A.2 Forces and momentum

1. It slows gradually as the force from the stick is used up. — A student who thinks the stick gives the puck a store of 'force' that runs out picks this. A force exists only while the stick touches the puck. Once contact ends there is no force from the stick, and with negligible friction nothing slows the puck.
2. It continues in a straight line at a constant speed. — Newton's first law: with no resultant force, the velocity of the puck is constant, so it moves in a straight line at constant speed. The stick exerted a force only while in contact; no force is needed to keep the puck moving.
3. It slows and stops, as moving bodies come to rest naturally. — A student who believes that rest is the natural state of every body picks this. Everyday objects stop because friction and drag act on them. Here these are negligible, so by Newton's first law the puck keeps its velocity.
4. It stops at once, since nothing now pushes it along its path. — A student who believes a moving body needs a force in its direction of motion picks this. Newton's first law says a force is needed to CHANGE velocity, not to maintain it. With no resultant force, the puck keeps moving.

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1. Its weight and a decreasing upward force from the throw — A student who thinks the ball carries the 'force of the throw' with it picks this. The hand's force acts only during contact. After release the ball carries momentum, not force, and that upward momentum decreases because the weight acts downwards.
2. Its weight and an upward force larger than its weight — A student who believes a body moving upwards must have an upward resultant force picks this. The ball is slowing down, so the resultant force is downwards. Its upward motion is due to its velocity, which needs no force to continue.
3. No force until it stops at the top and then starts to fall — A student who thinks gravity acts only on falling objects picks this. The Earth pulls on the ball throughout the flight; that is why the ball's upward velocity decreases steadily and reaches zero at the top.
4. Only its weight, the downward pull of the Earth — Once the ball leaves the hand, the only body still interacting with it is the Earth (air resistance is negligible). Its weight acts downwards throughout the flight, which is why the ball slows as it rises.

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1. 25 N — East–west: 30 N − 10 N = 20 N east. North: 15 N. These are perpendicular, so the resultant is √(20² + 15²) N = 25 N.
2. 55 N — A student who adds the magnitudes of all the forces gets 30 + 10 + 15 = 55 N. Forces are vectors: the west force opposes the east force, and the north force is perpendicular to both.
3. 35 N — A student who correctly finds 20 N east but then adds the perpendicular 15 N arithmetically gets 35 N. Perpendicular components combine by Pythagoras: √(20² + 15²) = 25 N.
4. 43 N — A student who adds the east and west forces as though they pointed the same way gets 40 N, and then √(40² + 15²) = 43 N. The 10 N force acts westwards, so it subtracts from the 30 N force.

Working Take east and north as positive. F_x = 30 N − 10 N = 20 N; F_y = 15 N. |F| = √(20² + 15²) N = √625 N = 25 N.

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1. A magnetic field exerts a force on a charged particle even when it is at rest. — A student who treats magnetic forces on charges like electric forces picks this. A magnetic field exerts a force only on a MOVING charge; a stationary charge in a magnetic field experiences no magnetic force.
2. An electric field exerts a force on a charged particle even when it is at rest. — An electric field exerts a force F_e on any charge in it, whether the charge is moving or stationary. This is one of the differences between the electric and the magnetic force.
3. A gravitational force acts on a body only where there is air around it. — A student who links gravity to the atmosphere picks this. Gravity is a field force that acts through empty space: the Sun's gravity holds the Earth in orbit across the vacuum between them.
4. An astronaut in orbit has no weight, as gravity does not act there. — A student who takes 'weightlessness' literally picks this. Gravity acts strongly in orbit; it provides the centripetal force that keeps the astronaut and spacecraft falling around the Earth together.

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1. When every collision between the bodies is elastic — A student who links momentum conservation to kinetic-energy conservation picks this. Momentum is conserved in inelastic collisions and explosions too; elastic collisions are special only because total kinetic energy is also conserved.
2. When the bodies exert no forces at all on one another — A student who does not distinguish internal from external forces picks this. The bodies in a collision exert large forces on each other, but these are equal and opposite, so they cancel in the total momentum.
3. When no resultant external force acts on the system — Forces between the bodies are internal and cancel in third-law pairs, so only an external resultant force can change the total momentum. With no resultant external force, the total momentum is constant, whatever happens inside the system.
4. When every body in the system moves at a constant speed — A student who equates constant speed with constant momentum picks this. Momentum is a vector: a body moving at constant speed along a curve has changing momentum. What matters is the resultant external force on the system.

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1. +3.0 N s — A student who assumes the change in momentum is in the ball's original direction of motion gets the magnitude right but the sign wrong. The wall pushes the ball away from itself, so the impulse is negative.
2. −0.6 N s — A student who uses speeds without signs calculates 0.060 × (20 − 30) = −0.6 N s. The final velocity is −20 m s⁻¹, because the ball moves away from the wall; momentum is a vector.
3. −1.8 N s — A student who counts only the stopping of the ball calculates 0.060 × (0 − 30) = −1.8 N s. The wall must also push the ball back out at 20 m s⁻¹, which adds a further −1.2 N s.
4. −3.0 N s — Impulse = change in momentum = m(v − u) = 0.060 kg × (−20 − 30) m s⁻¹ = −3.0 N s: 3.0 N s directed away from the wall.

Working J = Δp = m(v − u) with u = +30 m s⁻¹ and v = −20 m s⁻¹: J = 0.060 kg × (−20 − 30) m s⁻¹ = 0.060 × (−50) = −3.0 N s (3.0 N s away from the wall).

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1. The bodies bounce apart after colliding, instead of sticking together. — A student who equates 'elastic' with 'bouncing' picks this. Bodies can bounce apart and still lose kinetic energy; such a collision is inelastic. Only sticking together is ruled out for an elastic collision.
2. The total kinetic energy of the two bodies is the same before and after. — An elastic collision is defined by the conservation of total kinetic energy. Total momentum is conserved too, as it is in every collision of an isolated system.
3. At least one of the bodies is made of a stretchy material such as rubber. — A student using the everyday meaning of 'elastic' picks this. In physics the word refers to kinetic energy being conserved, whatever the materials; hard steel balls can collide almost elastically.
4. Momentum is conserved, which is not the case in an inelastic collision. — A student who links momentum conservation to kinetic-energy conservation picks this. Momentum is conserved in inelastic collisions as well. What distinguishes an elastic collision is the kinetic energy.

