The Workbook

(i) \(p = 2 \times 3 = 6\) kg m/s. (ii) \(p = 0.05 \times 400 = 20\) kg m/s. The light, fast bullet carries more momentum.

\(P = 2 \times 3 + 3 \times (-2) = 0\). \(v_{\text{cm}} = \dfrac{P}{M} = 0\): the centre of mass is at rest even though both bodies move.

\(v_{\text{cm}} = \dfrac{1 \times 4 + 3 \times 0}{4} = 1\) m/s, in the direction of the moving body.

1. \(P = 2 \times 4 - 1 \times 2 = 6\) kg m/s, east.

2. \(v_{\text{cm}} = \dfrac{6}{3} = 2\) m/s east.

3. Momentum is conserved: \(6 = 3v\), \(v = 2\) m/s east, the same as \(v_{\text{cm}}\). After sticking, the carts move with the centre of mass.

Before firing the total momentum is zero. After: \(5v + 0.01 \times 500 = 0\), so \(v = -1\) m/s: the gun recoils at \(1\) m/s.

1. \(2 \times 6 + 3v = 0\), so \(v = -4\) m/s: \(4\) m/s west.

2. \(K = \dfrac{1}{2} \times 2 \times 36 + \dfrac{1}{2} \times 3 \times 16 = 36 + 24 = 60\) J, all from the explosive.

3. The \(2\) kg piece gets \(36\) J, the \(3\) kg piece \(24\) J. Both have the same momentum \(12\) kg m/s, and \(K = \dfrac{p^2}{2m}\) is larger for the smaller mass.

Total momentum stays zero: \(1 \times 4\hat i + 1 \times 3\hat j + 2\vec v = 0\), so \(\vec v = -2\hat i - 1.5\hat j\) m/s. Speed \(= \sqrt{4 + 2.25} = 2.5\) m/s.

\(60v + 2 \times 15 = 0\), so \(v = -0.5\) m/s: the skater slides backward at \(0.5\) m/s.

1. \(1000 \times 20 = 2000v\), so \(v = 10\) m/s.

2. Before: \(\dfrac{1}{2} \times 1000 \times 400 = 2 \times 10^{5}\) J. After: \(\dfrac{1}{2} \times 2000 \times 100 = 10^{5}\) J. Lost: \(10^{5}\) J, half of the original.

1. Man relative to water \(=\) (man relative to boat) \(+\) (boat relative to water) \(= 1 - u\).

2. Total momentum stays zero: \(60(1 - u) - 90u = 0\), so \(60 = 150u\) and \(u = 0.4\) m/s.

3. \(1 - 0.4 = 0.6\) m/s forward. Check: \(60 \times 0.6 = 36 = 90 \times 0.4\).

The spring force is internal, so the total momentum stays zero: \(1 \times 6 + 3v = 0\), \(v = -2\) m/s, opposite to the \(1\) kg block. \(v_{\text{cm}} = 0\): the centre of mass stays where it was.

1. \(4 \times 20 = 2 \times 0 + 2v\), so \(v = 40\) m/s horizontally.

2. The burst is internal and gravity acts on the centre of mass exactly as it did on the shell. So the centre of mass continues along the original parabola and lands at \(Q\).

3. The pieces are equal, so the centre of mass is midway between them. One piece lies \(40\) m before \(Q\), so the other lands \(40\) m beyond \(Q\).

\(a_{\text{cm}} = \dfrac{F_{\text{ext}}}{M} = \dfrac{10}{5} = 2\) m/s\(^2\). The spring's forces are internal and cancel in the sum, so the stretching of the spring does not affect the centre of mass.

\(v_{\text{cm}} = \dfrac{1 \times 20 + 1 \times 0}{2} = 10\) m/s upward. Both balls have acceleration \(g\) downward, so \(a_{\text{cm}} = g = 10\) m/s\(^2\) downward.

1. Momentum stays zero: \(1 \times v_1 = 3 \times v_2\), so \(v_1 = 3v_2\).

2. Energy: \(\dfrac{1}{2} \times 1 \times 9v_2^2 + \dfrac{1}{2} \times 3 \times v_2^2 = 24\), so \(6v_2^2 = 24\), \(v_2 = 2\) m/s and \(v_1 = 6\) m/s.

3. \(K_1 : K_2 = 18 : 6 = 3 : 1\), the inverse of the mass ratio.

