Exam study guide
Cambridge AS Level Physics Dynamics Practice Questions
Dynamics is a foundational topic in Cambridge AS Level Physics, exploring the causes of motion and changes in motion. It builds upon kinematics by introducing concepts like force, mass, momentum, and impulse. Mastering these principles is crucial not only for the AS Level examination but also for understanding more complex physics topics. This collection of practice exercises is designed to help you solidify your understanding of these core concepts and apply them effectively to problem-solving.
Quick revision summary
- Newton's First Law: An object remains at rest or in uniform motion unless acted upon by a resultant force.
- Newton's Second Law: Resultant Force = mass × acceleration (F = ma). The acceleration is in the direction of the resultant force.
- Newton's Third Law: For every action, there is an equal and opposite reaction.
- Weight (W) = mass (m) × gravitational field strength (g).
- Momentum (p) = mass (m) × velocity (v). Momentum is a vector quantity.
- Impulse (I) = Force (F) × time (Δt) = change in momentum (Δp). Impulse is also a vector quantity.
- Principle of Conservation of Momentum: For a closed system, the total momentum before a collision/explosion is equal to the total momentum after, provided no external forces act.
Understanding Forces and Motion
The study of dynamics begins with Newton's three laws of motion, which govern how objects behave under the influence of forces. Newton's First Law introduces the concept of inertia, stating that objects resist changes to their state of motion. The Second Law, F=ma, quantifies the relationship between force, mass, and acceleration, allowing us to calculate how an object's motion changes. Newton's Third Law describes the nature of forces as interactions between two objects, always occurring in equal and opposite pairs.
Momentum and Impulse
Momentum is a measure of an object's 'quantity of motion' and is defined as the product of its mass and velocity. It is a crucial concept for understanding collisions and explosions. Impulse, defined as the change in momentum, is the product of the force applied and the time for which it acts. The principle of conservation of momentum is a powerful tool for analysing systems where external forces are negligible, stating that the total momentum of a system remains constant.
Practice questions
Question 1 (2 marks)
A spacecraft is travelling through space at a constant velocity, far from any significant gravitational fields. Its engines are off. Describe the resultant force acting on the spacecraft and explain your reasoning.
Answer: The resultant force acting on the spacecraft is zero.
Explanation: According to Newton's First Law of Motion, an object moving at a constant velocity (zero acceleration) will have no resultant force acting on it. Since the spacecraft is far from gravitational fields and its engines are off, there are no forces to change its motion, hence the resultant force is zero.
Question 2 (4 marks)
A 5.0 kg block is pulled along a rough horizontal surface by a rope. The tension in the rope is 30 N, acting horizontally. If the coefficient of kinetic friction between the block and the surface is 0.40, calculate the acceleration of the block. (Take g = 9.81 m s⁻²)
Answer: a = 2.04 m s⁻²
Explanation: First, calculate the normal force (N). Since the surface is horizontal and there's no vertical acceleration, N = Weight = mg = 5.0 kg * 9.81 m s⁻² = 49.05 N. Next, calculate the frictional force (F_f) = μ_k * N = 0.40 * 49.05 N = 19.62 N. The resultant force (F_net) = Applied Tension - Frictional force = 30 N - 19.62 N = 10.38 N. Using Newton's Second Law, F_net = ma, so 10.38 N = 5.0 kg * a. Therefore, a = 10.38 N / 5.0 kg = 2.076 m s⁻². Rounding to 3 significant figures, a = 2.08 m s⁻².
Question 3 (5 marks)
A 0.050 kg golf ball is hit off a tee. The club is in contact with the ball for 0.0018 s, and the ball leaves the tee at a speed of 60 m s⁻¹. Assuming the ball was initially at rest, (a) Calculate the impulse delivered to the golf ball. (b) Determine the average force exerted by the club on the ball.
