v_{t,i} = m_c v_{c,f} + m_t v_{t,f}$ [Stating conservation of momentum]
2. Substitute the given values: $(0.50 \text{ kg} \times 2.0 \text{ m/s}) + (0.80 \text{ kg} \times 0 \text{ m/s}) = (0.50 \text{ kg} \times -0.50 \text{ m/s}) + (0.80 \text{ kg} \times v_{t,f})$ [Substituting values, noting the negative sign for the rebound velocity]
3. Solve for $v_{t,f}$: $1.0 = -0.25 + 0.80v_{t,f}$ $1.25 = 0.80v_{t,f}$ $v_{t,f} = \frac{1.25}{0.80} = 1.5625 \text{ m/s} \approx 1.56 \text{ m/s}$ [Rearranging and solving for the final velocity of the truck]

How to earn full marks:

• Correctly apply the principle of conservation of momentum.
• Correct substitution of values, including the negative sign for the rebound velocity.
• Correct final answer with correct units: $\boxed{1.56 \text{ m/s}}$

Common Pitfall: Remember that momentum is a vector quantity, so direction matters! Make sure to include a negative sign for velocities in the opposite direction to your chosen positive direction. Also, always include the correct SI unit (kg m/s) in your final answer.

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#### Exam-Style Question 2 — Extended Response [8 marks]

**Question:**

A student investigates the impulse experienced by a ball when it bounces off a force sensor. The ball, of mass 0.20 kg, is dropped from a height of 1.8 m above the sensor. Assume air resistance is negligible.

(a) Calculate the speed of the ball just before it hits the force sensor. Assume $g = 10 \text{ m/s}^2$. [3]

(b) The force sensor records the force exerted on the ball during the impact. The graph below shows how the force varies with time.

<div class="diagram-placeholder"><span class="diagram-icon">📊</span><span class="diagram-text">A graph of Force (y-axis, units Newtons) against Time (x-axis, units seconds). The x-axis ranges from 0 to 0.1 s. The y-axis ranges from 0 to 50 N. The graph shows a curve that starts at (0,0), rises to a peak of approximately 40 N at approximately 0.02 s, and then decreases back to (0.1,0). The area under the curve is approximately triangular.</span></div>

Use the graph to estimate the impulse experienced by the ball during the collision. [3]

(c) Calculate the speed of the ball immediately after the collision. [2]

**Worked Solution:**

**(a)**
1.  Use the equation $v^2 = u^2 + 2as$ where $u=0$, $a=g=10 \text{ m/s}^2$, and $s=1.8 \text{ m}$.
    *[Using the kinematic equation to determine final velocity]*
2.  $v^2 = 0^2 + 2 \times 10 \text{ m/s}^2 \times 1.8 \text{ m} = 36 \text{ m}^2/\text{s}^2$
    *[Substituting values]*
3.  $v = \sqrt{36} = 6.0 \text{ m/s}$
    *[Taking the square root]*

**How to earn full marks:**
- Correctly state the appropriate kinematic equation.
- Correct substitution of values.
- Correct final answer with correct units: $\boxed{6.0 \text{ m/s}}$

**(b)**
1.  Impulse is equal to the area under the force-time graph.
    *[Stating the relationship between impulse and area under the F-t graph]*
2.  Estimate the area as a triangle: Area $\approx \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 0.1 \text{ s} \times 40 \text{ N} = 2.0 \text{ Ns}$
    *[Approximating the area of the graph as a triangle and calculating its area]*

**How to earn full marks:**
- State that impulse is the area under the force-time graph.
- Correctly estimate the area under the graph (allow a range of reasonable estimates, e.g., 1.8 - 2.2 Ns).
- Correct final answer with correct units: $\boxed{2.0 \text{ Ns}}$

**(c)**
1.  Impulse = change in momentum: $I = \Delta p = m(v_f - v_i)$
    *[Stating the impulse-momentum theorem]*
2.  $2.0 \text{ Ns} = 0.20 \text{ kg} (v_f - (-6.0 \text{ m/s}))$ *Remember to include the negative sign for initial downward velocity*
    *[Substituting values and recognizing that the initial velocity is negative since it's in the opposite direction to the final velocity]*
3.  $2.0 = 0.20v_f + 1.2$
    $0.8 = 0.20v_f$
    $v_f = \frac{0.8}{0.20} = 4.0 \text{ m/s}$
    *[Solving for the final velocity]*

**How to earn full marks:**
- Correctly relate impulse to change in momentum.
- Correct substitution of values, including the negative sign for the initial velocity.
- Correct final answer with correct units: $\boxed{4.0 \text{ m/s}}$

**Common Pitfall:** When calculating impulse from a force-time graph, remember that impulse is the *area* under the graph. Also, be careful with signs when using the impulse-momentum equation. Define a direction as positive, and make sure velocities in the opposite direction are negative.

#### Exam-Style Question 3 — Short Answer [6 marks]

**Question:**

A rocket of total mass 500 kg is launched vertically upwards. The rocket engines expel exhaust gases downwards at a rate of 5.0 kg/s and at a velocity of 800 m/s relative to the rocket.

(a) Explain how the expulsion of exhaust gases causes the rocket to accelerate upwards, in terms of Newton's Laws. [3]

(b) Calculate the magnitude of the thrust force exerted on the rocket by the exhaust gases. [3]

**Worked Solution:**

**(a)**
1. The rocket expels hot gases downwards. *[Identifying the initial action]*
2. This expulsion of gases exerts a force on the gases. *[Describing the action force]*
3. By Newton's Third Law, the gases exert an equal and opposite force on the rocket, pushing it upwards. *[Applying Newton's Third Law]*
4. This upward force (thrust) is greater than the weight of the rocket, so the rocket accelerates upwards, according to Newton's Second Law ($F = ma$). *[Linking to Newton's Second Law and resultant force]*

**How to earn full marks:**
- Mention the expulsion of gases downwards.
- State Newton's Third Law (or the principle of equal and opposite forces).
- Explain that the upward force on the rocket is what causes it to accelerate upwards, relating it to Newton's Second Law.