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1. 2.0 m s⁻¹ — A student who reasons that equal and opposite forces give equal speeds picks this. The forces and impulses are equal, so the MOMENTA are equal and opposite; the heavier skater therefore moves more slowly.
2. 1.5 m s⁻¹ — The total momentum is zero before the push, so it is zero after: 60 × v_A = 45 × 2.0, giving v_A = 1.5 m s⁻¹ in the opposite direction to B.
3. 1.7 m s⁻¹ — A student who assumes the skaters receive equal kinetic energies writes ½ × 60 × v² = ½ × 45 × 2.0² and gets 1.7 m s⁻¹. Momentum, not kinetic energy, is shared equally in magnitude; the lighter skater receives more kinetic energy.
4. 0.0 m s⁻¹ — A student who thinks the skater who does the pushing receives no force picks this. B pushes back on A with an equal force (third law), so A recoils; total momentum stays zero only if A moves.

Working Total momentum before = 0. After: m_A v_A + m_B v_B = 0 ⇒ 60 kg × v_A = −45 kg × 2.0 m s⁻¹ ⇒ v_A = −1.5 m s⁻¹, a speed of 1.5 m s⁻¹ opposite to B.

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1. 2.0 × 10² m s⁻² — A student who takes a to be inversely proportional to T, as v is, calculates 4π² × 0.25/0.050 = 2.0 × 10². Check the units: m ÷ s is not an acceleration; the period must be squared.
2. 7.9 × 10³ m s⁻² — A student who substitutes the diameter, 0.50 m, as r gets twice the correct acceleration. The equation uses the radius, 0.25 m.
3. 3.1 × 10¹ m s⁻² — A student who calculates 2πr/T = 31 gives the speed of the sock, 31 m s⁻¹, not its acceleration. The acceleration is v²/r = 31.4²/0.25 = 3.9 × 10³ m s⁻².
4. 3.9 × 10³ m s⁻² — r = 0.25 m. a = 4π²r/T² = 4π² × 0.25/(0.050)² = 3.9 × 10³ m s⁻², directed towards the axis of the drum.

Working r = 0.50 m/2 = 0.25 m. a = 4π²r/T² = 4π² × 0.25 m/(0.050 s)² = 9.87 m/2.5 × 10⁻³ s² = 3.9 × 10³ m s⁻². (Equivalently v = 2πr/T = 31.4 m s⁻¹ and a = v²/r = 3.9 × 10³ m s⁻².)

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1. Her velocity is constant, so the resultant force acting on her is zero. — A student who treats constant speed as constant velocity picks this. The direction of her motion keeps changing, so her velocity changes and a resultant force must act.
2. Her velocity changes, so a resultant force acts towards the centre. — Velocity is a vector. Her speed is constant but the direction of her velocity changes continuously, so she accelerates towards the centre and, by Newton's second law, a resultant force acts towards the centre.
3. A resultant force acts on her outwards, away from the centre of the track. — A student who believes in a real centrifugal force picks this. An outward resultant force would make her curve away from the centre. The resultant force is inwards; the outward 'feeling' is her tendency to continue in a straight line.
4. A resultant force acts on her forwards, along her direction of motion. — A student who thinks motion needs a force along it picks this. A forward resultant force would increase her speed, but her speed is constant. The resultant is perpendicular to her velocity, towards the centre.

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1. The normal force exerted on the book by the table — A student who pairs any two equal and opposite forces picks this. The normal force and the weight both act on the book and come from different interactions (contact and gravitational). They balance because the book is in equilibrium, which is Newton's first law, not the third.
2. The contact force exerted on the table by the book — A student who thinks the book's push on the table is its weight picks this. That push is a contact force: it forms a third-law pair with the table's normal force on the book. The weight is gravitational, so its partner is gravitational too: the book's pull on the Earth.
3. The gravitational force exerted on the Earth by the book — A third-law pair is the two forces of ONE interaction, here the gravitational interaction between the book and the Earth. The Earth pulls the book down; the book pulls the Earth up with a force of equal magnitude. The two forces are of the same type and act on different bodies.
4. No force, because the book is too small to pull the Earth — A student who thinks a small body cannot exert a force on a massive one picks this. The book pulls the Earth with a force equal in magnitude to its weight. The Earth's resulting acceleration is negligible because its mass is enormous, not because the force is absent.

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1. 38 N — The lamp is in translational equilibrium, so the resultant force is zero. The horizontal components of the tensions cancel. Vertically, 2T sin 40° = mg, so T = (5.0 × 9.8 N)/(2 × 0.643) = 38 N.
2. 25 N — A student who assumes each cable carries half the weight calculates 49 N ÷ 2 = 25 N. That is true only for vertical cables. Here only the vertical component of each tension, T sin 40°, supports the lamp, so the tension must be larger.
3. 32 N — A student who uses the cosine for the vertical component writes 2T cos 40° = mg and gets 32 N. The angle is measured from the horizontal, so the vertical component is T sin 40°.
4. 49 N — A student who takes the tension in a supporting cable to equal the weight gives mg = 49 N. The two cables share the load, and each supports it only through its vertical component, so the tension must be found from the equilibrium condition.

Working Translational equilibrium: resultant force = 0. Horizontal components cancel by symmetry. Vertical: 2T sin 40° = mg ⇒ T = mg/(2 sin 40°) = (5.0 kg × 9.8 m s⁻²)/(2 × 0.6428) = 49 N/1.286 = 38 N.

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1. The truck's force is larger, since the truck has more mass. — A student who thinks a more massive body exerts a larger force picks this. The masses decide the accelerations, not the forces: the car's acceleration is ten times the truck's because its mass is ten times smaller, while the forces are equal.
2. They are equal in magnitude and opposite in direction. — The two forces are the two sides of one interaction between the car and the truck, so by Newton's third law they are equal in magnitude and opposite in direction, whatever the masses and speeds.
3. The car's force is larger, since the car is moving faster. — A student who thinks the faster, more active body exerts the larger force picks this. Speed does not decide the size of an interaction force. Newton's third law applies whatever the speeds: the forces are equal in magnitude.
4. The truck's force is larger, since the car's velocity changes more. — A student who judges force by its effect picks this. The car's velocity does change more, but because its mass is smaller (a = F/m), not because it receives a larger force. The forces are equal in magnitude.

