On the ice rink at Gulmarg, a girl and her older brother stand still, face to face, palms together. They push. Both glide backwards, away from each other, the girl faster than her brother. Neither the push nor its duration was measured, yet the ratio of their speeds is fixed in advance.
When nothing outside pushes
The last lesson proved that only the net external force changes the total momentum of a system: \(\vec F_{\text{ext}} = d\vec P/dt\). Now suppose the external forces add to zero. Then \(d\vec P/dt = 0\), and the total momentum cannot change at all. The parts of the system may push, pull, collide or explode. They trade momentum among themselves, but the sum stays fixed.
If the net external force on a system is zero, the total momentum of the system stays constant. Its value at any instant equals its value at any other instant, whatever the internal forces do.
A system with no net external force is called an isolated system. Few real systems are perfectly isolated. The skaters feel gravity and the push of the ice, but these two are vertical and balance. Friction on smooth ice is small. So in the horizontal direction the two skaters come very close to an isolated system.
Block A of \(3\,\text{kg}\) and block B of \(1\,\text{kg}\) rest on a smooth floor with a compressed spring between them. The thread holding them is cut. After the spring has pushed them apart, which block has the larger momentum?
Play the figure above on each chip. Before the thread is cut the total momentum is zero, and so it stays zero. The spring pushes A and B with equal and opposite forces for the same time. So A and B receive equal and opposite momenta, \(6\,\text{kg m/s}\) each at the end. Block A always leaves at \(2\,\text{m/s}\), and B at \(2\), \(4\) or \(6\,\text{m/s}\) as its mass falls.
\[ 0 = m_A v_A + m_B v_B \quad\Rightarrow\quad \frac{v_B}{v_A} = -\frac{m_A}{m_B} \]
The marker C in the figure is the centre of mass. The total momentum is \(M\vec v_{\text{cm}}\), and it stays zero, so C never moves. The blocks fly apart around a point that stays exactly where it was. Seen from C, the two momenta are always equal and opposite. That is true for any two bodies, pushed apart or not.
Equal momenta do not mean equal kinetic energies. With \(K = p^2/2m\), the same \(p\) gives more energy to the lighter body. With B of \(1\,\text{kg}\) in the figure, A gets \(6\,\text{J}\) and B gets \(18\,\text{J}\). The spring's stored energy is shared in the inverse ratio of the masses.
The girl on the rink has a mass of \(40\,\text{kg}\) and her brother \(60\,\text{kg}\). After the push the girl glides at \(1.5\,\text{m/s}\). Ignore friction. Find the brother's velocity.
The two start at rest, so the total momentum is zero before and after the push. Take the girl's direction as positive.
\(40 \times 1.5 + 60\,v = 0\), so \(v = -60/60 = -1.0\,\text{m/s}\). The brother glides at \(1.0\,\text{m/s}\) in the opposite direction.
The ratio of the speeds, \(1.5 : 1.0\), is the inverse of the ratio of the masses, \(40 : 60\). The strength of the push fixes the speeds, but not their ratio.
Two skaters of \(50\,\text{kg}\) and \(60\,\text{kg}\) stand at rest on smooth ice and push each other apart. The \(50\,\text{kg}\) skater moves off at \(2.4\,\text{m/s}\). Find the speed of the other skater, in m/s.
In 1644 René Descartes wrote that God keeps the total quantity of motion in the world constant. He measured it as the size of a body times its speed, with no regard to direction. His rules for collisions, built on that idea, failed against experiment. In 1668 and 1669 John Wallis and Christiaan Huygens studied collisions for the Royal Society in London. They showed that the quantity that stays fixed must count direction, so that motions in opposite directions subtract. That quantity is the momentum of this chapter, a vector.
One component at a time
The law \(\vec F_{\text{ext}} = d\vec P/dt\) is a vector equation, so it holds for each component separately. Along \(x\), \(F_{\text{ext},x} = dP_x/dt\). Suppose every external force is vertical, like gravity and the push of a smooth floor. Then \(F_{\text{ext},x} = 0\), and \(P_x\) is conserved, even while \(P_y\) changes. This one idea solves a large family of problems.
The floor in Fig 2.1 is smooth, so it can push only vertically. The horizontal momentum is zero before the throw and zero after it. The ball takes \(16\,\text{kg m/s}\) forward, so the boy and the trolley take \(16\,\text{kg m/s}\) backward. They recoil at \(16/80 = 0.2\,\text{m/s}\). The vertical \(12\,\text{kg m/s}\) is not balanced by anything moving down. The floor supplies it, by pushing harder on the wheels while the boy throws.
In Fig 2.1, suppose the boy throws the ball straight up instead. What happens to the trolley?
A \(60\,\text{kg}\) boy stands on a \(40\,\text{kg}\) trolley at rest on a smooth floor. He throws a \(5\,\text{kg}\) bag at \(8\,\text{m/s}\), at \(60°\) above the horizontal. Find the recoil speed of the boy and the trolley, in m/s.
Conserved for a brief instant
A shell that bursts in mid-air is not an isolated system, because gravity acts on it. Yet momentum is conserved across the burst. The reason is time. The burst lasts a few thousandths of a second. The impulse of gravity in that time is \(Mg\,\Delta t\), which is tiny. The internal blast forces are enormous, and they cancel in pairs.
For a \(10\,\text{kg}\) shell moving at \(20\,\text{m/s}\), the momentum is \(200\,\text{kg m/s}\). If the burst takes \(2\,\text{ms}\), gravity delivers \(10 \times 10 \times 0.002 = 0.2\,\text{N s}\). That is one part in a thousand. The same reasoning covers two cars that collide on a rough road. The contact forces between the cars are huge. The road's friction acts for only the few milliseconds of the crash, so its impulse can be ignored.
Suppose internal forces act for a very short time and are much larger than the external forces. Then the total momentum just after the event equals the total momentum just before it. Ordinary external forces, such as gravity and friction, are ignored for that short interval.
Think of the external force as a slow leak and the internal forces as a sudden shuffle. Over the whole flight of a shell, gravity changes its momentum a great deal. Over the instant of the burst, the leak has no time to act. So the momentum just before and just after are equal, and gravity acts alone again for the rest of the flight.
For which of these systems can the total momentum be taken as conserved across the event described?
A \(6\,\text{kg}\) shell moving horizontally at \(20\,\text{m/s}\) bursts at the top of its flight into two pieces of \(2\,\text{kg}\) and \(4\,\text{kg}\). The \(2\,\text{kg}\) piece drops from rest just after the burst. Find the speed of the \(4\,\text{kg}\) piece just after the burst, in m/s.
Now return to the rink at Gulmarg. The ice pushes up and gravity pulls down, and these balance. Along the ice nothing pushes, so the horizontal momentum stays zero. The girl and her brother must take equal and opposite momenta. Their speeds are in the inverse ratio of their masses, \(60 : 40\). Once the girl's speed is known, her brother's follows, however hard they pushed.
- Saying momentum is conserved because "no force acts".Forces do act. Momentum is conserved when the external forces add to zero along the direction chosen.
- Expecting equal kinetic energies when the momenta are equal.\(K = p^2/2m\). With equal momenta the lighter body gets more kinetic energy.
- Conserving vertical momentum when the floor or the ground pushes.The floor is external. Conserve only the component along which no external force acts.