On a railway platform a porter lets go of a loaded handcart, and it rolls slowly towards the edge. At the same moment a cricket ball comes towards you from a gentle throw. You could stop the ball with one hand. The handcart, moving far more slowly, needs both hands and a firm stance. Speed alone does not decide how hard a body is to stop.
Momentum, recalled
The chapter on Accelerated Objects defined the momentum of a body as its mass times its velocity, \(\vec p = m\vec v\). It showed that the impulse of the net force equals the change in momentum, \(\vec J = \vec F_{\text{avg}}\,\Delta t = \Delta\vec p\). Momentum is a vector along the velocity, measured in kg m/s. A heavy slow body and a light fast body can carry the same momentum.
A \(60\,\text{kg}\) sprinter runs at \(10\,\text{m/s}\). A \(1500\,\text{kg}\) car creeps forward at \(0.4\,\text{m/s}\). Which has the greater momentum?
The second law as Newton wrote it
Newton stated his second law with momentum, not with acceleration. The net force on a body equals the rate at which its momentum changes. Write the momentum as \(m\vec v\) and differentiate the product. The derivation below shows where the familiar form \(\vec F = m\vec a\) comes from, and what it leaves out.
\[ \vec F = \dfrac{d\vec p}{dt} = \dfrac{d(m\vec v)}{dt} \]
Why. the second law in Newton's form, with \(\vec p = m\vec v\).
\[ \dfrac{d(m\vec v)}{dt} = m\dfrac{d\vec v}{dt} + \vec v\dfrac{dm}{dt} \]
Why. the product rule, since both the mass and the velocity may change with time.
\[ \vec F = m\vec a + \vec v\dfrac{dm}{dt} \]
Why. the rate of change of velocity is the acceleration, \(\vec a = d\vec v/dt\).
\[ \vec F = m\vec a \text{ when } \dfrac{dm}{dt} = 0 \]
Read it. for a body whose mass does not change, the second term vanishes.
The net force on a body equals the rate of change of its momentum, \(\vec F = d\vec p/dt\). For a body of constant mass this is \(\vec F = m\vec a\).
On a graph of momentum against time, the slope at any instant is \(dp/dt\). So the slope is the net force at that instant. A steep stretch means a large force, and a flat stretch means no net force at all. Fig 1.1 shows a momentum that grows faster and faster, because the force on the body keeps increasing.
The momentum of a body along a line is \(p = 4t^2 + 3t\), with \(p\) in kg m/s and \(t\) in seconds. Find the net force on it at \(t = 2\,\text{s}\), in newtons.
A cricket bat gives the clearest example of the average form. The ball arrives from the bowler, and the bat sends it back the other way in about two thousandths of a second. The change in momentum is large, because the direction reverses. Divided by so short a time, it gives a force thousands of times the weight of the ball.
A \(0.16\,\text{kg}\) cricket ball reaches the bat at \(30\,\text{m/s}\) and leaves along the same line in the opposite direction at \(40\,\text{m/s}\). The contact lasts \(2\,\text{ms}\). Find the average force of the bat on the ball, in newtons.
\(\vec F = m\vec a\) is the constant-mass case of the second law. A rocket that throws out gas, or a belt that collects sand, changes its mass as it moves. For such bodies the term \(\vec v\,dm/dt\) matters, and the simple form gives wrong answers. Lessons 5 and 6 of this chapter handle them.
Now return to the platform. A handcart of \(200\,\text{kg}\) rolling at \(0.5\,\text{m/s}\) carries \(100\,\text{kg m/s}\). A cricket ball of \(0.16\,\text{kg}\) at \(30\,\text{m/s}\) carries only \(4.8\,\text{kg m/s}\). To stop either in one second, your hands must remove all its momentum in that second. The cart needs an average force of \(100\,\text{N}\), and the ball only \(4.8\,\text{N}\).
Momentum of a system
A system is any set of bodies we choose to study together: two carts, a gun and its bullet, a family on a boat. Its total momentum \(\vec P\) is the sum of the momenta of all its parts. The sum is a vector sum. Two carts that move towards each other with equal and opposite momenta give a total of zero, even though both are moving.
\[ \vec P = m_1\vec v_1 + m_2\vec v_2 + \dots + m_n\vec v_n = \sum_i m_i\vec v_i \]
The chapter on the Centre of Mass defines the centre of mass of a system and its velocity, \(\vec v_{\text{cm}} = \sum m_i\vec v_i / M\), where \(M\) is the total mass. Multiply both sides by \(M\). The right side becomes the total momentum. So a whole system carries the momentum of a single particle of mass \(M\) moving with its centre of mass.
