Resolving Forces on Inclined Planes, Projectile Motion SUVAT & Elastic/Inelastic Collisions
Classical mechanics rests on vector resolution, static equilibrium, kinematics under uniform acceleration (SUVAT), and Newton's three laws of motion. Whether analysing a block sliding down an inclined plane at angle θ (where weight resolves into mg sin θ parallel to the slope and mg cos θ perpendicular to the slope) or a 2D projectile in flight (where horizontal velocity u cos θ remains constant while vertical velocity u sin θ accelerates downward at g = 9.81 m s⁻²), separating orthogonal components is the golden rule. Also, Newton's second law expressed as F = Δ(mv) / Δt links the area under a force-time graph directly to impulse, while the conservation of linear momentum governs every collision and explosion. Examiners regularly test multi-support bridge moments, projectile range/time-of-flight derivations, and proving whether a collision is elastic or inelastic.
- Resolving Vectors & Forces on an Inclined Plane
- Any vector F at angle θ to the horizontal resolves into two perpendicular components: F_x = F cos θ (adjacent to θ) and F_y = F sin θ (opposite to θ). For an object of mass m on a slope inclined at angle θ to the horizontal: 1. Component of weight acting down (parallel to) the slope = mg sin θ. 2. Component of weight acting perpendicular (into) the slope = mg cos θ (balanced by the normal contact force R = mg cos θ if no other vertical forces act). By Newton's Second Law down a slope with friction F_f: mg sin θ - F_f = ma.
- Moments, Couples & Conditions for Static Equilibrium
- The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force: Moment = F × d sin θ (units: N m). A couple is a pair of equal and opposite parallel forces acting along different lines, producing pure rotation (zero resultant linear force) with Torque of a couple = F × d (where d is the perpendicular separation between the two forces). For a rigid body to be in static equilibrium, two conditions must both be met: 1. Resultant force in any direction is zero (ΣF_x = 0 and ΣF_y = 0, closed vector triangle). 2. Principle of Moments: Sum of clockwise moments about any point equals sum of anticlockwise moments about that same point.
- Uniform Acceleration (SUVAT) & Motion Graphs
- For constant acceleration a only: 1. v = u + at; 2. s = ½(u + v)t; 3. s = ut + ½at²; 4. v² = u² + 2as. Graphical rules: 1. Displacement-Time (s-t) graph: gradient = velocity v. 2. Velocity-Time (v-t) graph: gradient = acceleration a; area under graph = displacement s. 3. Acceleration-Time (a-t) graph: area under graph = change in velocity Δv.
Resolving Forces on Inclined Planes, Projectile Motion SUVAT & Elastic/Inelastic Collisions: Wave Equation (v = fλ) & Force Vector (F = ma) Simulator
Adjust physical variables on the sliders below to see real-time changes in wave geometry, resultant forces, and acceleration.
F_net = 80 N − 25 N = 55 NAcceleration: a = F_net ÷ m = 55 ÷ 10 = 5.50 m/s²Key Concepts & Exam Definitions (6)
- Resolving Vectors & Forces on an Inclined Plane
- Any vector F at angle θ to the horizontal resolves into two perpendicular components: F_x = F cos θ (adjacent to θ) and F_y = F sin θ (opposite to θ). For an object of mass m on a slope inclined at angle θ to the horizontal: 1. Component of weight acting down (parallel to) the slope = mg sin θ. 2. Component of weight acting perpendicular (into) the slope = mg cos θ (balanced by the normal contact force R = mg cos θ if no other vertical forces act). By Newton's Second Law down a slope with friction F_f: mg sin θ - F_f = ma.
- Moments, Couples & Conditions for Static Equilibrium
- The moment of a force about a pivot is the product of the force and the perpendicular distance from the pivot to the line of action of the force: Moment = F × d sin θ (units: N m). A couple is a pair of equal and opposite parallel forces acting along different lines, producing pure rotation (zero resultant linear force) with Torque of a couple = F × d (where d is the perpendicular separation between the two forces). For a rigid body to be in static equilibrium, two conditions must both be met: 1. Resultant force in any direction is zero (ΣF_x = 0 and ΣF_y = 0, closed vector triangle). 2. Principle of Moments: Sum of clockwise moments about any point equals sum of anticlockwise moments about that same point.
- Uniform Acceleration (SUVAT) & Motion Graphs
- For constant acceleration a only: 1. v = u + at; 2. s = ½(u + v)t; 3. s = ut + ½at²; 4. v² = u² + 2as. Graphical rules: 1. Displacement-Time (s-t) graph: gradient = velocity v. 2. Velocity-Time (v-t) graph: gradient = acceleration a; area under graph = displacement s. 3. Acceleration-Time (a-t) graph: area under graph = change in velocity Δv.
