Dynamics: Particles, Planar Rigid Bodies, Energy and Momentum

This chapter answers the second half of Section XE3.1, Mechanics of Rigid Bodies, of XE3 Solid Mechanics, one of the nine optional sections of the Engineering Sciences (XE) paper. The syllabus text is: “particle kinematics and dynamics; dynamics of inter-connected and/or constrained rigid bodies under planar motion; systems that conserve energy and/or momentum.” It teaches particle motion in Cartesian, polar and normal-tangential components, Newton’s second law with work-energy and impulse-momentum, the instantaneous centre, rolling without slipping, moments of inertia, F = ma with M = Iα for connected bodies, collisions with the coefficient of restitution, and angular momentum. The numericals are projectile and curved-path motion, a sphere rolling down a slope, a pulley with mass, and an impact on a hinged rod. Take g = 9.81 m/s² throughout, and treat strings as light and inextensible and pulleys as smooth-bearing unless the stem says otherwise.

1. Particle kinematics: rectilinear, projectile and curvilinear motion

In rectilinear motion v = dx/dt and a = dv/dt = v dv/dx; the last form is the one to use when acceleration is given as a function of position. For constant acceleration v = u + at, s = ut + ½at², v² = u² + 2as. A projectile (no air resistance, g constant) has a_x = 0 and a_y = −g, so the horizontal velocity stays u cos α while the vertical one is u sin α − gt. For launch and landing at the same level the time of flight is 2u sin α/g, the range is R = u² sin 2α/g (greatest at 45°) and the greatest height is H = u² sin²α/(2g). A body thrown horizontally with speed u from a height h lands after t = √(2h/g), independent of u, at horizontal distance u·t.

For curvilinear motion use components fitted to the path. In normal-tangential components a_t = dv/dt changes the speed and a_n = v²/ρ (ρ is the radius of curvature, directed towards the centre of curvature) changes the direction, so |a| = √(a_t² + a_n²). In polar components, v_r = ṙ and v_θ = rθ̇, while a_r = r̈ − rθ̇² and a_θ = rθ̈ + 2ṙθ̇. The term −rθ̇² is the centripetal part, and 2ṙθ̇ is the Coriolis term that appears when r changes while the line turns.

Worked example. A ball is launched at 20 m/s at 30° above the horizontal from level ground. R = 20² sin 60°/9.81 = 400(0.8660)/9.81 = 35.3 m and H = 400(0.25)/(2 × 9.81) = 5.10 m. A car on a circular bend of radius 30 m at 12 m/s speeding up at 2 m/s² has a_n = 144/30 = 4.8 m/s² and a_t = 2 m/s², so |a| = √(4.8² + 2²) = 5.2 m/s².

⚠️ Speed versus velocity
A particle moving at constant speed on a curve still accelerates, with a_n = v²/ρ and a_t = 0. In polar form, a_r is not r̈ alone: forgetting the −rθ̇² term is the standard slip. At the top of a projectile’s path the velocity is horizontal, not zero, and the acceleration is still g downward.

2. Particle kinetics: Newton, work-energy and impulse-momentum

Newton’s second law, ΣF = ma, is written along convenient axes: for a particle on a circle, ΣF_n = mv²/ρ and ΣF_t = m dv/dt. Two integrals of it avoid solving for the time history. Work-energy: the work of all forces from state 1 to 2 equals the change of kinetic energy, W₁₋₂ = ½mv₂² − ½mv₁². Work of a force is ∫F·dr; weight does mg·(drop in height), a spring does −½k(δ₂² − δ₁²), and a normal force or the tension of a taut string on a body moving along the surface or arc does none. If only weight and springs work, energy is conserved: ½mv² + mgh + ½kδ² is constant. Friction and impact dissipate energy. Impulse-momentum: ∫ΣF dt = mv₂ − mv₁, a vector equation; for a system with no net external force the total momentum is constant.

Worked example (sphere on a curve). A bead of mass 0.2 kg slides without friction from rest at A, 1.5 m above the lowest point B of a vertical circular wire of radius R = 1 m. Energy: ½v² = 9.81(1.5), so v_B = 5.42 m/s. At B the wire pushes up, N − mg = mv²/R, so N = 0.2(9.81) + 0.2(29.43)/1 = 7.85 N. The speed depends only on the drop in height, not on the shape of the wire.