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1. Weight, normal force, tension, friction, push of box on floor — A student who draws both forces of a third-law pair on one diagram picks this. The push of the box on the floor acts ON THE FLOOR, so it belongs on the floor's free-body diagram, not the box's.
2. Weight, normal force, tension, friction, forward force of motion — A student who thinks a moving body needs a force of its own in its direction of motion picks this. There is no 'force of motion': the box moves at constant velocity because the forward component of the tension balances friction.
3. Weight, normal force, rope tension and friction from the floor — A free-body diagram shows every force that other bodies exert ON the box: the Earth (weight F_g), the floor (normal force F_N and friction F_f) and the rope (tension T). Nothing else touches the box or exerts a field force on it.
4. Weight, tension and friction; the rigid floor exerts no push up — A student who thinks a rigid floor is just an obstacle picks this. The floor is compressed slightly and pushes up on the box with the normal force F_N. The box presses on the floor (that is why friction acts), so F_N must appear on its free-body diagram.

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1. 60 N — A student who thinks the applied force passes undiminished through A to B picks this. Part of the 60 N accelerates A itself; B receives only the force needed to give it the common acceleration.
2. 30 N — A student who shares the force equally between the two blocks picks this. Both blocks have the same acceleration, so the force on each is proportional to its mass; the heavier block B needs more than half.
3. 25 N — A student who takes the force on B to be the resultant force on A calculates m_A a = 5.0 × 5.0 = 25 N. That is the resultant on A (60 N − 35 N). The force on B is m_B a.
4. 35 N — Treat A and B as one system: a = 60 N/12 kg = 5.0 m s⁻². On B's free-body diagram the only horizontal force is the push from A, so it equals m_B a = 7.0 kg × 5.0 m s⁻² = 35 N.

Working System A + B: a = F/(m_A + m_B) = 60 N/(5.0 + 7.0) kg = 5.0 m s⁻². Block B alone: F_AB = m_B a = 7.0 kg × 5.0 m s⁻² = 35 N. Check with block A: 60 N − 35 N = 25 N = 5.0 kg × 5.0 m s⁻².

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1. 7.8 N — A student who takes the normal force to equal the weight uses F_N = 39.2 N and gets 7.8 N. On an incline only the component of the weight perpendicular to the plane, mg cos 25°, is balanced by F_N.
2. 7.1 N — Perpendicular to the plane there is no acceleration, so F_N = mg cos 25° = 4.0 × 9.8 × 0.906 = 35.5 N. The block slides, so F_f = μ_d F_N = 0.20 × 35.5 N = 7.1 N.
3. 3.3 N — A student who uses mg sin 25° for the normal force gets 3.3 N. The angle between the weight and the normal to the plane is 25°, so the perpendicular component is mg cos 25°.
4. 8.7 N — A student who resolves the normal force instead of the weight writes F_N cos 25° = mg, so F_N = 43.3 N and F_f = 8.7 N. Perpendicular to the plane, F_N balances the component mg cos 25°, so F_N is smaller than mg.

Working F_N = mg cos θ = 4.0 kg × 9.8 m s⁻² × cos 25° = 39.2 N × 0.9063 = 35.5 N. Sliding, so F_f = μ_d F_N = 0.20 × 35.5 N = 7.1 N.

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1. 40 N, equal and opposite to the worker's push — F_N = mg = 118 N, so the maximum static friction is μ_s F_N = 59 N. The 40 N push is less than this, so the crate stays at rest, and equilibrium requires friction of 40 N opposite to the push: F_f ≤ μ_s F_N is an inequality.
2. 59 N, which is the value of μ_s F_N for the crate — A student who treats μ_s F_N as the static friction itself picks this. It is only the maximum. A friction force of 59 N against a 40 N push would give the crate a resultant force and it would accelerate backwards, which does not happen.
3. 47 N, the value of μ_d F_N for the crate — A student who uses the dynamic coefficient picks this. μ_d applies only while the crate slides. This crate is stationary, so it experiences static friction, which balances the push.
4. 0 N, as friction acts only when a crate slides — A student who thinks friction needs motion picks this. If no friction acted, the 40 N push would be a resultant force and the crate would accelerate. It stays at rest, so static friction of 40 N must act.

Working F_N = mg = 12 kg × 9.8 m s⁻² = 117.6 N. Maximum static friction = μ_s F_N = 0.50 × 117.6 N = 58.8 N. Push 40 N < 58.8 N, so the crate stays at rest; equilibrium gives F_f = 40 N, opposite to the push.

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1. +1.2 N — A student who drops the minus sign gets +1.2 N. The magnitude is right, but a stretched spring pulls back towards its natural length, opposite to the extension; the minus sign in F_H = −kx shows this.
2. −5.2 N — A student who uses the stretched length, 0.130 m, as x gets −5.2 N. Hooke's law uses the extension, the change from the natural length: 0.030 m.
3. −120 N — A student who substitutes the extension in centimetres, 3.0, gets −120 N. With k in N m⁻¹ the extension must be in metres: 0.030 m.
4. −1.2 N — The extension is x = 13.0 cm − 10.0 cm = 3.0 cm = 0.030 m. F_H = −kx = −40 N m⁻¹ × 0.030 m = −1.2 N: a force of 1.2 N back towards the natural length.

Working x = 13.0 cm − 10.0 cm = 3.0 cm = 0.030 m (positive, in the direction of stretching). F_H = −kx = −(40 N m⁻¹)(0.030 m) = −1.2 N, i.e. 1.2 N directed opposite to the extension.

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1. 1.2 × 10⁻² m s⁻¹ — A student who leaves out buoyancy because the sphere sinks sets F_d equal to the whole weight. The glycerol displaced by the sphere pushes up on it whether it floats or sinks, so F_d = F_g − F_b.
2. 5.0 × 10⁻³ m s⁻¹ — A student who substitutes the diameter, 2.0 × 10⁻³ m, as r in 6πηrv gets half the correct speed. Stokes' law uses the radius, 1.0 × 10⁻³ m.
3. 1.0 × 10⁻² m s⁻¹ — At terminal velocity the resultant force is zero: F_g = F_b + F_d. F_b = ρVg = 1.3 × 10³ × 4.2 × 10⁻⁹ × 9.8 = 5.4 × 10⁻⁵ N, so F_d = 2.7 × 10⁻⁴ N. With r = 1.0 × 10⁻³ m, v = F_d/(6πηr) = 1.0 × 10⁻² m s⁻¹.
4. 1.0 × 10⁻⁵ m s⁻¹ — A student who substitutes the radius in millimetres, 1.0, into F_d = 6πηrv gets a speed 1000 times too small. The radius must be in metres, 1.0 × 10⁻³ m.

