The Workbook

\(e\) is the ratio of the speed of separation to the speed of approach along the line of impact. \(e = \dfrac{3}{6} = 0.5\).

Before: \(v = \sqrt{2 \times 10 \times 5} = 10\) m/s. After: \(v' = \sqrt{2 \times 10 \times 3.2} = 8\) m/s. The floor does not move, so \(e = \dfrac{8}{10} = 0.8\). In general \(e = \sqrt{h'/h}\).

\(v' = eu = 0.5 \times 8 = 4\) m/s. Height \(= \dfrac{v'^2}{2g} = \dfrac{16}{20} = 0.8\) m.

1. \(2 \times 6 = 6v\), so \(v = 2\) m/s.

2. Before: \(\dfrac{1}{2} \times 2 \times 36 = 36\) J. After: \(\dfrac{1}{2} \times 6 \times 4 = 12\) J. Lost: \(24\) J.

3. The blocks do not separate, so the separation speed is zero and \(e = 0\).

(i) Elastic: kinetic energy is conserved. (ii) Perfectly inelastic: the bodies move together and the maximum possible kinetic energy is lost. (iii) Inelastic: some kinetic energy is lost, but the bodies separate.

Equal masses in an elastic head-on collision exchange velocities. The first ball stops and the second moves off at \(3\) m/s.

1. \(v_1 = \dfrac{2 - 4}{6} \times 6 = -2\) m/s (it rebounds). \(v_2 = \dfrac{2 \times 2}{6} \times 6 = 4\) m/s.

2. Momentum: before \(12\), after \(2(-2) + 4(4) = 12\) kg m/s. Kinetic energy: before \(36\) J, after \(\dfrac{1}{2}(2)(4) + \dfrac{1}{2}(4)(16) = 4 + 32 = 36\) J.

1. Momentum: \(1(4) + 3(-2) = 1v_1 + 3v_2\), so \(v_1 + 3v_2 = -2\). Restitution: \(v_2 - v_1 = e(4 - (-2)) = 6\).

2. Adding: \(4v_2 = 4\), \(v_2 = 1\) m/s east. Then \(v_1 = 1 - 6 = -5\) m/s, that is \(5\) m/s west. Check the energy: \(8 + 6 = 14\) J before, \(12.5 + 1.5 = 14\) J after.

(i) With \(m_1 \gg m_2\): \(v_1 \approx u\) and \(v_2 \approx 2u\). The heavy ball is barely slowed and the light ball is thrown forward at twice the speed. (ii) With \(m_1 \ll m_2\): \(v_1 \approx -u\) and \(v_2 \approx 0\). The light ball rebounds with its speed unchanged, like a ball off a wall.

\(v_2 = \dfrac{2m_1u}{m_1 + m_2}\). Fraction \(= \dfrac{\frac{1}{2}m_2v_2^2}{\frac{1}{2}m_1u^2} = \dfrac{m_2}{m_1} \cdot \dfrac{4m_1^2}{(m_1 + m_2)^2} = \dfrac{4m_1m_2}{(m_1 + m_2)^2}\). For \(m_2 = 3m_1\): \(\dfrac{4 \times 3}{16} = \dfrac{3}{4}\).

\(v_1 = \dfrac{m - 2m}{3m}u = -\dfrac{u}{3}\): the neutron rebounds at a third of its speed. Fraction transferred \(= \dfrac{4 \times m \times 2m}{9m^2} = \dfrac{8}{9}\). Check: the neutron keeps \(\dfrac{1}{9}\) of its energy, since \((1/3)^2 = 1/9\).

1. \(v_{\text{cm}} = \dfrac{2 \times 5}{5} = 2\) m/s.

2. Momentum: \(2v_1 + 3v_2 = 10\). Restitution: \(v_2 - v_1 = 0.5 \times 5 = 2.5\). Then \(2v_1 + 3(v_1 + 2.5) = 10\), \(5v_1 = 2.5\), \(v_1 = 0.5\) m/s and \(v_2 = 3\) m/s.