⚠️ Which law to choose
Use work-energy when a speed is wanted after a known displacement, impulse-momentum when a time or an impact is involved, and F = ma only when a force or acceleration at one instant is wanted. Energy is not conserved when friction acts or in an inelastic impact, but momentum is still conserved in the impact if external impulses are negligible.

3. Planar rigid-body kinematics: relative velocity, instantaneous centre, rolling

For two points A and B of the same rigid body in planar motion, v_B = v_A + ω × r_B/A, so v_B/A = ω·AB and is perpendicular to AB. The acceleration is a_B = a_A + α × r_B/A − ω² r_B/A: a tangential part αr and a part ω²r towards A. The instantaneous centre (IC) is the point of the body (or of its extension) with zero velocity at that instant; every point moves at speed ω times its distance from the IC, perpendicular to the line to it. Find it as the intersection of the perpendiculars to two known velocity directions. A body that moves purely by translation has its IC at infinity (ω = 0). The IC is a velocity concept only: its acceleration is not zero.

Rolling without slipping. A wheel of radius R rolling on a fixed surface has its contact point at rest, so v_G = ωR and a_G = αR (centre moving in a straight line). The top point moves at 2v_G, and the contact point is the IC. The contact friction is static, does no work and is only as large as the motion needs. If the wheel slips, v_G ≠ ωR and kinetic friction μ_k N acts and does work.

⚠️ Acceleration of the contact point
The contact point has zero velocity but not zero acceleration: for a wheel rolling at constant speed it accelerates towards the centre with v²/R. So the IC may be used to find velocities, never accelerations. A bar whose ends slide along a wall and the floor has its IC at the corner of the rectangle formed by the wall, the floor and the bar.

4. Planar kinetics: moment of inertia, F = ma and M = Iα for connected bodies

The moment of inertia about an axis is I = ∫r² dm, and I = mk² defines the radius of gyration k. The parallel-axis theorem is I = I_G + md², where d is the distance between the axis through the centre of mass G and the parallel axis; it holds only when one of the two axes passes through G. Standard values (uniform bodies, axis through G unless stated): thin rod of length L, I_G = mL²/12 and about an end mL²/3; disc or solid cylinder mR²/2; thin ring mR²; solid sphere 2mR²/5; thin spherical shell 2mR²/3.

For planar motion the equations of motion of a rigid body of mass m are ΣF_x = ma_Gx, ΣF_y = ma_Gy and ΣM_G = I_Gα. About a fixed pivot O, ΣM_O = I_O α. Moments about the instantaneous centre or any other moving point are valid only in special cases, so take them about G or a fixed point. For connected bodies, draw a separate free-body diagram for each body, write one equation per degree of freedom, and add the kinematic constraint (an inextensible string gives equal speeds and accelerations along it; a non-slipping rope on a pulley gives a = αR).

Worked example (rolling). A uniform solid sphere (k² = 2R²/5) rolls without slipping down a slope of angle θ. Along the slope mg sin θ − F = ma, and about G, FR = I α = (2/5)mR²(a/R), so F = (2/5)ma. Hence a = g sin θ/(1 + 2/5) = (5/7)g sin θ, independent of mass and radius, and F = (2/7)mg sin θ. The surface must supply μ_s ≥ F/N = (2/7)tan θ. A solid cylinder gives a = (2/3)g sin θ and μ_s ≥ tan θ/3; for θ = 30° the cylinder has a = 3.27 m/s² and needs μ_s ≥ 0.19.

Worked example (pulley with mass). Blocks of 5 kg and 3 kg hang over a uniform disc pulley of mass 4 kg and radius R, the rope not slipping. The disc has I = ½MR², so I α/R = ½Ma = 2a. Treating the three bodies together gives (5 − 3)(9.81) = (5 + 3 + 2)a, so a = 1.96 m/s². The tensions differ: T₁ = 5(9.81 − 1.962) = 39.2 N and T₂ = 3(9.81 + 1.962) = 35.3 N, and their difference times R drives the pulley. The Atwood result a = (m₁ − m₂)g/(m₁ + m₂) holds only for a massless pulley.

5. Systems that conserve energy and momentum: collisions and angular momentum

For a rigid body, kinetic energy is T = ½mv_G² + ½I_Gω² (or ½I_Oω² about a fixed point O), so a rolling body’s energy is shared between translation and rotation: for a solid sphere released from rest at a drop h, mgh = ½mv²(1 + 2/5), and v = √(2gh/1.4) = √(10gh/7). Angular momentum about a point O is H_O = I_O ω for a body turning about fixed O, and r × mv for a particle. It is constant when the net external moment about O is zero, even if energy is lost, which makes it the tool for an impact on a hinged body.