Working Terminal velocity: F_g = F_b + F_d. F_b = ρVg = (1.3 × 10³ kg m⁻³)(4.2 × 10⁻⁹ m³)(9.8 m s⁻²) = 5.35 × 10⁻⁵ N. F_d = 3.2 × 10⁻⁴ N − 5.35 × 10⁻⁵ N = 2.67 × 10⁻⁴ N. r = 1.0 mm = 1.0 × 10⁻³ m. v = F_d/(6πηr) = 2.67 × 10⁻⁴ N/(6π × 1.4 Pa s × 1.0 × 10⁻³ m) = 2.67 × 10⁻⁴/2.64 × 10⁻² = 1.0 × 10⁻² m s⁻¹.

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1. It is larger on the steel sphere, as steel is the denser metal — A student who thinks buoyancy depends on the body's own density picks this. Buoyancy depends on the density of the FLUID and the volume displaced. The denser steel sphere has a greater weight, not a greater buoyancy.
2. It is larger on the aluminium sphere, as that one is deeper — A student who thinks buoyancy grows with depth, because pressure does, picks this. The pressure increases equally on the top and bottom of the sphere, so the difference between them, which is the buoyancy, does not change with depth.
3. It is zero on both spheres, as neither sphere would float — A student who links buoyancy only to floating picks this. Every immersed body displaces fluid and experiences buoyancy. These spheres sink when released because their weights exceed the buoyancy, not because the buoyancy is zero.
4. They are equal, as each sphere displaces the same volume of water — F_b = ρVg, where ρ is the density of the water and V is the volume displaced. Both spheres are fully immersed and have the same volume, so they displace equal volumes of the same fluid and experience equal buoyancy.

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1. Her weight is 686 N. — A student who treats weight as a fixed property keeps her Earth weight, 70.0 × 9.8 = 686 N. Weight depends on the local gravitational field strength, which is much smaller on the Moon: 70.0 × 1.62 = 113 N.
2. Her mass is reduced. — A student who thinks mass becomes smaller where gravity is weaker picks this. Mass is a property of the body and does not depend on location: she still has a mass of 70.0 kg. Only her weight changes, to 113 N.
3. Her weight is 113 N. — Mass does not depend on location, so it stays 70.0 kg. Weight is F_g = mg = 70.0 kg × 1.62 N kg⁻¹ = 113 N.
4. Her weight is 0.0 N. — A student who thinks gravity needs air, and so is absent on the airless Moon, gives zero weight. Gravity acts through a vacuum: the Moon's field strength is 1.62 N kg⁻¹, so her weight is 70.0 × 1.62 = 113 N.

Working Mass is independent of location: m = 70.0 kg on the Moon. Weight on the Moon: F_g = mg = 70.0 kg × 1.62 N kg⁻¹ = 113.4 N ≈ 113 N (on Earth it is 70.0 × 9.8 = 686 N).

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1. The ball's momentum increases, while the total momentum of the ball and the Earth stays constant. — The ball's weight is a resultant external force on the ball, so its momentum increases. The ball pulls the Earth up with an equal force (third law), so the Earth gains an equal and opposite momentum. For the ball–Earth system these forces are internal, so the total is constant.
2. The ball's momentum stays constant, because momentum is conserved in every situation. — A student who applies conservation to a single body picks this. Momentum is conserved only for a system with no resultant external force. The ball alone has a resultant force (its weight) on it, so its momentum increases as it speeds up.
3. The ball's momentum stays constant, because its gain in E_k equals its loss in E_p. — A student who merges energy and momentum conservation picks this. Energy conservation says nothing about momentum. The ball speeds up, so p = mv increases.
4. The ball's momentum increases, and the Earth's is unchanged, as the ball cannot pull it. — A student who thinks a small body cannot pull a large one picks this. The ball pulls the Earth with a force equal to the ball's weight, so the Earth gains momentum equal and opposite to the ball's; its velocity change is tiny only because its mass is huge.

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1. 200 N s — A student who uses the applied force alone calculates 40.0 N × 5.00 s = 200 N s. That is the impulse of the push; friction delivers an impulse in the opposite direction, so the resultant impulse is smaller.
2. 125 N s — The resultant force is 40.0 N − 15.0 N = 25.0 N forwards. The impulse of the resultant force is J = FΔt = 25.0 N × 5.00 s = 125 N s.
3. 275 N s — A student who adds the magnitudes of the push and the friction, as if forces were scalars, uses 55.0 N. Friction opposes the push, so the resultant force is their difference, 25.0 N.
4. 5.0 N s — A student who divides the resultant force by the time gets 25.0 ÷ 5.00 = 5.0. Impulse is force MULTIPLIED by time; the unit N s itself shows this.

Working Resultant force F = 40.0 N − 15.0 N = 25.0 N (forwards). J = FΔt = 25.0 N × 5.00 s = 125 N s.

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1. Each egg receives the same impulse, but the foam acts for longer, so the average force is smaller. — Both eggs arrive with the same velocity and end at rest, so each undergoes the same change in momentum and receives the same impulse. Since F = Δp/Δt, the foam's longer stopping time gives a smaller average force, too small to break the shell.
2. The foam reduces the egg's change in momentum, so the average force exerted on it is smaller. — A student who thinks cushions reduce the momentum change picks this. Each egg goes from the same velocity to rest, so Δp is identical. The foam changes the TIME over which the change happens, not the change itself.
3. The foam absorbs the force of the falling egg, so less force is left to act on the egg's shell. — A student who treats force as something an egg carries and a cushion soaks up picks this. A force is an interaction: the foam exerts a force on the egg to stop it. That force is smaller because it acts for longer.
4. The foam absorbs the egg's kinetic energy, so unlike the concrete it exerts no force on the egg. — A student who sees the foam as only receiving the impact, not pushing back, picks this. The foam does absorb the kinetic energy, but only by exerting a force on the egg over a distance; without a force the egg would not stop. The force is smaller, not zero.