3. Before: \(\dfrac{1}{2} \times 2 \times 25 = 25\) J. After: \(\dfrac{1}{2}(2)(0.25) + \dfrac{1}{2}(3)(9) = 0.25 + 13.5 = 13.75\) J. Lost: \(11.25\) J.

\(\mu = \dfrac{1 \times 2}{3} = \dfrac{2}{3}\) kg, \(u_{\text{rel}} = 6\) m/s. \(\Delta K = \dfrac{1}{2} \times \dfrac{2}{3} \times 36 \times (1 - 0.25) = 12 \times 0.75 = 9\) J.

1. Speed of impact \(= \sqrt{2 \times 10 \times 5} = 10\) m/s. After the bounce: \(0.5 \times 10 = 5\) m/s.

2. \(h_1 = \dfrac{25}{20} = 1.25\) m, which is \(e^2h = 0.25 \times 5\).

3. Time up and down at \(5\) m/s: \(t = \dfrac{2 \times 5}{10} = 1\) s.

Rebound speed \(= 0.4 \times 10 = 4\) m/s. \(|\Delta p| = 0.5(10 + 4) = 7\) kg m/s. Kinetic energy goes from \(25\) J to \(4\) J, so the fraction lost is \(1 - e^2 = 1 - 0.16 = 0.84\).

1. \(0.01 \times 400 = (0.01 + 1.99)v = 2v\), so \(v = 2\) m/s.

2. After the impact energy is conserved: \(\dfrac{1}{2} \times 2 \times 4 = 2 \times 10 \times h\), so \(h = 0.2\) m.

3. Before: \(\dfrac{1}{2} \times 0.01 \times 160000 = 800\) J. After: \(4\) J. Lost: \(796\) J, nearly all of it, as heat and deformation.

1. At maximum compression both blocks move together: \(2 \times 4 = 8v\), \(v = 1\) m/s.

2. Energy: \(\dfrac{1}{2} \times 2 \times 16 = \dfrac{1}{2} \times 8 \times 1 + \dfrac{1}{2} \times 600 \times x^2\), so \(16 = 4 + 300x^2\) and \(x = 0.2\) m.

The spring gives back all its stored energy, so both momentum and kinetic energy are conserved overall. Using the elastic formulae with \(m_1 = 2\), \(m_2 = 6\), \(u = 4\): \(v_1 = \dfrac{2 - 6}{8} \times 4 = -2\) m/s, \(v_2 = \dfrac{2 \times 2}{8} \times 4 = 2\) m/s. Check: \(-4 + 12 = 8\) kg m/s and \(4 + 12 = 16\) J.

\(\dfrac{1}{2} \times 1 \times 36 = \dfrac{1}{2} \times 900 \times x^2\), so \(x^2 = 0.04\) and \(x = 0.2\) m. The spring returns all the energy, so the block leaves at \(6\) m/s.

At maximum compression both move at \(v = \dfrac{1 \times 4}{2} = 2\) m/s. Energy: \(8 = \dfrac{1}{2} \times 2 \times 4 + \dfrac{1}{2} \times 200 \times x^2\), so \(4 = 100x^2\) and \(x = 0.2\) m.

While the first block is faster than the second, the gap between them is still closing and the spring is still being squeezed. Once the second block is faster, the spring is opening out. The compression is therefore greatest at the instant the relative velocity is zero.

Momentum: \(2 \times 3 = 2 \times 1 + 1 \times v_B\), so \(v_B = 4\) m/s. \(e = \dfrac{4 - 1}{3 - 0} = 1\). Kinetic energy: before \(9\) J, after \(1 + 8 = 9\) J. The collision is elastic.

Momentum: \(v_1 + v_2 = 6\). Restitution: \(v_2 - v_1 = 0.5 \times 6 = 3\). So \(v_2 = 4.5\) m/s and \(v_1 = 1.5\) m/s, both in the original direction.

Momentum: \(v = \dfrac{mu}{m + M}\). Final kinetic energy: \(K' = \dfrac{1}{2}(m + M)v^2 = \dfrac{m^2u^2}{2(m + M)} = \dfrac{m}{m + M}K\), where \(K = \dfrac{1}{2}mu^2\). Fraction lost \(= 1 - \dfrac{m}{m + M} = \dfrac{M}{m + M}\).