- 2D Projectile Motion (Independence of Horizontal & Vertical Motion)
- For a projectile launched at speed u at angle θ above the horizontal (neglecting air resistance): Horizontal and vertical motions are completely independent (linked only by shared time t): 1. Horizontal motion (a_x = 0): horizontal velocity is constant, v_x = u cos θ, and horizontal range x = (u cos θ) t. 2. Vertical motion (a_y = -g = -9.81 m s⁻²): u_y = u sin θ; at maximum height H, vertical velocity v_y = 0 (so 0 = (u sin θ)² - 2gH → H = u² sin²θ / 2g, and time to peak t_peak = u sin θ / g). Note: Air resistance reduces both the maximum height H and horizontal range R, making the trajectory asymmetrical (steeper descent).
- Newton's Laws, Linear Momentum (p = mv) & Impulse (FΔt = Δ(mv))
- Linear momentum p = mv (vector, units: kg m s⁻¹ or N s). Newton's Second Law: the resultant force is equal to the rate of change of momentum, F = Δ(mv) / Δt (reduces to F = ma when mass m is constant, or F = v(Δm/Δt) for a continuous fluid jet/rocket). Impulse = FΔt = Δ(mv) = mv - mu, which equals the area under a Force-Time (F-t) graph. Safety features (crumple zones, airbags, crash mats) increase the impact time Δt for a fixed change in momentum Δ(mv), thereby reducing the peak impact force F on the passenger.
- Conservation of Momentum & Elastic vs Inelastic Collisions
- Principle of Conservation of Linear Momentum: in any collision or explosion, total momentum before equals total momentum after, provided NO external resultant forces act on the system (m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂; remember to assign '+' and '-' signs for opposite directions). Momentum is always conserved in both elastic and inelastic collisions: 1. Elastic Collision: total kinetic energy (Σ½mv²) IS also conserved. 2. Inelastic Collision: total kinetic energy is not conserved (some KE is transferred to thermal, sound, and deformation energy; total energy is still conserved).
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
For a block of mass m on a slope inclined at angle θ to the horizontal, what are the components of weight parallel and perpendicular to the slope?
State the two conditions required for a rigid object to be in static equilibrium.
At the highest point of a projectile's trajectory (launched at speed u at angle θ), what are its vertical velocity, horizontal velocity, and acceleration?
How does air resistance change the trajectory of a projectile compared to motion in a vacuum?
What does the area under a Force-Time (F-t) graph represent, and what are its two equivalent SI units?
What is conserved in an inelastic collision versus an elastic collision in an isolated system?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
A uniform horizontal bridge beam AB of length 12.0 m and weight 45.0 kN is supported by two vertical piers at each end, A and B. A lorry of weight 80.0 kN is at a point 3.0 m from pier A. Calculate the upward normal reaction forces exerted by pier A and pier B on the bridge. (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Since the bridge is uniform, its weight (45.0 kN) acts through its centre of mass at the midpoint, 6.0 m from pier A [1 mark]. 2. Take moments about pier A (eliminating R_A) and apply the Principle of Moments (Sum of clockwise moments = Sum of anticlockwise moments): (80.0 kN × 3.0 m) + (45.0 kN × 6.0 m) = R_B × 12.0 m [1 mark]. 3. Solve for R_B: 240 + 270 = 510 kN m = 12.0 R_B → R_B = 510 ÷ 12.0 = 42.5 kN [1 mark]. 4. Use vertical force equilibrium (R_A + R_B = total downward weight): R_A = (80.0 + 45.0) - 42.5 = 125.0 - 42.5 = 82.5 kN [1 mark].
- Q2
A golf ball is struck from horizontal ground with an initial velocity of 42.0 m s⁻¹ at an angle of 35.0° above the horizontal. Neglecting air resistance (g = 9.81 m s⁻²), calculate: (a) the initial horizontal and vertical components of velocity, (b) the maximum height reached by the ball, and (c) the total horizontal distance (range) travelled before hitting the ground. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. (a) Horizontal component u_x = 42.0 cos(35.0°) = 34.40 m s⁻¹; Vertical component u_y = 42.0 sin(35.0°) = 24.09 m s⁻¹ [1 mark]. 2. (b) At maximum height, v_y = 0 m s⁻¹ and a_y = -9.81 m s⁻²; substitute into v_y² = u_y² + 2a_y s_y: 0 = (24.09)² + 2(-9.81)H [1 mark]. 3. Solve for maximum height H: H = 580.3 ÷ 19.62 = 29.6 m [1 mark]. 4. (c) Find time to reach maximum height using v_y = u_y + a_y t: 0 = 24.09 - 9.81 t_peak → t_peak = 2.456 s, so total time of flight T = 2 × 2.456 = 4.91 s [1 mark]. 5. Calculate horizontal range R = u_x × T = 34.40 m s⁻¹ × 4.911 s = 169 m [1 mark].