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1. 0.0 N — A student who applies F = ma to the belt's constant velocity gets zero. F = ma assumes constant mass. Here new sand arrives all the time and must be accelerated from rest to 2.5 m s⁻¹, so the sand's horizontal momentum increases at 2.0 kg m s⁻², which requires a horizontal force of 2.0 N.
2. 1.0 N — A student who assumes all the belt's work becomes kinetic energy of the sand writes Fv = ½(Δm/Δt)v² and gets 1.0 N. While the sand slips before matching the belt's speed, friction dissipates as much energy as the sand gains, so the momentum method must be used.
3. 2.0 N — The mass on the belt is changing, so F = Δp/Δt must be used. Each second, 0.80 kg of sand gains a horizontal velocity of 2.5 m s⁻¹: F = v(Δm/Δt) = 2.5 × 0.80 = 2.0 N.
4. 7.8 N — A student who thinks the force needed to move sand is its weight, and treats the 0.80 kg arriving each second as the load, gets 0.80 × 9.8 = 7.8 N. The weight is vertical and is supported by the belt's normal force; the horizontal force depends only on the horizontal momentum given to the sand.

Working F = Δp/Δt. Horizontal momentum gained per second = (Δm/Δt) × v = 0.80 kg s⁻¹ × 2.5 m s⁻¹ = 2.0 kg m s⁻² = 2.0 N. (F = ma would give zero because the belt's velocity is constant, but the mass being moved is not constant.)

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1. 1.5 m s⁻¹ — A student who averages the two initial velocities, (3.0 + 0) ÷ 2, gets 1.5 m s⁻¹. That works only for equal masses; momentum conservation weights each velocity by its mass.
2. 4.5 m s⁻¹ — A student who divides the momentum by the mass of the stationary trolley alone, 3.6 ÷ 0.80, gets 4.5 m s⁻¹, faster than the trolley that hit it. The trolleys move together, so the momentum is shared by the total mass, 2.0 kg.
3. 2.3 m s⁻¹ — A student who assumes kinetic energy is conserved writes ½ × 1.2 × 3.0² = ½ × 2.0 × v² and gets 2.3 m s⁻¹. Trolleys that stick together collide inelastically; momentum, not kinetic energy, gives the final velocity.
4. 1.8 m s⁻¹ — Momentum is conserved: 1.2 × 3.0 = (1.2 + 0.80)v, so v = 3.6 ÷ 2.0 = 1.8 m s⁻¹.

Working p before = 1.2 kg × 3.0 m s⁻¹ + 0 = 3.6 kg m s⁻¹. p after = (1.2 + 0.80) kg × v. v = 3.6/2.0 = 1.8 m s⁻¹, in the original direction.

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1. It is elastic: the total kinetic energy is 4.0 J before and after. — Momentum: before 0.50 × 4.0 = 2.0 kg m s⁻¹; after 0.50 × (−2.0) + 1.5 × 2.0 = 2.0 kg m s⁻¹, so the data are consistent. Kinetic energy: before ½ × 0.50 × 4.0² = 4.0 J; after 1.0 J + 3.0 J = 4.0 J. Total E_k is unchanged, so the collision is elastic.
2. It is impossible: the momentum after is 4.0 kg m s⁻¹, not 2.0. — A student who adds the momenta as positive numbers gets 1.0 + 3.0 = 4.0 kg m s⁻¹. The 0.50 kg ball moves backwards, so its momentum is −1.0 kg m s⁻¹, and the total is 2.0 kg m s⁻¹, as before.
3. It is inelastic, because the kinetic energy of the 0.50 kg ball decreased. — A student who expects each ball to keep its own kinetic energy picks this. In a collision kinetic energy passes from one body to another; elastic means the TOTAL is unchanged, and here it is 4.0 J before and after.
4. It is impossible, as the momentum of the 0.50 kg ball was not conserved. — A student who expects each body's momentum to be conserved picks this. Each ball receives an impulse from the other, so each ball's momentum changes; it is the TOTAL momentum that is conserved, and it is here, so the collision is possible.

Working Take the initial direction as positive. p before = 0.50 × 4.0 = 2.0 kg m s⁻¹; p after = 0.50 × (−2.0) + 1.5 × (+2.0) = −1.0 + 3.0 = 2.0 kg m s⁻¹ (conserved). E_k before = ½ × 0.50 × 4.0² = 4.0 J; E_k after = ½ × 0.50 × 2.0² + ½ × 1.5 × 2.0² = 1.0 J + 3.0 J = 4.0 J. Total E_k unchanged ⇒ elastic.

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1. 6.0 J — E_k before = ½ × 0.50 × 6.0² = 9.0 J. E_k after = ½ × 1.5 × 2.0² = 3.0 J. Kinetic energy transferred to internal energy and sound = 9.0 − 3.0 = 6.0 J.
2. 0.0 J — A student who thinks kinetic energy is conserved in every collision picks this. Total energy is conserved, but in an inelastic collision some kinetic energy becomes internal energy and sound: here E_k falls from 9.0 J to 3.0 J.
3. 9.0 J — A student who thinks sticking together destroys all the kinetic energy picks this. The combined trolleys still move at 2.0 m s⁻¹, carrying the original momentum, so they keep 3.0 J of kinetic energy.
4. 4.0 J — A student who substitutes the change in speed into ½mv² calculates ½ × 0.50 × (6.0 − 2.0)² = 4.0 J. Kinetic energy depends on v², so the energies before and after must be calculated separately and subtracted.

Working Momentum check: 0.50 × 6.0 = 3.0 kg m s⁻¹ = 1.5 × 2.0. E_k before = ½ × 0.50 kg × (6.0 m s⁻¹)² = 9.0 J. E_k after = ½ × 1.5 kg × (2.0 m s⁻¹)² = 3.0 J. Transferred = 9.0 J − 3.0 J = 6.0 J.

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1. Their total momentum has increased, as has their total kinetic energy. — A student who thinks motion from rest means new momentum picks this. Momentum is a vector: the fragments fly off in different directions and their momenta add to zero, the value before the explosion.
2. Their total momentum is zero, and so is their total kinetic energy. — A student who treats kinetic energy as a vector picks this. Kinetic energy ½mv² is a scalar and is never negative, so the fragments' kinetic energies add; only momenta can cancel.
3. Their total momentum is zero and their total kinetic energy has increased. — The explosive forces are internal, so the total momentum stays zero: the fragments' momenta cancel as vectors. Chemical energy is transferred to kinetic energy, so the total kinetic energy, zero before, has increased.
4. Their total momentum is zero, and the total energy of the system has risen. — A student who thinks explosions create energy picks this. The kinetic energy comes from the chemical energy stored in the explosive; the total energy of the system is unchanged.

