Equilibrium momentum and impulse

A body is in equilibrium when the vector sum of all forces acting on it is zero: $\sum \vec F = 0$. From Newton's second law $\vec F_{\text{net}} = m\vec a$ it follows at once that in equilibrium the acceleration is zero and the velocity is constant: the body either stays at rest or moves uniformly in a straight line. The momentum of a body $\vec p = m\vec v$ is the measure of quantity of motion, and in equilibrium it also does not change.

The change of momentum is set by the impulse of a force $\vec J = \vec F \Delta t$: the longer a force acts, the larger the momentum change. That is why, for the same change of velocity, a soft pad acts for a longer time and produces a smaller force than a rigid wall does. Newton's second law in impulse form is written as $\vec F_{\text{net}} = \Delta \vec p / \Delta t$.

For a system of bodies, the external forces deliver the total impulse to the system, while internal forces only redistribute momentum among the bodies. If the external forces balance, the total momentum of the system is conserved, even though the momentum of each individual body may change.

A cart of mass $m = 0.50\ \text{kg}$ moves along a horizontal track with velocity $v_0 = 3.0\ \text{m/s}$. A constant force $F = 2.0\ \text{N}$ directed opposite to the motion acts on it for $\Delta t = 0.40\ \text{s}$. Find the impulse of the force and the final velocity of the cart.

A ball of mass $m = 0.16\ \text{kg}$ approaches a wall at $12\ \text{m/s}$ and rebounds at $10\ \text{m/s}$. The contact time is $\Delta t = 0.020\ \text{s}$. Find the change of the ball's momentum and the average force exerted on it by the wall.

Two carts on a frictionless horizontal track ($m_1 = 1.0\ \text{kg}$, $m_2 = 2.0\ \text{kg}$) are at rest, and a compressed spring between them expands. Afterwards cart 1 moves to the left at $3.0\ \text{m/s}$. Find the velocity of cart 2, the impulse received by each cart, and explain why the total momentum of the system did not change even though the momentum of each cart did.

1. Open the PhET simulation Collision Lab (Explore 1D mode). Set elasticity to 100 % and turn friction off.
2. Set two carts, for example $m_1 = 1.0\ \text{kg}$ and $m_2 = 2.0\ \text{kg}$, and give them velocities. Before the collision record $p_1$, $p_2$, and the sum $p_1 + p_2$.
3. After the collision measure the momenta again and compare the sum with the initial one. State a conclusion about conservation of momentum when the external forces balance.
4. Turn friction on as an external force and repeat the experiment: show that now the total momentum changes, and that its change equals the impulse of the external force.
5. Switch to Forces and Motion: Basics, apply two forces to the crate, and achieve the condition $\sum \vec F = 0$; check that the velocity then stays constant, which means the body is in equilibrium.

All observations are made in the simulation: real collisions and throwing of solid bodies are unnecessary and potentially hazardous.

Show that with balanced external forces the momentum of a body or system does not change, while an unbalanced force changes the momentum by $\vec J = \vec F \Delta t$ over a time $\Delta t$.

A two-panel diagram. Left: a crate on a table with force arrows whose vector sum is zero, next to it an unchanging vector $\vec p$; the caption reads $\sum \vec F = 0 \Rightarrow \vec p = \text{const}$. Right: a graph of force versus time $F(t)$ with a constant nonzero value and a shaded area under the graph equal to the impulse $\vec J$; next to it an arrow for the momentum change $\Delta \vec p$.

A $0.15\ \mathrm{kg}$ baseball flies at $30\ \mathrm{m/s}$. A player catches it and the glove moves backward $0.10\ \mathrm{m}$ while the ball stops. The stopping time is about $0.0067\ \mathrm{s}$; the ball's momentum changes from $p_i=mv=0.15\cdot30=4.5\ \mathrm{kg\cdot m/s}$ to $p_f=0$.

This uses the impulse-momentum theorem: $\vec J=\int \vec F\,dt=\Delta\vec p$; the same change $\Delta p=-4.5\ \mathrm{kg\cdot m/s}$ gives an average force $F\approx |\Delta p|/\Delta t\approx 6.7\times10^2\ \mathrm{N}$. Pulling the glove back increases $\Delta t$, so the force on the hand is smaller.

A $70\ \mathrm{kg}$ passenger rides at $15\ \mathrm{m/s}$ ($54\ \mathrm{km/h}$). In a crash the passenger must lose momentum $p=70\cdot15=1050\ \mathrm{kg\cdot m/s}$. Without a seatbelt the stop takes $0.050\ \mathrm{s}$; with an airbag and crumple zone it takes $0.25\ \mathrm{s}$.

The impulse-momentum theorem $F_{\mathrm{avg}}=\Delta p/\Delta t$ shows the same $\Delta p=1050\ \mathrm{kg\cdot m/s}$ gives $F_{\mathrm{avg}}\approx2.1\times10^4\ \mathrm{N}$ in $0.050\ \mathrm{s}$ but only $4.2\times10^3\ \mathrm{N}$ in $0.25\ \mathrm{s}$. This is why restraints extend the time of the impulse.

A $0.20\ \mathrm{kg}$ phone falls from a $0.80\ \mathrm{m}$ table. At impact its speed is $v=\sqrt{2gh}=\sqrt{2\cdot9.81\cdot0.80}\approx3.96\ \mathrm{m/s}$, so its downward momentum is $p=mv\approx0.79\ \mathrm{kg\cdot m/s}$. On a carpet it stops in $0.050\ \mathrm{s}$; on tile it stops in $0.0020\ \mathrm{s}$.

From $\vec J=\Delta\vec p$, the average resultant force is $F_{\mathrm{avg}}\approx|\Delta p|/\Delta t$: about $16\ \mathrm{N}$ on carpet and about $4.0\times10^2\ \mathrm{N}$ on tile. After stopping, $\sum\vec F=0$, so the normal force equals the phone's weight, $mg\approx1.96\ \mathrm{N}$.