In a direct central collision the line of impact is the common normal. Momentum along it is conserved, m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. The coefficient of restitution is e = (v₂ − v₁)/(u₁ − u₂), the ratio of speed of separation to speed of approach; e = 1 is elastic (kinetic energy conserved) and e = 0 is perfectly plastic (the bodies move together). Components perpendicular to the line of impact are unchanged for smooth bodies. A ball dropped from height h onto a fixed floor rebounds to e²h.

Worked example (collision). A 2 kg block at 6 m/s strikes a 1 kg block at rest, e = 0.5. Momentum: 2v₁ + v₂ = 12, restitution: v₂ − v₁ = 0.5(6) = 3. So v₁ = 3 m/s and v₂ = 6 m/s. Kinetic energy falls from 36 J to 9 + 18 = 27 J, a loss of 9 J. Worked example (impact on a rod). A uniform rod of mass 3 kg and length 1.2 m hangs from a frictionless hinge at one end. A 0.5 kg clay lump moving horizontally at 8 m/s strikes the free end and sticks. Angular momentum about the hinge is conserved (the hinge force has no moment about it): 0.5(8)(1.2) = [3(1.2)²/3 + 0.5(1.2)²]ω = 2.16ω, so ω = 2.22 rad/s. Linear momentum is not conserved, because the hinge exerts an impulse.

⚠️ What is conserved where
In an inelastic impact kinetic energy is lost, so never write energy conservation across the impact; it can be used afterwards, for example for the swing of the rod. For the rod, momentum conservation is wrong but angular momentum about the hinge is right. In a collision the sign of the relative velocities matters: e is a positive ratio of separation to approach speed.

Key takeaways

  • Rectilinear: a = v dv/dx. Projectile: R = u² sin 2α/g and H = u² sin²α/(2g) at equal levels. Curved path: a_n = v²/ρ and a_t = dv/dt; polar a_r = r̈ − rθ̇², a_θ = rθ̈ + 2ṙθ̇.
  • Pick work-energy for speed after a displacement, impulse-momentum for impacts and times, and F = ma for instantaneous forces. Normal forces and taut strings do no work on a body moving along the surface.
  • The instantaneous centre has zero velocity, not zero acceleration. In rolling without slipping v_G = ωR, a_G = αR, and static friction does no work; a solid sphere has a = (5/7)g sin θ, a solid cylinder (2/3)g sin θ.
  • I = I_G + md² (one axis through G); rod mL²/12 centre, mL²/3 end; disc mR²/2; ring mR²; sphere 2mR²/5. For connected bodies write a free-body diagram for each, then add the constraint; tensions differ across a massive pulley.
  • Collisions: momentum conserved along the line of impact; e = separation speed/approach speed; kinetic energy is lost unless e = 1; a ball dropped from h rebounds to e²h. Angular momentum about a hinge or fixed point survives when linear momentum does not.
  • Exam traps: omitting −rθ̇² in a_r, using the Atwood formula with a massive pulley, using energy across an inelastic impact, taking moments about a moving point, and using the IC for accelerations.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A particle moves at constant speed v along a circular arc of radius ρ. Which statement about its acceleration is correct?

    1. It is zero, because the speed does not change and so no force is needed to keep the particle moving
    2. It is v²/ρ directed along the tangent, in the direction of motion, so the particle keeps gaining speed
    3. The tangential component is zero and the normal component is v²/ρ towards the centre of curvature
    4. The normal component is zero and the tangential component is v²/ρ, directed backwards along the path
    Show answer

    Answer: C — The tangential component is zero and the normal component is v²/ρ towards the centre of curvature

    The tangential component dv/dt is zero because the speed is constant, but the direction of the velocity keeps changing, which needs a normal acceleration v²/ρ pointing towards the centre of curvature. Saying that the acceleration is zero because the speed is constant is the tempting error; acceleration is the rate of change of the velocity vector, not of the speed. The v²/ρ term is normal to the path, not tangential.