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1. Friction from the road acts towards the centre and is the resultant force. — On a flat road the only horizontal force towards the centre is friction between the tyres and the road. With air resistance negligible, no force is needed along the path at constant speed, so this friction is the resultant force and provides the centripetal force mv²/r.
2. Friction acts towards the centre and is balanced by a centrifugal force. — A student who believes in a real outward force picks this. If friction were balanced, the resultant would be zero and the car would travel in a straight line. There is no outward force on the car; the resultant is inwards.
3. Friction and a separate centripetal force both act towards the centre. — A student who treats centripetal force as an extra force picks this. 'Centripetal force' is the name for the resultant towards the centre, and here that resultant IS the friction. It is not a second force.
4. A forward driving force along the bend is the resultant force on the car. — A student who thinks motion needs a resultant force in its direction picks this. At constant speed there is no resultant force along the path. The resultant is towards the centre, perpendicular to the velocity, and changes only the car's direction.

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1. 3.9 N — A student who takes the tension alone to be mv²/r gets 3.9 N. The weight acts downwards, away from the centre, so the tension must supply mv²/r AND balance the weight.
2. 1.0 N — A student who uses T + mg = mv²/r, the equation for the TOP of a circle, gets 1.0 N. At the lowest point the weight acts away from the centre, so T − mg = mv²/r.
3. 6.9 N — At the lowest point the centre of the circle is above the ball. The resultant force towards the centre is T − mg = mv²/r, so T = 0.30 × 9.8 + 0.30 × 2.8²/0.60 = 2.9 + 3.9 = 6.9 N.
4. 2.9 N — A student who takes the tension to equal the weight gives mg = 2.9 N. That applies to a ball hanging at rest; a ball moving through the lowest point of its arc accelerates towards the centre, so T is greater than mg.

Working At the lowest point, taking towards the centre (upwards) as positive: T − mg = mv²/r. T = mg + mv²/r = 0.30 × 9.8 N + 0.30 × (2.8)²/0.60 N = 2.94 N + 3.92 N = 6.86 N ≈ 6.9 N.

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1. 17.6 m s⁻¹ — A student who substitutes the diameter as r gets twice the correct speed. In one revolution the rim travels a circumference, 2πr = π × diameter = 2.20 m.
2. 1.40 m s⁻¹ — A student who divides the radius by the period gets 0.350 ÷ 0.250 = 1.40 m s⁻¹. In one period the point travels round the whole circumference, 2πr, not a distance r.
3. 25.1 m s⁻¹ — A student who stops at ω = 2π/T = 25.1 has found the angular velocity, in rad s⁻¹. The linear speed is v = ωr = 25.1 × 0.350 = 8.80 m s⁻¹.
4. 8.80 m s⁻¹ — r = 0.350 m. ω = 2π/T = 25.1 rad s⁻¹, so v = ωr = 2πr/T = 2π × 0.350/0.250 = 8.80 m s⁻¹.

Working r = 0.700 m/2 = 0.350 m. ω = 2π/T = 2π/0.250 s = 25.1 rad s⁻¹. v = ωr = 25.1 rad s⁻¹ × 0.350 m = 8.80 m s⁻¹ (= 2πr/T).

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1. Two arrows: the weight downwards and a normal force of equal size from the floor upwards. — The normal force equals the weight only when the vertical acceleration is zero. Here the person accelerates upwards, so the floor must push with more than the weight: the two arrows cannot be equal.
2. Two arrows: the weight downwards and a larger normal force from the floor upwards. — Only two bodies act on the person: the Earth (weight, down) and the floor (normal force, up). Since the person accelerates upwards the resultant is upwards, so the normal-force arrow must be longer than the weight arrow: F_N − mg = ma.
3. Three arrows: the weight, the normal force, and an upward force of the lift's motion. — There is no separate force of motion. The only upward push on the person comes from the floor, and it is that push exceeding the weight that produces the upward acceleration. A free-body diagram shows forces, not motion.
4. Three arrows: the weight, the normal force, and the person's push down on the floor. — The person's push on the floor acts on the floor, not on the person. A free-body diagram of the person shows only the forces acting ON the person; the third-law partner of the normal force belongs on a diagram of the floor.

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1. It is the product of the maximum force exerted on the ball and the total time of contact. — Using the peak force as if it acted throughout overestimates the impulse. The force rises and falls, so the impulse is the average force multiplied by the time, which is the area under the force–time graph.
2. It is the product of the average resultant force on the ball and the time of contact. — J = FΔt, where F is the AVERAGE resultant force over the contact time Δt. Because the force varies, its average over the contact must be used; the impulse then equals the change in momentum of the ball.
3. It is the largest value reached by the resultant force on the ball during the contact. — Impulse is not a force. It is force multiplied by time (unit N s), and it equals the change in momentum of the ball. The large brief force produces the impulse; it is not the impulse itself.
4. It is the average resultant force on the ball divided by the total time of contact. — Dividing force by time gives a quantity with unit N s⁻¹, which is not impulse. Impulse is force MULTIPLIED by time, unit N s, and a longer contact at the same average force delivers a larger impulse.

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1. It stays constant, because a constant thrust produces a constant acceleration. — F = ma gives a constant acceleration only if the mass is constant. A rocket loses mass continuously as it ejects fuel, so the same thrust produces a larger acceleration later in the burn; this is why the momentum form F = Δp/Δt is needed.
2. It is zero, because there is nothing in empty space for the exhaust gas to push against. — The rocket does not push on space; it pushes on its own exhaust gas. The rocket exerts a backward force on the gas and the gas exerts an equal forward force on the rocket (Newton's third law). No surrounding medium is required.
3. It decreases, because the force that the engine gives the rocket is gradually used up. — A force is not something a body stores and uses up. As long as the engine ejects gas, the gas exerts a forward force on the rocket; and because the rocket is getting lighter, that constant force produces an increasing, not decreasing, acceleration.
4. It increases, because the mass of the rocket keeps decreasing as fuel is used up. — The thrust is constant but the rocket's mass falls as fuel is ejected. With a changing mass F = ma cannot be used with one fixed m; at each instant a = F/m with the current mass, so the acceleration rises steadily while the engine fires.