A \(2\,\text{kg}\) ball moves east at \(4\,\text{m/s}\). A \(3\,\text{kg}\) ball moves north at \(2\,\text{m/s}\). Find the total momentum of the pair and the velocity of their centre of mass.
The momenta are \(2 \times 4 = 8\,\text{kg m/s}\) east and \(3 \times 2 = 6\,\text{kg m/s}\) north. They are at right angles, so the size of the sum is \(\sqrt{64 + 36} = 10\,\text{kg m/s}\).
The direction is \(\tan\theta = 6/8 = 0.75\), so \(\theta = 37°\) north of east.
The total mass is \(5\,\text{kg}\), so \(v_{\text{cm}} = P/M = 10/5 = 2\,\text{m/s}\), in the same direction as \(\vec P\).
A \(1\,\text{kg}\) ball moves east at \(6\,\text{m/s}\). A \(2\,\text{kg}\) ball moves west at \(3\,\text{m/s}\). Which statement about the pair is right?
What changes the total momentum
Every force on a part of a system is one of two kinds. An internal force is exerted by another part of the same system. An external force comes from something outside it. Take a system of two bodies, 1 and 2. Body 2 pushes on body 1 with \(\vec F_{12}\), and body 1 pushes back on body 2 with \(\vec F_{21}\). Newton's third law says these two are equal and opposite.
\[ \dfrac{d\vec p_1}{dt} = \vec F_{1,\text{ext}} + \vec F_{12} \]
Why. the second law for body 1: the external force on it plus the force from body 2.
\[ \dfrac{d\vec p_2}{dt} = \vec F_{2,\text{ext}} + \vec F_{21} \]
Why. the second law for body 2, in the same way.
\[ \dfrac{d(\vec p_1 + \vec p_2)}{dt} = \vec F_{1,\text{ext}} + \vec F_{2,\text{ext}} + (\vec F_{12} + \vec F_{21}) \]
Why. add the two equations.
\[ \dfrac{d\vec P}{dt} = \vec F_{\text{ext}} \]
Read it. \(\vec F_{12} + \vec F_{21} = 0\) by the third law. What remains on the right is the net external force.
The net external force on a system equals the rate of change of its total momentum, \(\vec F_{\text{ext}} = d\vec P/dt\). Internal forces, however large, cannot change the total momentum.
The proof works the same way for any number of bodies, because the internal forces always come in pairs. This is why you cannot move a car by pushing on its dashboard from the driver's seat. Your push on the dashboard and the dashboard's push on you are both inside the system. The car moves only when something outside, such as the road, pushes on it.
Blocks of \(3\,\text{kg}\) and \(2\,\text{kg}\) rest on a smooth floor, joined by a light spring. A steady horizontal force of \(10\,\text{N}\) pulls the \(2\,\text{kg}\) block. The spring stretches and the blocks jerk back and forth relative to each other. Find the acceleration of the centre of mass and the total momentum after \(3\,\text{s}\).
Take the two blocks and the spring as the system. Gravity and the normal forces are vertical and balance. The spring force is internal.
\(\vec F_{\text{ext}} = M\vec a_{\text{cm}}\) and \(\vec F_{\text{ext}} = d\vec P/dt\), with only the \(10\,\text{N}\) pull as the net external force.
\(a_{\text{cm}} = 10/5 = 2\,\text{m/s}^2\). The momentum grows by \(10\,\text{kg m/s}\) every second, so after \(3\,\text{s}\) it is \(30\,\text{kg m/s}\).
The centre of mass accelerates at \(2\,\text{m/s}^2\) and the total momentum after \(3\,\text{s}\) is \(30\,\text{kg m/s}\). This holds however the blocks jerk relative to each other.
Two children on roller skates, on a smooth floor, pull on the two ends of a rope. What happens to the total momentum of the two children and the rope?
A driver sits in a stalled car and pushes hard on the steering wheel. Explain, with the second law for a system, why the car does not move forward.
Take the driver and the car together as the system. The driver's push on the wheel and the wheel's push back on the driver are an internal third-law pair, so they add to zero. The total momentum changes only through the net external force, and the push adds nothing to it. So the centre of mass of the driver and the car cannot start to move.
A loaded truck of \(10\) tonnes moves along a highway at \(36\,\text{km/h}\). Estimate its momentum in kg m/s.
- Comparing momenta by speed alone, or by mass alone.Multiply. A slow heavy body often carries more momentum than a fast light one.
- Adding the sizes of momenta that point in different directions.Momentum is a vector. Add components, or place the arrows head to tail.
- Counting internal forces in \(\vec F_{\text{ext}} = d\vec P/dt\).Draw the system boundary first. Forces between parts inside it cancel in pairs and never enter.