Exam study guide
Cambridge IGCSE Physics Momentum Practice Questions
Momentum is a fundamental concept in physics, crucial for understanding how objects interact during collisions and explosions. For Cambridge IGCSE Physics students, mastering the principles of momentum, impulse, and the conservation of momentum is vital for tackling a wide range of problems and achieving excellent results in their examinations. This resource provides targeted practice questions with detailed explanations to solidify your understanding.
Quick revision summary
- Momentum (p) is the product of mass (m) and velocity (v): p = mv. Unit: kg m/s. It is a vector quantity.
- Impulse (I) is the change in momentum: I = Δp = mv - mu. It is also the product of force (F) and time (Δt): I = FΔt. Unit: N s or kg m/s.
- Principle of Conservation of Momentum: For a closed system, the total momentum before a collision or explosion is equal to the total momentum after, provided no external forces act. (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂)
Understanding Momentum
Momentum is a measure of the 'quantity of motion' an object possesses. It depends directly on two factors: the object's mass and its velocity. A heavier object moving at the same speed as a lighter object will have more momentum. Similarly, an object moving faster will have more momentum than the same object moving slowly. Mathematically, momentum (p) is defined as the product of mass (m) and velocity (v): p = mv. The standard unit for momentum is kilogram metres per second (kg m/s). It's crucial to remember that momentum is a vector quantity, meaning it has both magnitude (size) and direction. Therefore, when dealing with momentum problems, the direction of motion must always be considered, often by assigning positive and negative signs to velocities in opposite directions.
Impulse and Force
When a force acts on an object for a certain duration, it causes a change in the object's momentum. This effect is known as impulse. Impulse (I) is defined as the product of the force (F) applied and the time interval (Δt) over which it acts: I = FΔt. According to Newton's second law of motion in terms of momentum, impulse is also equal to the change in momentum (Δp) of the object: I = Δp = mv - mu, where 'u' is the initial velocity and 'v' is the final velocity. This relationship, FΔt = mv - mu, is incredibly important as it links force, time, mass, and velocity. The unit for impulse can be expressed as Newton seconds (N s) or, equivalently, kilogram metres per second (kg m/s), matching the unit for momentum. Understanding impulse helps explain various safety features, such as crumple zones in cars or airbags, which work by increasing the time over which a collision force acts, thereby reducing the magnitude of the force for the same change in momentum.
The Principle of Conservation of Momentum
One of the most powerful principles in physics is the conservation of momentum. It states that for a closed system, where no external forces act, the total momentum before an interaction (like a collision or explosion) is equal to the total momentum after the interaction. In simpler terms, momentum is neither created nor destroyed, only transferred between objects within the system. For a system of two objects, this can be expressed as: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, where m₁ and m₂ are the masses of the objects, u₁ and u₂ are their initial velocities, and v₁ and v₂ are their final velocities. It is absolutely essential to correctly assign positive and negative signs to velocities based on their direction, as momentum is a vector. This principle applies to both elastic collisions (where kinetic energy is conserved) and inelastic collisions (where kinetic energy is not conserved, often converted to heat or sound), as well as to explosions where objects move apart.
Practice questions
Question 1 (2 marks)
A car of mass 1200 kg travels at a speed of 25 m/s. Calculate its momentum.
Answer: 30000 kg m/s
Explanation: Momentum (p) is calculated using the formula p = mv. Given mass (m) = 1200 kg and velocity (v) = 25 m/s. p = 1200 kg × 25 m/s = 30000 kg m/s. Remember to include the correct SI units for your final answer.
Question 2 (3 marks)
Explain how an airbag in a car helps to reduce injury to a driver during a collision.
Answer: An airbag increases the time duration over which the driver's momentum changes.
Explanation: During a collision, the driver's body undergoes a rapid change in momentum. The impulse (change in momentum) is equal to the force applied multiplied by the time over which it acts (FΔt = Δp). By inflating, an airbag increases the time (Δt) taken for the driver's body to come to rest. For a fixed change in momentum (Δp), increasing Δt results in a decrease in the average force (F) exerted on the driver, thereby reducing the severity of injury.
Question 3 (4 marks)
A 5 kg trolley moving at 3 m/s collides with a stationary 2 kg trolley. After the collision, the two trolleys stick together. Calculate the common velocity of the combined trolleys immediately after the collision.
Answer: 2.14 m/s
Explanation: This problem involves the principle of conservation of momentum. Let m₁ = 5 kg, u₁ = 3 m/s; m₂ = 2 kg, u₂ = 0 m/s. After collision, they stick together, so the combined mass is (m₁ + m₂) = 7 kg and they move with a common final velocity 'v'. Total momentum before collision = m₁u₁ + m₂u₂ = (5 kg × 3 m/s) + (2 kg × 0 m/s) = 15 kg m/s. Total momentum after collision = (m₁ + m₂)v = (5 kg + 2 kg)v = 7v. By conservation of momentum: 15 = 7v. v = 15 / 7 ≈ 2.14 m/s. The common velocity is 2.14 m/s in the initial direction of the 5 kg trolley.
Question 4 (2 marks)
State the SI unit for momentum and impulse, and explain why they are equivalent.
Answer: Both momentum and impulse have the SI unit kg m/s (kilogram metres per second), which is equivalent to N s (Newton second).
Explanation: Momentum (p = mv) has base units of mass (kg) multiplied by velocity (m/s), giving kg m/s. Impulse (I = FΔt) has units of force (N) multiplied by time (s), giving N s. To show equivalence, recall that from Newton's Second Law, Force (F) = mass (m) × acceleration (a), so 1 Newton = 1 kg m/s². Therefore, N s = (kg m/s²) × s = kg m/s. This confirms that the units are indeed equivalent, reflecting that impulse is defined as the change in momentum.