Answer: (a) Impulse = 3.0 N s (b) Average force = 1670 N
Explanation: (a) Impulse (I) is equal to the change in momentum (Δp). Since the ball starts from rest, its initial momentum is 0. Δp = final momentum - initial momentum = mv_f - mv_i = (0.050 kg * 60 m s⁻¹) - 0 = 3.0 N s. (b) Impulse (I) = Average Force (F_avg) × time (Δt). So, F_avg = I / Δt = 3.0 N s / 0.0018 s = 1666.6... N. Rounding to 3 significant figures, F_avg = 1670 N.
Question 4 (2 marks)
A 15 kg object is suspended by two ropes. One rope makes an angle of 45° with the ceiling, and the other makes an angle of 60° with the ceiling. Which statement about the tensions in the ropes is correct?
- A. The tension in the rope making 45° with the ceiling is greater than the tension in the rope making 60°.
- B. The tension in the rope making 60° with the ceiling is greater than the tension in the rope making 45°.
- C. The tensions in both ropes are equal.
- D. More information is needed to compare the tensions.
Answer: A
Explanation: Let T₁ be the tension in the rope at 45° and T₂ be the tension in the rope at 60°. For vertical equilibrium, the sum of the vertical components of the tensions must balance the weight of the object: T₁ sin(45°) + T₂ sin(60°) = mg. For horizontal equilibrium, the horizontal components must balance: T₁ cos(45°) = T₂ cos(60°). From the horizontal equilibrium: T₁ * (1/√2) = T₂ * (1/2), so T₁ = T₂ * (1/√2) / (1/2) = T₂ * (2/√2) = T₂ * √2. This means T₁ is greater than T₂. Generally, the rope that is more vertical (makes a smaller angle with the vertical, or a larger angle with the horizontal) carries a smaller tension, assuming symmetrical setup. Here, 45° to the ceiling is less 'vertical' (more horizontal) than 60° to the ceiling, so it will carry a larger tension.
Question 5 (6 marks)
Two masses, M₁ = 4.0 kg and M₂ = 6.0 kg, are connected by a light, inextensible string. M₁ rests on a smooth horizontal table, and the string passes over a smooth pulley at the edge, with M₂ hanging vertically. (a) Draw labelled free-body diagrams for both M₁ and M₂. (b) Calculate the acceleration of the system. (c) Determine the tension in the string. (Take g = 9.81 m s⁻²)
Answer: (a) Diagram for M₁: Tension (right), Normal force (up), Weight (down). Diagram for M₂: Tension (up), Weight (down). (b) a = 5.89 m s⁻² (c) T = 23.6 N
Explanation: (a) Free-body diagram for M₁: Forces are Tension (T) to the right, Normal force (N) upwards, and Weight (M₁g) downwards. Free-body diagram for M₂: Forces are Tension (T) upwards and Weight (M₂g) downwards. (b) For M₁ (horizontal motion): T = M₁a (Equation 1) For M₂ (vertical motion): M₂g - T = M₂a (Equation 2) Substitute (1) into (2): M₂g - M₁a = M₂a M₂g = (M₁ + M₂)a a = M₂g / (M₁ + M₂) = (6.0 kg * 9.81 m s⁻²) / (4.0 kg + 6.0 kg) = 58.86 / 10.0 = 5.886 m s⁻². Rounding to 3 significant figures, a = 5.89 m s⁻². (c) Substitute 'a' back into Equation 1: T = M₁a = 4.0 kg * 5.886 m s⁻² = 23.544 N. Rounding to 3 significant figures, T = 23.5 N.
Question 6 (3 marks)
Explain why a boxer 'rolls with the punch' to reduce the impact of a blow, applying physics principles related to impulse.
Answer: By 'rolling with the punch', the boxer increases the time duration (Δt) over which the impact force acts. Since impulse (change in momentum, Δp) is constant for a given punch (as the punch delivers a certain amount of momentum), and Impulse = FΔt, increasing Δt results in a decrease in the average force (F) exerted on the boxer's head, thus reducing the damaging effect of the blow.
Explanation: This directly applies the impulse-momentum theorem (FΔt = Δp). The punch delivers a specific change in momentum to the boxer's head. By moving their head backwards (rolling), the boxer effectively extends the time interval during which the force acts. Since the impulse is constant, a longer interaction time means the average force experienced is reduced, lessening the chance of injury.