The component along the surface is unchanged, because a smooth surface exerts no force along itself. The normal component reverses and its size is multiplied by \(e\): \(v_n' = e\,v_n\).

1. Along the floor: \(8\) m/s, unchanged. Normal: \(0.4 \times 15 = 6\) m/s, now upward.

2. Speed \(= \sqrt{64 + 36} = 10\) m/s. \(\tan\alpha = \dfrac{6}{8}\), so \(\alpha = 37^\circ\) to the floor.

3. Only the normal momentum changes: \(J = 0.1 \times (15 + 6) = 2.1\) N s, normal to the floor.

Normal component \(10\cos 37^\circ = 8\) m/s reverses; tangential \(6\) m/s stays. The ball leaves at \(10\) m/s at \(37^\circ\) to the normal, on the other side. \(|\Delta p| = 0.2 \times (8 + 8) = 3.2\) kg m/s.

Normal after: \(0.75 \times 8 = 6\) m/s. Tangential: \(6\) m/s. Speed \(= \sqrt{36 + 36} = 6\sqrt2 \approx 8.5\) m/s, at \(45^\circ\) to the normal.

1. \(R_1 = \dfrac{u^2\sin 90^\circ}{g} = \dfrac{400}{10} = 40\) m.

2. Just before landing: \(10\sqrt2\) m/s horizontal and \(10\sqrt2\) m/s down. After: horizontal \(10\sqrt2\) m/s, vertical \(0.5 \times 10\sqrt2 = 5\sqrt2\) m/s up.

3. \(R_2 = \dfrac{2u_xu_y}{g} = \dfrac{2 \times 10\sqrt2 \times 5\sqrt2}{10} = 20\) m, which is \(eR_1\).

A smooth surface exerts only a normal force. During the bounce there is no force along the surface, so no impulse and no change of the tangential momentum.

The line of impact is the common normal at the point of contact, which for spheres is the line joining their centres. The impulsive force acts along it, so \(e\) applies to the velocity components along the line of impact. Components perpendicular to it do not change.

1. Along the line: \(10\cos 37^\circ = 8\) m/s. Perpendicular: \(10\sin 37^\circ = 6\) m/s.

2. Along the line the collision is a 1D elastic collision of equal masses: they exchange the \(8\) m/s. \(B\) moves at \(8\) m/s along the line of impact. \(A\) keeps only its perpendicular \(6\) m/s.

3. \(A\) moves perpendicular to the line and \(B\) along it, so the angle between them is \(90^\circ\).

Resolve the moving sphere's velocity into \(u\cos\theta\) along the line of impact and \(u\sin\theta\) perpendicular to it. Along the line, equal masses exchange velocities: the second sphere takes \(u\cos\theta\) and the first is left with zero. Perpendicular components do not change. So the first sphere moves perpendicular to the line and the second along it: at right angles.

Along the line: \(v_A + v_B = 8\) and \(v_B - v_A = 0.5 \times 8 = 4\), so \(v_B = 6\) m/s, \(v_A = 2\) m/s. \(A\) has \(2\) m/s along and \(6\) m/s perpendicular, speed \(\sqrt{40} = 2\sqrt{10}\) m/s. Energy: before \(50\) J; after \(\dfrac{1}{2}(40) + \dfrac{1}{2}(36) = 38\) J. Lost: \(12\) J.

1. Along the line \(A\) has \(5 \times 0.8 = 4\) m/s. With \(e = 0\) the two move together along the line: \(2 \times 4 = 5v\), \(v = 1.6\) m/s. \(B\) moves at \(1.6\) m/s along the line of impact.

2. \(A\) has \(1.6\) m/s along the line and its unchanged \(5 \times 0.6 = 3\) m/s perpendicular. Speed \(= \sqrt{2.56 + 9} = \sqrt{11.56} = 3.4\) m/s.