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1. The two pieces move apart with equal speeds, as the forces on them were equal. — Equal forces for the same time give equal and opposite CHANGES IN MOMENTUM, not equal speeds. Since momentum is mv, the piece with one third of the mass needs three times the speed to carry the same magnitude of momentum.
2. The lighter piece moves away three times as fast as the heavier piece does. — Total momentum was zero and remains zero, so m v₁ = 3m v₂ in opposite directions: v₁ = 3v₂. The equal and opposite forces during the explosion act for the same time and give equal-magnitude momenta, so the lighter piece must move faster.
3. The two pieces carry equal kinetic energies, as the energy is shared out equally. — The pieces have equal magnitudes of momentum p, and E_k = p²/2m, so the lighter piece has three times the kinetic energy of the heavier one. Momentum, not kinetic energy, is shared equally in magnitude.
4. The total momentum of the pieces is greater than that of the body beforehand. — The two momenta are equal in magnitude and opposite in direction, so they add to zero, exactly as before the explosion. The forces of the explosion are internal to the system and cannot change its total momentum.

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1. Doubling the speed while keeping the radius the same — Doubling v does not double a; a = v²/r depends on the SQUARE of the speed, so the acceleration becomes four times as large.
2. Doubling the radius, with the speed unchanged — At a fixed speed a = v²/r, so doubling the radius HALVES the acceleration: a gentler curve needs less inward acceleration. It is at fixed angular velocity, a = ω²r, that a larger radius means a larger acceleration.
3. Halving the radius while keeping the speed the same — a = v²/r, so at a fixed speed the acceleration is inversely proportional to the radius: halving r doubles a. A tighter circle at the same speed needs a larger inward acceleration.
4. Halving the time period, with the radius unchanged — a = 4π²r/T² depends on the square of the period. Halving T at fixed r multiplies the acceleration by four, not two: the body covers the same circle in half the time, so its speed doubles and v² is four times larger.

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1. It points directly towards the centre, whatever the speed of the car is doing. — A resultant force that is exactly perpendicular to the velocity can only change the direction of motion, not the speed. Since the car is speeding up, there must also be a component along the velocity.
2. It points directly along the direction of motion, because the car is speeding up. — A force along the velocity alone would make the car speed up in a straight line. To keep changing direction the car also needs a component of force towards the centre of the circle.
3. It is directed away from the centre, as the car tends to slide outwards. — No outward force acts on the car. Its tendency to continue in a straight line (inertia) is what makes it seem to move outwards; the road's friction supplies the inward force that keeps it on the circle.
4. It has a component towards the centre and a component along the direction of motion. — The centripetal component, mv²/r, changes the direction of the velocity; a tangential component, along the velocity, changes its magnitude. Both are needed when a body moves in a circle with increasing speed, so the resultant points between the centre and the tangent.

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1. 9.4 rad s⁻¹ — The disc makes 1.5 revolutions per second, and each revolution is 2π rad, so ω = 2π × 1.5 = 9.4 rad s⁻¹. Equivalently T = 2.0/3.0 s and ω = 2π/T.
2. 1.5 rad s⁻¹ — 1.5 is the number of revolutions per second (the frequency), not the angle turned per second. Each revolution is 2π radians, so ω = 2π × 1.5 = 9.4 rad s⁻¹.
3. 540 rad s⁻¹ — 3.0 × 360° / 2.0 s = 540 is the rate in degrees per second. Angular velocity in v = ωr and a = ω²r must be in radians per second: 540° s⁻¹ × π/180 = 9.4 rad s⁻¹.
4. 1.4 rad s⁻¹ — 2πr/T = 2π × 0.15 / 0.67 = 1.4 is the linear speed of the mark in m s⁻¹, not the angular velocity. Every point on the disc shares the same ω = 2π/T; only v depends on the radius.

Working Frequency f = 3.0 rev / 2.0 s = 1.5 rev s⁻¹, so T = 1/f = 0.67 s. ω = 2π/T = 2πf = 2π × 1.5 = 9.42 rad s⁻¹ ≈ 9.4 rad s⁻¹. (The radius is not needed for ω; the linear speed of the mark would be v = ωr = 9.42 × 0.15 = 1.4 m s⁻¹.)

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1. The diagram is correct as drawn: F_N acts vertically so that it balances W. — 'Normal' means perpendicular to the surface, not vertical. On a slope F_N is tilted by 30° from the vertical and has magnitude mg cos 30°; it is F_N and F_f together, not F_N alone, that balance the weight.
2. F_f should be removed from the diagram, because the block is not sliding. — Static friction acts on a stationary body whenever something tends to make it slide. Without friction the block would accelerate down the slope, so F_f (up the slope) is a real force on it and belongs on the diagram.
3. An arrow for the push of the block on the slope is missing from the diagram. — The push of the block on the slope acts on the slope, not on the block. A free-body diagram shows only the forces acting on the chosen body, so that force must not appear.
4. F_N is wrongly drawn: it should be perpendicular to the surface of the slope. — The normal force is the component of the contact force perpendicular to the surface, so on a slope it is tilted from the vertical by the angle of the slope. W and F_f are correctly drawn; only the direction of F_N needs changing.

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1. 5.00 N — 50.0 sin 30.0° = 25.0 N is the VERTICAL component of the tension (which helps to support the sledge). The component along the snow, adjacent to the 30.0° angle, is 50.0 cos 30.0° = 43.3 N.
2. 23.3 N — The vertical forces balance (55.0 + 50.0 sin 30.0° = 80.0). Horizontally, the component of the tension is 50.0 cos 30.0° = 43.3 N forwards and friction is 20.0 N backwards, so the resultant is 43.3 − 20.0 = 23.3 N.
3. 63.3 N — Friction acts backwards, opposite to the horizontal component of the tension, so it must be subtracted: 43.3 − 20.0, not 43.3 + 20.0. Forces along one line are added with signs.
4. 30.0 N — 50.0 − 20.0 treats the whole tension as acting along the snow. The rope is at 30.0° to the horizontal, so only its component 50.0 cos 30.0° = 43.3 N acts along the direction of motion.

Working Vertical: F_N + T sin 30.0° − W = 55.0 + 25.0 − 80.0 = 0, so the vertical forces balance. Horizontal: T cos 30.0° − F_f = 50.0 × 0.866 − 20.0 = 43.3 − 20.0 = 23.3 N. The resultant is 23.3 N, horizontal, in the direction of travel.

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1. 32 N s — 40 N × 0.80 s is the area of the rectangle enclosing the triangle. The force is only at its peak for an instant; the impulse is the area UNDER the line, ½ × 0.80 × 40 = 16 N s.
2. 40 N s — 40 N is the maximum force read from the graph, not the impulse. Impulse is force multiplied by time, the area under the F–t graph, and here it is 16 N s.
3. 16 N s — The impulse is the area between the line and the time axis: a triangle of base 0.80 s and height 40 N, so J = ½ × 0.80 × 40 = 16 N s. This is the average force (20 N) multiplied by the time.
4. 50 N s — 40 / 0.80 = 50 divides the force by the time, which gives N s⁻¹, not N s. Impulse is force multiplied by time: the area under the graph, 16 N s.