During the collision the forces between the bodies are an action-reaction pair. Their impulses on the two bodies are equal and opposite, so the total momentum cannot change. Kinetic energy is not a vector and has no such pairing; part of it can be converted into heat, sound or deformation.

The ball alone is not an isolated system: the wall pushes on it. Momentum is conserved for the ball plus the wall and the Earth. The wall gains \(10\) kg m/s, but its enormous mass makes its velocity change unnoticeable.

1. Yes: the ice is smooth, so no external horizontal force acts; the pushes are internal.

2. No: gravity is an external force on the ball and its momentum grows steadily.

3. Nearly: the collision force is far larger than friction, so over the brief impact friction's impulse is negligible.

The explosion lasts a very short time \(\Delta t\). Gravity's impulse \(mg\,\Delta t\) in that time is tiny compared with the large internal impulses. So the momentum just after the explosion equals the momentum just before. Later, gravity changes the momentum in the usual way.

\(80v = 4 \times 10\), so \(v = 0.5\) m/s opposite to the throw. \(t = \dfrac{10}{0.5} = 20\) s.

\(0.02 \times 300 = (0.02 + 2.98)v = 3v\), so \(v = 2\) m/s.

\(50 \times 1.4 = 70v\), so \(v = 1\) m/s in the opposite direction.

\(3 \times 4 = 1 \times 0 + 2v\), so \(v = 6\) m/s in the original direction.

In the centre-of-mass frame the velocities are \(v_1 - v_{\text{cm}}\) and \(v_2 - v_{\text{cm}}\). Total momentum there: \(m_1(v_1 - v_{\text{cm}}) + m_2(v_2 - v_{\text{cm}}) = (m_1v_1 + m_2v_2) - (m_1 + m_2)v_{\text{cm}} = 0\), by the definition of \(v_{\text{cm}}\). So \(p_1' = -p_2'\).

\(F = v_{\text{rel}}\dfrac{dm}{dt}\), directed opposite to the ejected mass. For the hose: \(F = 20 \times 2 = 40\) N, pushing the hose backward.

Mass leaving per second \(= 10 \times 0.02 = 0.2\) kg/s. \(F = 500 \times 0.2 = 100\) N.

Momentum arriving per second \(= 0.5 \times 5 = 2.5\) kg m/s each second, all destroyed at the roof. \(F = 2.5\) N downward.

1. \(\dfrac{dm}{dt} = \rho A v = 1000 \times 2 \times 10^{-4} \times 10 = 2\) kg/s.

2. The normal velocity \(10\) m/s is destroyed: \(F = v\dfrac{dm}{dt} = 10 \times 2 = 20\) N.

3. The normal velocity is reversed, so the change is \(20\) m/s per kg: \(F = 20 \times 2 = 40\) N.

Speed on arrival \(= \sqrt{2 \times 10 \times 5} = 10\) m/s. Extra force \(= v\dfrac{dm}{dt} = 10 \times 0.2 = 2\) N. Later the weight of the collected sand adds to it.

Newton's second law in the form \(F = \dfrac{dp}{dt}\). In each second a mass \(\dfrac{dm}{dt}\) arrives with velocity \(v\) and is brought to rest, so momentum \(v\dfrac{dm}{dt}\) is destroyed per second. The wall must supply that rate of change of momentum as a force.

1. \(F = v\dfrac{dm}{dt} = 3 \times 2 = 6\) N.

2. \(P = Fv = 6 \times 3 = 18\) W.

3. \(\dfrac{dK}{dt} = \dfrac{1}{2}\dfrac{dm}{dt}v^2 = \dfrac{1}{2} \times 2 \times 9 = 9\) W. The other \(9\) W is lost as heat while the sand slips to the belt speed.

1. \(F = 2 \times 5 = 10\) N.

2. \(P = Fv = 10 \times 2 = 20\) W.

3. The sand gains \(\dfrac{1}{2} \times 5 \times 4 = 10\) J per second. The remaining \(10\) W becomes heat through friction between sand and belt.

Force \(F = \mu v\), so the motor's power is \(P = Fv = \mu v^2\). The sand's kinetic energy rises at \(\dfrac{1}{2}\mu v^2\) per second. The difference, \(\dfrac{1}{2}\mu v^2\), is half the motor's work, and it is dissipated as heat as each grain is dragged to the belt speed.