  2. A ball is thrown from level ground at 20 m/s at 45° above the horizontal. Neglect air resistance and take g = 9.81 m/s². At the highest point of its path:

    1. Its velocity is horizontal at 20 cos 45° = 14.1 m/s and its acceleration is 9.81 m/s² downward
    2. Its velocity is zero and its acceleration is also zero, because the ball momentarily stops
    3. Its velocity is zero for an instant, but its acceleration is 9.81 m/s² downward
    4. Its velocity is horizontal at 14.1 m/s and its acceleration is zero at the top of the path
    Show answer

    Answer: A — Its velocity is horizontal at 20 cos 45° = 14.1 m/s and its acceleration is 9.81 m/s² downward

    Only the vertical velocity component vanishes at the top; the horizontal component u cos α = 14.1 m/s is unchanged because a_x = 0. Gravity acts throughout, so the acceleration stays 9.81 m/s² downward. The answer with zero velocity and zero acceleration confuses the vertical component with the full velocity, and the answer with zero acceleration forgets that gravity does not switch off at the top.

  3. A stone is thrown horizontally at 15 m/s from the top of a cliff 20 m above level ground. Neglect air resistance and take g = 9.81 m/s². The horizontal distance from the foot of the cliff to the landing point is ____ m, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 30.29

    Vertical motion from rest: 20 = ½(9.81)t², so t = √(2 × 20/9.81) = 2.0193 s. The horizontal velocity stays 15 m/s, so the distance is 15 × 2.0193 = 30.29 m. The launch speed does not affect the time of fall.

  4. A small block of mass m is released from rest at the same height h on two smooth fixed tracks: a straight incline and a curved track with a dip, both ending at the same lower level. Ignoring air resistance, which statement about the block at the end of each track is correct?

    1. The speed is larger on the straight incline, because its path is shorter
    2. The speed is larger on the curved track, because the block travels through a dip and gathers more speed
    3. The speeds are equal and the times taken are equal, because both tracks drop through the same height
    4. The speeds are equal, √(2gh), but the times taken are in general different
    Show answer

    Answer: D — The speeds are equal, √(2gh), but the times taken are in general different

    The normal force from a smooth fixed track does no work, so only gravity works and ½mv² = mgh on both tracks: the final speed is √(2gh) whatever the shape. The times depend on the speed at every intermediate point and so differ in general. Concluding that the shorter path gives more speed confuses distance with the drop in height, and equal speeds do not imply equal times.

  5. A car moves on a circular bend of radius 30 m. At the instant considered its speed is 12 m/s and is increasing at 2 m/s². The magnitude of its total acceleration is ____ m/s², to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 5.20

    a_n = v²/ρ = 144/30 = 4.8 m/s² and a_t = 2 m/s², perpendicular to each other, so |a| = √(4.8² + 2²) = √27.04 = 5.20 m/s².

  6. Which of the following statements about the dynamics of a particle are correct?

    1. The work done by weight depends only on the vertical drop of the particle, not on the path followed
    2. The normal force of a smooth fixed surface on a particle sliding along it does no work
    3. The impulse of the net force over a time interval equals the change of the particle’s linear momentum, a vector equation
    4. A particle moving at constant speed on a curved path has constant kinetic energy and therefore zero net force
    Show answer

    Answer: A — The work done by weight depends only on the vertical drop of the particle, not on the path followed; B — The normal force of a smooth fixed surface on a particle sliding along it does no work; C — The impulse of the net force over a time interval equals the change of the particle’s linear momentum, a vector equation

    Weight is a conservative force, so its work is mg times the drop. A normal force on a smooth surface is perpendicular to the displacement and does no work. The impulse-momentum relation ∫ΣF dt = mv₂ − mv₁ is a vector equation. Constant speed gives constant kinetic energy, but the net force is not zero on a curved path: the normal component mv²/ρ is needed to turn the velocity, and it does no work.

  7. A wheel rolls without slipping on a horizontal road, with its centre moving at 4 m/s. The speed of the highest point of the wheel relative to the road is:

    1. 4 m/s, the same as the centre
    2. 8 m/s, twice the speed of the centre
    3. Zero, as for the contact point
    4. 5.66 m/s, the speed of the end of a horizontal diameter
    Show answer

    Answer: B — 8 m/s, twice the speed of the centre

    The contact point is the instantaneous centre, so the speed of any point is ω times its distance from it. The centre is at distance R, giving v_G = ωR = 4 m/s, and the top is at 2R, giving 2ωR = 8 m/s. Zero belongs to the contact point only, and 5.66 m/s (4√2) is the speed at the end of the horizontal diameter, at distance √2 R from the contact point.