Working Impulse = area under the F–t graph = area of the triangle = ½ × base × height = ½ × 0.80 s × 40 N = 16 N s. Equivalently the average force is ½ × 40 = 20 N and J = 20 N × 0.80 s = 16 N s.

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1. A exerts a larger force on B than B exerts on A, because A was the moving trolley. — The force on each trolley is the gradient of its momentum–time line. The two lines have gradients of the same magnitude (4.0 kg m s⁻¹ in 0.05 s) and opposite sign, so the forces are equal and opposite, whichever trolley was moving.
2. The impulses on A and B are equal and opposite, because the force pair acts for the same time. — Impulse equals change in momentum. From the graph Δp_A = 2.0 − 6.0 = −4.0 kg m s⁻¹ and Δp_B = +4.0 kg m s⁻¹ over the same 0.05 s, so the two impulses are equal in size and opposite in sign, as Newton's third law requires; the total momentum stays at 6.0 kg m s⁻¹.
3. The collision must be elastic, because the total momentum is the same before and after. — Momentum is conserved in every collision, elastic or not, because the forces are internal. Whether kinetic energy is conserved cannot be read from a momentum–time graph without the masses, so no conclusion about elasticity follows.
4. Momentum is not conserved, because the momentum of A falls during the collision. — Conservation applies to the TOTAL momentum of the system, not to each trolley. A loses 4.0 kg m s⁻¹ and B gains exactly 4.0 kg m s⁻¹, so the total is 6.0 kg m s⁻¹ throughout.

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1. 1.2 kg — Reading A's change as 4.0 − 1.0 = 3.0 m s⁻¹ ignores the sign: A moves BACKWARDS at 1.0 m s⁻¹ after the collision, so its velocity changes by 5.0 m s⁻¹, not 3.0. Momentum is a vector.
2. 2.0 kg — A's velocity changes from +4.0 to −1.0 m s⁻¹, so its momentum changes by 1.0 × (−1.0 − 4.0) = −5.0 kg m s⁻¹. B gains +5.0 kg m s⁻¹ while reaching 2.5 m s⁻¹, so m_B = 5.0 / 2.5 = 2.0 kg.
3. 1.6 kg — 4.0 / 2.5 assumes all of A's initial momentum is transferred to B. A keeps some momentum (in the negative direction), so B must gain more than 4.0 kg m s⁻¹: m_B v_B = 4.0 + 1.0 = 5.0 kg m s⁻¹.
4. 2.4 kg — Equating the kinetic energy lost by A to that gained by B assumes kinetic energy is conserved, which is not true in general and is not implied by the graph. The mass follows from conservation of momentum, which holds in every collision.

Working From the graph: u_A = 4.0 m s⁻¹, v_A = −1.0 m s⁻¹ (A rebounds), u_B = 0, v_B = 2.5 m s⁻¹. Conservation of momentum: m_A u_A = m_A v_A + m_B v_B → 1.0 × 4.0 = 1.0 × (−1.0) + m_B × 2.5 → m_B = 5.0 / 2.5 = 2.0 kg.

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1. 25 N — mg cos 30° = 25.5 N is the component of the weight PERPENDICULAR to the slope, which is balanced by F_N. The component along the slope, which friction must balance, is mg sin 30° = 14.7 N.
2. 18 N — μ_s F_N = 0.70 × 25.5 = 17.8 N is the MAXIMUM static friction, not its actual value. For a body at rest F_f ≤ μ_s F_N; the actual friction is whatever equilibrium needs, mg sin 30° = 14.7 N.
3. 15 N — The block is at rest, so the forces along the slope balance: F_f = mg sin 30° = 3.0 × 9.8 × 0.50 = 14.7 N ≈ 15 N. Static friction takes whatever value (up to μ_s F_N = 17.8 N) is needed for equilibrium.
4. 21 N — 0.70 × 3.0 × 9.8 = 21 N takes friction to be at its maximum μ_s F_N AND uses F_N = mg. On a slope F_N = mg cos 30° = 25.5 N, and for a block at rest F_f is not μ_s F_N but whatever equilibrium needs: mg sin 30° = 14.7 N.

Working The block is in equilibrium, so along the slope F_f = mg sin 30° = 3.0 × 9.8 × 0.50 = 14.7 N ≈ 15 N (up the slope). Check it is possible: F_N = mg cos 30° = 25.5 N, so the maximum static friction is μ_s F_N = 0.70 × 25.5 = 17.8 N, which exceeds 14.7 N — the block can indeed stay at rest.

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1. The diagram is correct: T is balanced by W and F_c together. — If T were balanced the resultant would be zero and the ball would move in a straight line, not a circle. At the lowest point the ball accelerates upwards, towards the centre, so T must exceed W; no outward force acts on the ball.
2. F_c should be removed; only T and W act on the ball. — The only bodies acting on the ball are the string (tension T, upwards along the string towards the centre) and the Earth (weight W, downwards). Their resultant, T − W upwards, IS the centripetal force: T − W = mv²/r. There is no additional force, inward or outward.
3. F_c is right but should point upwards, towards the centre. — 'Centripetal force' is the name given to the resultant of the real forces towards the centre, not an extra force. Here T − W already provides it; a third arrow towards the centre would count the centripetal force twice.
4. T should equal W in size, since the ball is not falling. — Tension equals weight only for a body with no vertical acceleration. At the lowest point the ball's velocity is changing direction, with an acceleration v²/r upwards, so the string must pull with more than the weight: T = W + mv²/r.

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1. Path B — The ball has no 'memory' of circular motion. It kept curving only because the string pulled it towards the centre at every instant; with the string gone the direction of its velocity no longer changes, so it moves in a straight line.
2. Path C — No outward force acts on the ball, so it cannot be flung radially outwards. At the moment the string breaks its velocity is along the tangent, and with no force it simply keeps that velocity.
3. Path A — Once the string breaks no horizontal force acts on the ball, so by Newton's first law it continues with the velocity it had at that instant: in a straight line along the tangent, at constant speed. The tension had only been changing its direction.
4. Path D — The centripetal force is not a separate force belonging to the ball; it was the tension in the string. When the string breaks there is no longer any force towards the centre, so the ball cannot curve inwards.

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