The belt and the sand on it have momentum. Each new grain arrives with no horizontal momentum and must be given some by the belt. With no motor, the total horizontal momentum stays fixed while the moving mass grows, so the speed falls.

1. Thrust \(= u\dfrac{dm}{dt} = 2000 \times 50 = 10^{5}\) N.

2. \(a = \dfrac{F - mg}{m} = \dfrac{10^{5} - 5 \times 10^{4}}{5000} = 10\) m/s\(^2\).

1. Before: \(mv\). After: rocket \((m - dm)(v + dv)\) and gas \(dm\,(v - u)\), since the gas moves at \(u\) backward relative to the rocket.

2. No external force, so \(mv = (m - dm)(v + dv) + dm\,(v - u)\). Expanding and dropping \(dm\,dv\): \(0 = m\,dv - u\,dm\), so \(m\,dv = u\,dm\). Dividing by \(dt\): \(m\dfrac{dv}{dt} = u\dfrac{dm}{dt}\), so the thrust is \(u\dfrac{dm}{dt}\).

(i) \(v = 2000\ln 2 \approx 2000 \times 0.7 = 1400\) m/s. (ii) \(v = 2000\ln 4 = 2 \times 1400 = 2800\) m/s. Each halving of the mass adds the same \(1400\) m/s.

Thrust must at least equal the weight: \(u\dfrac{dm}{dt} = mg\), so \(\dfrac{dm}{dt} = \dfrac{4000 \times 10}{2000} = 20\) kg/s.

Mass at burnout \(= 1000 - 10 \times 50 = 500\) kg. \(v = 1000\ln 2 - 10 \times 50 \approx 700 - 500 = 200\) m/s.

In free space each kilogram of gas ejected gives the rocket the same impulse, \(u\) per kilogram relative to the rocket, whenever it is ejected. So only the total mass ejected matters, through \(\ln(m_0/m)\). With gravity, the loss \(gt\) grows with the burn time, so a faster burn reaches a higher speed.

1. The falling part is in free fall: \(v = \sqrt{2gx} = \sqrt{10} \approx 3.2\) m/s.

2. \(\dfrac{dm}{dt} = \lambda v = \sqrt{10} \approx 3.2\) kg/s.

3. Impact force \(= v\dfrac{dm}{dt} = \lambda v^2 = 2\lambda gx = 2 \times 1 \times 10 \times 0.5 = 10\) N.

4. Add the weight of the landed part, \(\lambda xg = 5\) N: total \(15\) N, three times the weight of the landed part.

Just before the last link lands, \(x = 1\) m and \(v^2 = 2g\). Force \(= \lambda xg + \lambda v^2 = 3\lambda xg = 3 \times 2 \times 1 \times 10 = 60\) N. The weight of the chain is \(20\) N, so the table feels three times the weight.

Chain leaves the heap at \(\lambda v = 1\) kg/s and is brought from rest to \(2\) m/s. \(F = v\dfrac{dm}{dt} = \lambda v^2 = 0.5 \times 4 = 2\) N. Power \(= Fv = 4\) W. Kinetic energy gained per second \(= \dfrac{1}{2}(1)(4) = 2\) W. The other \(2\) W becomes heat in the jerks as each link starts moving.

1. Each second \(3\) kg of sand must be brought to \(2\) m/s: \(F = \mu v = 3 \times 2 = 6\) N.

2. With \(F\) removed, horizontal momentum is conserved: \(30 \times 2 = 40v\), so \(v = 1.5\) m/s.

The cart keeps its speed. The leaking sand leaves with the same horizontal velocity as the cart, so it carries away momentum in proportion to its mass. There is no relative velocity and hence no thrust on the cart.

Mass arriving per second \(= 0.01 \times 500 \times 2 = 10\) kg/s. \(F = 20 \times 10 = 200\) N. Pressure \(= \dfrac{200}{2} = 100\) Pa.

The engine raises the air's speed by \(500 - 200 = 300\) m/s. Thrust \(= \dfrac{dm}{dt}\,\Delta v = 50 \times 300 = 15000\) N \(= 15\) kN.

Momentum given to the water per second is \(\mu v\), so the force is \(F = \mu v\). The kinetic energy given per second is \(\dfrac{1}{2}\mu v^2\), so that is the power. The product \(Fv = \mu v^2\) is not the power. The water does not move at \(v\) while the force acts on it. It is accelerated from rest, at an average speed of \(\dfrac{v}{2}\).