Question 5 (4 marks)
A ball of mass 0.15 kg travelling at 10 m/s hits a wall and rebounds with a speed of 8 m/s in the opposite direction. If the collision lasts for 0.02 seconds, calculate the average force exerted by the wall on the ball.
Answer: 135 N
Explanation: First, calculate the change in momentum (Δp). Define the initial direction as positive. Initial velocity (u) = +10 m/s Final velocity (v) = -8 m/s (opposite direction) Mass (m) = 0.15 kg Change in momentum (Δp) = mv - mu = (0.15 kg × -8 m/s) - (0.15 kg × 10 m/s) Δp = -1.2 kg m/s - 1.5 kg m/s = -2.7 kg m/s. Now use the impulse-momentum theorem: F = Δp / Δt. Time of collision (Δt) = 0.02 s. Average force (F) = -2.7 kg m/s / 0.02 s = -135 N. The magnitude of the average force exerted by the wall on the ball is 135 N. The negative sign indicates the force acts in the opposite direction to the ball's initial motion.
Question 6 (1 marks)
Which of the following statements correctly defines the principle of conservation of momentum?
- A) Momentum is always conserved in all types of collisions.
- B) The total momentum of a system remains constant only if external forces act upon it.
- C) The total momentum of a closed system remains constant, provided no external forces act.
- D) Momentum is only conserved in elastic collisions.
Answer: C) The total momentum of a closed system remains constant, provided no external forces act.
Explanation: The principle of conservation of momentum states that the total momentum of a closed system remains constant, provided no net external forces act upon it. This applies to both elastic and inelastic collisions. Options A and D are incorrect as momentum is conserved even in inelastic collisions. Option B is incorrect as external forces would change the total momentum, violating the principle.
Question 7 (3 marks)
A stationary 6 kg rifle fires a 0.015 kg bullet at a velocity of 400 m/s. Calculate the recoil velocity of the rifle.
Answer: -1.0 m/s
Explanation: This is an explosion-type problem, so we apply the principle of conservation of momentum. Initially, both the rifle and bullet are stationary, so the total initial momentum is 0. Let m_b = 0.015 kg (bullet mass), v_b = +400 m/s (bullet velocity after firing). Let m_r = 6 kg (rifle mass), v_r = ? (recoil velocity of rifle). Total momentum before firing = 0. Total momentum after firing = (m_b × v_b) + (m_r × v_r). By conservation of momentum: 0 = (0.015 kg × 400 m/s) + (6 kg × v_r) 0 = 6 kg m/s + 6 kg × v_r 6 kg × v_r = -6 kg m/s v_r = -1.0 m/s. The negative sign indicates that the rifle recoils in the opposite direction to the bullet's motion.
Common mistakes
- Forgetting to assign directions: Momentum is a vector quantity. Many students forget to use positive and negative signs for velocities when objects move in opposite directions, especially in conservation of momentum problems. Always define a positive direction at the start of your calculation.
- Incorrect units: Using grams instead of kilograms for mass, or kilometres per hour instead of metres per second for velocity, will lead to incorrect answers. Ensure all quantities are converted to their standard SI units (kg, m, s) before calculation.
- Confusing momentum with kinetic energy: While both depend on mass and velocity, momentum is a vector (p=mv) and kinetic energy is a scalar (KE=½mv²). Kinetic energy is often not conserved in inelastic collisions, but momentum always is (in a closed system with no external forces).
Exam tips
- Draw Diagrams: For collision and explosion problems, always sketch a simple diagram showing the objects before and after the interaction, with arrows indicating velocities and directions. This helps in visualising the problem and correctly assigning positive and negative signs to velocities.
- State Formulas and Show Working: Clearly write down the formula you are using (e.g., p=mv, FΔt=Δp, m₁u₁+m₂u₂=m₁v₁+m₂v₂). Substitute values and show each step of your calculation. Showing your working can allow method marks to be awarded even if a later arithmetic error occurs.
- Include Units: Always write the correct SI units with your final numerical answer. Incorrect or missing units can lead to a loss of marks, even if the numerical value is correct.
- Define Positive Direction: For problems involving direction (which includes most momentum questions), explicitly state which direction you are taking as positive at the beginning of your solution. This clarifies your sign convention and prevents confusion with negative signs in your answers.
Why this matters for exams
Momentum can be assessed through multiple-choice, theory, calculation, and data-handling questions. A strong understanding of momentum, impulse, conservation of momentum, and direction conventions will help you solve both conceptual and numerical problems accurately.
Frequently asked questions
What is the difference between momentum and kinetic energy?
Momentum (p=mv) is a vector quantity that describes an object's 'quantity of motion' and direction. Kinetic energy (KE=½mv²) is a scalar quantity that describes the energy an object possesses due to its motion. Momentum is conserved in all types of collisions (in a closed system with no external forces), while kinetic energy is only conserved in elastic collisions.
When is the principle of conservation of momentum applicable?
The principle of conservation of momentum applies to a closed system (one where no mass or energy transfers with the surroundings) where no net external forces act on the system. This includes both elastic and inelastic collisions, as well as explosions where objects move apart.
Is impulse a scalar or vector quantity?
Momentum is conserved for a system when the net external force on the system is zero. This applies to collisions and explosions when external forces are negligible during the interaction.
How does momentum relate to Newton's Second Law?
Newton's Second Law can be stated as 'the rate of change of momentum of an object is directly proportional to the resultant force acting on it, and is in the direction of the resultant force.' Mathematically, this is expressed as F = Δp/Δt, which can be rearranged to FΔt = Δp. This equation directly links force and the time over which it acts (impulse) to the change in an object's momentum.