Question 7 (3 marks)
A force-time graph for a 0.50 kg object shows the force increasing uniformly from 0 N to 10 N over 0.20 s, then remaining constant at 10 N for 0.10 s, and finally decreasing uniformly to 0 N over another 0.20 s. Calculate the total impulse delivered to the object.
Answer: Total Impulse = 3.0 N s
Explanation: The impulse is the area under the force-time graph. The graph describes a trapezium with parallel bases (time durations) and a height (force). Break the area into three parts: 1. Triangle (0 to 0.20 s): Area = 0.5 × base × height = 0.5 × 0.20 s × 10 N = 1.0 N s. 2. Rectangle (0.20 s to 0.30 s): Area = base × height = 0.10 s × 10 N = 1.0 N s. 3. Triangle (0.30 s to 0.50 s): Area = 0.5 × base × height = 0.5 × (0.50 - 0.30) s × 10 N = 0.5 × 0.20 s × 10 N = 1.0 N s. Total Impulse = 1.0 N s + 1.0 N s + 1.0 N s = 3.0 N s.
Common mistakes
- Confusing Newton's Third Law action-reaction pairs with forces in equilibrium. Action-reaction forces act on *different* bodies, while forces in equilibrium act on the *same* body and sum to zero.
- Forgetting the vector nature of force, velocity, momentum, and impulse. This often leads to errors in calculations involving changes in direction, such as rebound collisions where the change in momentum is (final momentum - initial momentum) and signs must be consistently applied.
- Incorrectly identifying or resolving forces on free-body diagrams, especially when dealing with inclined planes or forces at angles. Always ensure all relevant forces (weight, normal force, tension, friction, applied force) are included and resolved into perpendicular components correctly.
Exam tips
- Always draw clear, labelled free-body diagrams for each object in a system. This helps visualise all forces acting and ensures you don't miss any components when resolving.
- Consistently define a positive direction for vector quantities (force, velocity, acceleration, momentum) at the start of each problem. Stick to this convention throughout your calculations to avoid sign errors.
- Pay close attention to keywords like 'smooth' (no friction), 'light' (mass negligible, tension uniform), 'inextensible' (constant length, same acceleration), and 'constant velocity' (zero resultant force) as these simplify problems significantly.
Why this matters for exams
Dynamics forms a crucial cornerstone of the Cambridge AS Level Physics syllabus. It is frequently assessed through a variety of question types, including calculations involving Newton's Laws, momentum, and impulse, as well as conceptual explanations of force interactions and conservation principles. A strong grasp of dynamics is essential for tackling more advanced topics like work, energy, power, and circular motion, which often integrate these fundamental concepts.
Frequently asked questions
What is the difference between mass and weight?
Mass is a measure of the amount of matter in an object and is a scalar quantity, constant regardless of location. Weight, on the other hand, is the gravitational force acting on an object's mass (Weight = mass × gravitational field strength) and is a vector quantity, varying with the local gravitational field.
When should I apply the principle of conservation of momentum?
The principle of conservation of momentum should be applied when analyzing interactions (like collisions or explosions) within a closed system where no external resultant forces are acting. It's particularly useful when forces involved are internal and hard to quantify directly, or when the time of interaction is very short.
How do I deal with friction in AS Level Physics problems?
Friction is a force that opposes relative motion or attempted motion between surfaces in contact. You'll typically encounter kinetic friction (when objects are sliding) or static friction (when objects are at rest but have a tendency to slide). The magnitude of kinetic friction (F_k) is given by F_k = μ_k * N, where μ_k is the coefficient of kinetic friction and N is the normal force. Always include friction as a force opposing motion in your free-body diagrams and equations.
What is impulse and how does it relate to momentum?
Impulse is the product of the force applied to an object and the time duration for which it acts (Impulse = FΔt). Critically, the impulse delivered to an object is equal to the change in its momentum (Δp). This relationship, FΔt = Δp, is known as the impulse-momentum theorem and is very useful for problems involving forces that vary over time or short, intense interactions.