**(b)**
1. Thrust force is equal to the rate of change of momentum of the exhaust gases: $F = \frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$
    *[Stating the relationship between thrust and rate of change of momentum]*
2.  Since the velocity of the exhaust gases is constant relative to the rocket, $F = v \frac{\Delta m}{\Delta t}$
    *[Simplifying the equation]*
3.  $F = 800 \text{ m/s} \times 5.0 \text{ kg/s} = 4000 \text{ N}$
    *[Substituting values and calculating the force]*

**How to earn full marks:**
- Correctly relate thrust to the rate of change of momentum.
- Correct substitution of values.
- Correct final answer with correct units: $\boxed{4000 \text{ N}}$

**Common Pitfall:** Make sure you understand how Newton's Third Law applies to rocket propulsion. The rocket exerts a force on the exhaust gases, and the exhaust gases exert an equal and opposite force on the rocket. Also, remember that thrust is the rate of change of momentum of the exhaust gases.

#### Exam-Style Question 4 — Extended Response [9 marks]

**Question:**

Two trolleys, A and B, are moving towards each other on a horizontal track. Trolley A has a mass of 2.0 kg and is moving to the right at 3.0 m/s. Trolley B has a mass of 1.0 kg and is moving to the left at 4.0 m/s. The trolleys collide and stick together.

(a) Calculate the total momentum of the system (trolleys A and B) before the collision. State the direction of the momentum. [3]

(b) Calculate the velocity (magnitude and direction) of the combined trolleys after the collision. [3]

(c) Calculate the total kinetic energy of the system before the collision. [2]

(d) Explain why the kinetic energy of the system is *not* conserved in this collision. [1]

**Worked Solution:**

**(a)**
1.  Momentum of trolley A: $p_A = m_A v_A = 2.0 \text{ kg} \times 3.0 \text{ m/s} = 6.0 \text{ kg m/s}$ (to the right)
    *[Calculating the momentum of trolley A]*
2.  Momentum of trolley B: $p_B = m_B v_B = 1.0 \text{ kg} \times (-4.0 \text{ m/s}) = -4.0 \text{ kg m/s}$ (to the left) *Remember to include the negative sign for a direction opposite to A*
    *[Calculating the momentum of trolley B, noting the direction is opposite]*
3.  Total momentum: $p_{total} = p_A + p_B = 6.0 \text{ kg m/s} - 4.0 \text{ kg m/s} = 2.0 \text{ kg m/s}$ (to the right)
    *[Adding the momenta, taking direction into account]*

**How to earn full marks:**
- Correctly calculate the momentum of trolley A.
- Correctly calculate the momentum of trolley B, including the correct sign.
- Correctly calculate the total momentum with the correct direction: $\boxed{2.0 \text{ kg m/s} \text{ to the right}}$

**(b)**
1.  Apply the principle of conservation of momentum: $p_{total, before} = p_{total, after}$
    *[Stating the conservation of momentum]*
2.  $2.0 \text{ kg m/s} = (m_A + m_B) v_{final} = (2.0 \text{ kg} + 1.0 \text{ kg}) v_{final}$
    *[Substituting values]*
3.  $v_{final} = \frac{2.0 \text{ kg m/s}}{3.0 \text{ kg}} = 0.67 \text{ m/s}$ (to the right)
    *[Solving for the final velocity]*

**How to earn full marks:**
- Correctly apply the principle of conservation of momentum.
- Correct substitution of values.
- Correct final answer with correct units and direction: $\boxed{0.67 \text{ m/s} \text{ to the right}}$

**(c)**
1.  Kinetic energy of trolley A: $KE_A = \frac{1}{2} m_A v_A^2 = \frac{1}{2} \times 2.0 \text{ kg} \times (3.0 \text{ m/s})^2 = 9.0 \text{ J}$
    *[Calculating the kinetic energy of trolley A]*
2.  Kinetic energy of trolley B: $KE_B = \frac{1}{2} m_B v_B^2 = \frac{1}{2} \times 1.0 \text{ kg} \times (-4.0 \text{ m/s})^2 = 8.0 \text{ J}$
    *[Calculating the kinetic energy of trolley B]*
3.  Total kinetic energy: $KE_{total} = KE_A + KE_B = 9.0 \text{ J} + 8.0 \text{ J} = 17.0 \text{ J}$
    *[Adding the kinetic energies]*

**How to earn full marks:**
- Correctly calculate the kinetic energy of trolley A.
- Correctly calculate the kinetic energy of trolley B.
- Correct final answer with correct units: $\boxed{17.0 \text{ J}}$

**(d)**
1. The collision is inelastic. Some kinetic energy is converted into other forms of energy, such as heat and sound, during the collision. *[Explaining why kinetic energy is not conserved in an inelastic collision]*

**How to earn full marks:**
- State that the collision is inelastic and that kinetic energy is converted into other forms of energy (e.g., heat, sound).

**Common Pitfall:** Remember the difference between elastic and inelastic collisions. In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision (like this one, where the trolleys stick together), momentum is conserved, but kinetic energy is not. Some of the kinetic energy is transformed into other forms of energy, like heat or sound.