  8. A thin uniform ring and a uniform solid cylinder of the same radius, mass and material surface roll without slipping down the same rough incline of angle θ, each starting from rest. The ratio a_cylinder/a_ring of their centre accelerations is:

    1. 1, because for a rolling body the acceleration is independent of mass and radius
    2. 3/2, because the cylinder has half the moment of inertia of the ring and so rolls half as slowly
    3. 4/3, because a = (2/3)g sin θ for the cylinder and a = (1/2)g sin θ for the ring
    4. 2, because the cylinder has half the rotational inertia and therefore doubles the acceleration
    Show answer

    Answer: C — 4/3, because a = (2/3)g sin θ for the cylinder and a = (1/2)g sin θ for the ring

    For rolling without slipping a = g sin θ/(1 + k²/R²). The ring has k² = R², so a = ½g sin θ; the solid cylinder has k² = R²/2, so a = (2/3)g sin θ. The ratio is (2/3)/(1/2) = 4/3. Both accelerations are independent of mass and radius, but they depend on the mass distribution through k²/R², so the ratio is not 1. Halving the moment of inertia does not double a, because the translational inertia m also appears in the denominator.

  9. A uniform solid sphere starts from rest and rolls without slipping down a rough slope, descending through a vertical height of 2 m. Take g = 9.81 m/s² and I = (2/5)mR² about the centre. Its centre speed at the end is ____ m/s, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 5.29

    Static friction does no work, so mgh = ½mv² + ½Iω² = ½mv²(1 + 2/5) with ω = v/R. Then v = √(2gh/1.4) = √(2 × 9.81 × 2/1.4) = √28.03 = 5.29 m/s. A frictionless block would reach √(2gh) = 6.26 m/s.

  10. A uniform thin rod of mass 3 kg and length 2 m rotates in a plane about a fixed axis perpendicular to the rod, passing through a point 0.5 m from the centre of the rod. Its moment of inertia about this axis is:

    1. 1.00 kg·m², the value about the centre of the rod
    2. 1.75 kg·m², centre value plus md²
    3. 4.00 kg·m², the value about an end of the rod
    4. 0.75 kg·m², from md² alone with d = 0.5 m
    Show answer

    Answer: B — 1.75 kg·m², centre value plus md²

    I_G = mL²/12 = 3(4)/12 = 1.00 kg·m². The parallel-axis theorem adds md² = 3(0.25) = 0.75, so I = 1.75 kg·m². The centre value alone ignores the offset of the axis, md² alone ignores the rod’s own I_G, and 4.00 kg·m² (mL²/3) is the end value, correct only for an axis through the end.

  11. Two blocks of masses 6 kg and 2 kg hang from the ends of a light inextensible rope passing over a pulley that is a uniform disc of mass 8 kg and radius R, mounted on a smooth fixed axle. The rope does not slip on the pulley. Take g = 9.81 m/s². The acceleration of the blocks is ____ m/s², to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 3.27

    The pulley has I = ½MR², and with a = αR its effective inertia in the linear equation is I/R² = 4 kg. For the whole system (6 − 2)(9.81) = (6 + 2 + 4)a, so a = 39.24/12 = 3.27 m/s². The massless-pulley formula would give (4/8)(9.81) = 4.91 m/s², which is too large.

  12. A wheel rolls without slipping, at an instant, on a fixed horizontal surface with its centre moving at speed v_G in a straight line, radius R. Which of the following are correct?

    1. The static friction at the contact point does no work on the wheel
    2. The contact point has zero velocity and, whatever the motion, zero acceleration as well
    3. The speed of the highest point is 2v_G
    4. The angular speed of the wheel is ω = v_G/R
    Show answer

    Answer: A — The static friction at the contact point does no work on the wheel; C — The speed of the highest point is 2v_G; D — The angular speed of the wheel is ω = v_G/R

    The contact point has zero velocity, so friction there does no work; v_G = ωR gives ω = v_G/R; the top point at distance 2R from the instantaneous centre moves at 2ωR = 2v_G. The contact point has zero velocity but not zero acceleration: for a wheel rolling steadily it accelerates towards the centre of the wheel at v²/R, so the statement of zero acceleration is incorrect.