- Q3
A crate of mass 25.0 kg slides down a rough ramp inclined at 28.0° to the horizontal. A constant frictional force of 65.0 N opposes the motion. Calculate the normal contact force exerted by the ramp on the crate and the acceleration of the crate down the ramp (g = 9.81 m s⁻²). (4 marks)
Click “Check Mark Scheme” to compare your answer, 1. Resolve the crate's weight (mg = 25.0 × 9.81 = 245.25 N) perpendicular to the ramp: Normal contact force R = mg cos(28.0°) = 245.25 × 0.8829 = 217 N (216.5 N) [1 mark]. 2. Resolve the weight parallel to (down) the ramp: F_parallel = mg sin(28.0°) = 245.25 × 0.4695 = 115.1 N [1 mark]. 3. Calculate the resultant force down the ramp: F_net = mg sin(28.0°) - F_friction = 115.1 - 65.0 = 50.1 N [1 mark]. 4. Apply Newton's Second Law (a = F_net ÷ m): a = 50.1 N ÷ 25.0 kg = 2.00 m s⁻² [1 mark].
- Q4
A trolley A of mass 1.50 kg moving to the right at 4.00 m s⁻¹ collides head-on with a trolley B of mass 2.50 kg moving to the left at 2.00 m s⁻¹. After the collision, trolley A rebounds to the left at 1.25 m s⁻¹. Calculate the velocity of trolley B after the collision, and determine by calculation whether the collision is elastic or inelastic. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Take movement to the right as positive (+): Initial momentum p_initial = (1.50 × +4.00) + (2.50 × -2.00) = +6.00 - 5.00 = +1.00 kg m s⁻¹ [1 mark]. 2. Apply conservation of linear momentum (p_final = p_initial): (1.50 × -1.25) + 2.50 v_B = +1.00 → -1.875 + 2.50 v_B = +1.00 [1 mark]. 3. Solve for v_B: 2.50 v_B = +2.875 → v_B = +1.15 m s⁻¹ (1.15 m s⁻¹ to the right) [1 mark]. 4. Calculate total initial kinetic energy: KE_initial = ½(1.50)(4.00)² + ½(2.50)(-2.00)² = 12.00 + 5.00 = 17.00 J [1 mark]. 5. Calculate total final kinetic energy: KE_final = ½(1.50)(-1.25)² + ½(2.50)(1.15)² = 1.172 + 1.653 = 2.83 J; since KE_final < KE_initial (14.17 J lost as heat/sound), the collision is inelastic [1 mark].
Common Exam Questions (4)
When an object reaches terminal velocity while falling through air, how are the energy transfers and Newton's First Law explained?
At terminal velocity, upward viscous drag equals downward weight (F_drag = mg), so the resultant force is zero and acceleration is zero (constant velocity by Newton's First Law). Because speed is constant, kinetic energy stays constant; all of the gravitational potential energy lost (ΔE_p = mgΔh) is transferred directly into internal thermal energy of the air and falling object via work done against air resistance.
In Newton's Third Law, why isn't 'weight downwards and normal reaction upwards on a book resting on a table' an action-reaction pair?
A genuine Newton's Third Law pair must: 1. Be the exact same type of force, and 2. Act on two different objects. Weight (Earth's gravitational pull on the book) and Normal Reaction (electrostatic contact push of table on the book) act on the same object and are different force types. The true Third Law partner to Earth pulling down on the book is the book's gravity pulling UP on the Earth.
How do you calculate the force exerted on a flat wall by a horizontal water jet from a hosepipe of cross-sectional area A and speed v (assuming the water does not rebound)?
In 1 second, a cylinder of water of volume A × v and mass m = ρAv strikes the wall and loses all its horizontal velocity v. Therefore, Force F = Δ(mv) / Δt = (ρAv) × v = ρAv² (if the water rebounds elastically at speed v, Δv = 2v so F = 2ρAv²).
Why do I have to be careful when a projectile is launched from the top of a cliff of height h above the sea?
When taking upwards as positive (+), the sea at the bottom of the cliff is below the launch origin, so the final vertical displacement is negative (s_y = -h, not +h!) while acceleration is a_y = -9.81 m s⁻².
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