1. \(A\) after: \((6 \times 0.6,\ 6 \times 0.8) = (3.6,\ 4.8)\) m/s. Momentum: \(B\) gets \((10 - 3.6,\ -4.8) = (6.4,\ -4.8)\) m/s, speed \(8\) m/s at \(\tan^{-1}(4.8/6.4) = 37^\circ\) below the \(x\) axis.

2. Before: \(\dfrac{1}{2}(1)(100) = 50\) J. After: \(\dfrac{1}{2}(36) + \dfrac{1}{2}(64) = 18 + 32 = 50\) J. Elastic. The paths are at \(90^\circ\), as expected.

Momentum \((3, 4)\) kg m/s shared by \(2\) kg: \(v = (1.5, 2)\), speed \(2.5\) m/s at \(\tan^{-1}(2/1.5) = 53^\circ\) to the \(3\) m/s direction. Energy before \(\dfrac{1}{2}(9 + 16) = 12.5\) J, after \(\dfrac{1}{2}(2)(6.25) = 6.25\) J. Lost: \(6.25\) J.

Momentum \((6, 6)\) kg m/s shared by \(3\) kg: \(v = (2, 2)\) m/s, speed \(2\sqrt2 \approx 2.8\) m/s at \(45^\circ\) to \(x\).

1. \(v = \sqrt{2 \times 10 \times 5} = 10\) m/s, vertically down.

2. Along the incline: \(10\cos 45^\circ = 5\sqrt2\) m/s. Normal, into the surface: \(5\sqrt2\) m/s.

3. With \(e = 1\) the normal part reverses; the tangential part stays. Down-slope \(5\sqrt2\) plus outward-normal \(5\sqrt2\) add to \(10\) m/s horizontal, directed away from the incline.

Equal masses in an elastic collision with one at rest move at right angles, so \(u^2 = v_1^2 + v_2^2\): \(16 = 5.76 + v_2^2\), \(v_2 = 3.2\) m/s. The angle between the paths is \(90^\circ\).

Before: \(v_x = v\cos\theta\), \(v_y = v\sin\theta\). After: \(v_x\) unchanged, \(v_y' = e\,v_y\). \(\tan\theta' = \dfrac{v_y'}{v_x} = \dfrac{e\,v\sin\theta}{v\cos\theta} = e\tan\theta\).

Each bounce multiplies the vertical velocity by \(e\) and the range by \(e\): \(R_2 = eR_1\), \(R_3 = e^2R_1\), and so on. Total \(= R_1(1 + e + e^2 + \dots) = \dfrac{R_1}{1 - e}\). With \(R_1 = 40\) m and \(e = 0.5\): \(80\) m.

A smooth wall cannot exert friction, so its force on the ball is purely normal. Impulse is force times time, hence along the normal too. Momentum along the wall is therefore unchanged, and only the normal component of the velocity changes.

Momentum \((4, 4)\) kg m/s for \(2\) kg: \(v = (2, 2)\) m/s, that is \(2\sqrt2\) m/s at \(45^\circ\) to each original direction.

First ball after: \((3 \times 0.6,\ 3 \times 0.8) = (1.8,\ 2.4)\) m/s. Second ball: \((5 - 1.8,\ -2.4) = (3.2,\ -2.4)\) m/s, speed \(4\) m/s at \(37^\circ\) on the other side of the line. Check: \(9 + 16 = 25\), elastic.

The object ball leaves along the line of impact, so the line of impact is at \(60^\circ\) to the cue ball's velocity. Object ball: \(4\cos 60^\circ = 2\) m/s. Cue ball keeps the perpendicular part: \(4\sin 60^\circ = 2\sqrt3 \approx 3.5\) m/s, at \(90^\circ\) to the object ball.

Normal component \(= 10\sin 37^\circ = 6\) m/s, reversed by the elastic bounce. \(J = 0.2 \times (6 + 6) = 2.4\) N s, normal to the floor.

Along the line of impact the equal masses exchange velocities: \(B\) moves off at \(u\cos\theta\) and \(A\) keeps \(u\sin\theta\) perpendicular to the line. \(K_A = \dfrac{1}{2}mu^2\sin^2\theta\) and \(K_B = \dfrac{1}{2}mu^2\cos^2\theta\); their sum is the original energy.