  13. A uniform rod of mass m and length L is hinged at one end to a frictionless pivot and released from rest in the horizontal position. At the instant of release the vertical reaction at the hinge is:

    1. mg/2 upward, because each half of the rod is supported by half the weight
    2. mg upward, because the rod is at rest at that instant and must be in equilibrium
    3. Zero, because the rod is in free fall at the instant of release
    4. mg/4 upward, because the centre accelerates downward at 3g/4
    Show answer

    Answer: D — mg/4 upward, because the centre accelerates downward at 3g/4

    At release ω = 0, so only the moment of the weight about the hinge acts: mg(L/2) = (mL²/3)α, giving α = 3g/(2L). The centre of mass then accelerates downward at αL/2 = 3g/4. Vertical force balance mg − R = m(3g/4) gives R = mg/4. The rod is not in equilibrium merely because its velocity is zero; its acceleration is not zero. Free fall would need α = 2g/L, which the hinge prevents.

  14. In a direct central collision of two bodies the coefficient of restitution is e = 0. Which description of the outcome is correct?

    1. The bodies move off together with one common velocity; momentum is conserved
    2. The bodies exchange their velocities, as for equal masses in an elastic impact in a straight line
    3. The bodies separate with the same relative speed with which they had approached each other
    4. The lighter body stops dead and the heavier body carries away all of the momentum
    Show answer

    Answer: A — The bodies move off together with one common velocity; momentum is conserved

    e is the ratio of the speed of separation to the speed of approach, so e = 0 means zero relative velocity after impact: the bodies share one velocity, and the momentum is conserved because no external impulse acts. Exchange of velocities and separation at the approach speed describe e = 1. Stopping of the lighter body is not implied by any value of e.

  15. A 2 kg block moving at 6 m/s collides head-on with a 1 kg block at rest on a smooth horizontal surface, and the two stick together. The kinetic energy lost in the impact is:

    1. 36 J, because all the kinetic energy of the moving block is lost
    2. 24 J, the kinetic energy of the combined mass after the impact
    3. 18 J, half of the initial kinetic energy, since the blocks end up sharing the motion equally
    4. 12 J, the difference between the energy before and after
    Show answer

    Answer: D — 12 J, the difference between the energy before and after

    Momentum: 2(6) = 3v, so v = 4 m/s. Kinetic energy before is ½(2)(36) = 36 J and after is ½(3)(16) = 24 J, so the loss is 36 − 24 = 12 J. Taking 24 J is quoting the energy that remains rather than the energy lost, and 36 J would need the combined mass to stop, which would violate momentum conservation.

  16. A uniform rod of mass 2 kg and length 1 m hangs at rest from a frictionless hinge at its upper end. A bullet of mass 0.05 kg, treated as a particle, strikes the lower end horizontally at 100 m/s and embeds there. Take I = mL²/3 for the rod about the hinge. The angular speed of the rod with the bullet just after impact is ____ rad/s, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 6.98

    The hinge force has no moment about the hinge, so angular momentum about it is conserved: m_b v L = (ML²/3 + m_b L²)ω. Then 0.05(100)(1) = (2/3 + 0.05)ω, so ω = 5/0.71667 = 6.98 rad/s. Conserving linear momentum would be wrong, because the hinge applies an impulse during the impact.

  17. Which of the following statements about collisions and conserved quantities are correct?

    1. When a ball is dropped from height h onto a fixed floor with restitution coefficient e, it rebounds to a height e²h
    2. In a collision between two free bodies with no external impulse, the total linear momentum is conserved whatever the value of e
    3. When a lump of clay sticks to a hinged rod, the angular momentum about the hinge is conserved across the impact
    4. Kinetic energy is conserved in every collision, because energy cannot be destroyed
    Show answer

    Answer: A — When a ball is dropped from height h onto a fixed floor with restitution coefficient e, it rebounds to a height e²h; B — In a collision between two free bodies with no external impulse, the total linear momentum is conserved whatever the value of e; C — When a lump of clay sticks to a hinged rod, the angular momentum about the hinge is conserved across the impact

    The speed after the floor impact is e times the speed before, and the rise height goes as the square of speed, giving e²h. Momentum is conserved in free collisions for any e. The hinge force has no moment about the hinge, so angular momentum about it is conserved. Total energy is conserved, but in an inelastic impact the lost kinetic energy goes to heat, sound and deformation, so the kinetic energy alone is not conserved unless e = 1.