momentum impulse and centre of mass

Momentum of a particle is the vector $\vec p = m\vec v$; its direction follows the velocity, and the total momentum of a system is the vector sum $\vec P = \sum_i m_i\vec v_i$ (for a rigid body of mass $M$ this gives $\vec P = M\vec v_C$).

Newton's second law in momentum form reads $\vec F = \dfrac{d\vec p}{dt}$ and reduces to $\vec F = m\vec a$ when the mass is constant.

The impulse of a force is $\vec J = \int_{t_1}^{t_2}\vec F\,dt$, numerically equal to the area under the $F$–$t$ graph; the impulse–momentum theorem gives $\vec J = \Delta\vec p = m\vec v_2 - m\vec v_1$.

The centre of mass is the mass-weighted mean position $\vec r_C = \dfrac{1}{M}\sum_i m_i \vec r_i$ (for a continuous body, $\vec r_C = \dfrac{1}{M}\int \vec r\,dm$).

Internal forces act in Newton's-third-law pairs, so they cancel in the sum, and the centre of mass obeys $M\vec a_C = \sum \vec F^{ext}$ — it moves exactly like a single particle of mass $M$ under the total external force.

Consequently, when the total external force is zero, $\vec P$ and $\vec v_C$ stay constant — a useful tool for explosions and collisions and for checking rigid-body equilibrium conditions, in which the weight acts at the centre of mass.

A cart of mass $2.0\ \mathrm{kg}$ moves in a straight line at $3.0\ \mathrm{m/s}$.

(a) Find its momentum.

(b) A constant force $F = 20\ \mathrm{N}$ acts along the motion for $0.50\ \mathrm{s}$. Find the impulse of this force and the final speed of the cart.

A ball of mass $0.15\ \mathrm{kg}$ hits a wall moving at $20\ \mathrm{m/s}$ perpendicular to the wall and bounces straight back at $15\ \mathrm{m/s}$. The contact lasts $0.020\ \mathrm{s}$. Take the direction away from the wall as positive.

Find the change of the ball's momentum, the average force from the wall, and explain how the answer follows from the area under the $F$–$t$ graph.

Three particles are given: $m_1 = 1.0\ \mathrm{kg}$ at $\vec r_1 = (0,\,0)\ \mathrm{m}$ with $\vec v_1 = (2,\,0)\ \mathrm{m/s}$; $m_2 = 2.0\ \mathrm{kg}$ at $\vec r_2 = (3,\,0)\ \mathrm{m}$ with $\vec v_2 = (-1,\,0)\ \mathrm{m/s}$; $m_3 = 3.0\ \mathrm{kg}$ at $\vec r_3 = (3,\,4)\ \mathrm{m}$ with $\vec v_3 = (1,\,1)\ \mathrm{m/s}$.

(a) Find the centre of mass of the system.

(b) Find the total momentum and the velocity of the centre of mass.

(c) The particles are pieces of a firework that explodes. Describe the subsequent motion of the centre of mass if air resistance is neglected.

Self-study with the PhET Collision Lab simulation (or an equivalent collision model). Work through the steps in order.

1. Set two pucks: $m_1 = 1.0\ \mathrm{kg}$, $m_2 = 2.0\ \mathrm{kg}$, elasticity $100\%$. Give them initial velocities, run the model, and record the velocities before and after the collision. Compute $\vec P = m_1\vec v_1 + m_2\vec v_2$ before and after — the vector must come out the same.
2. Repeat with elasticity $0\%$. Momentum is again conserved while kinetic energy is not; compare both changes quantitatively.
3. Switch on the centre-of-mass display (marker and velocity vectors). Compare the simulation marker with your own calculation $\vec r_C = \dfrac{m_1\vec r_1 + m_2\vec r_2}{m_1 + m_2}$ from the position readouts.
4. A safe check with everyday objects: tape a small coin to one end of a $30\ \mathrm{cm}$ ruler and balance the ruler on a finger to locate the centre of mass. Compare the position you find with $x_C = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}$, where $m_1$ is the ruler's mass and $m_2$ the coin's mass. Work on a table; no sharp tools and no falling weights are needed.

To make visible that internal forces cancel in pairs, so only the total external force moves the centre of mass, and that the impulse of a force is the area under its $F$–$t$ graph.

A three-panel sketch.

• Left: two blocks joined by a spring, with arrows for the equal and opposite internal forces $\vec F_{12} = -\vec F_{21}$ and the labelled centre of mass $\vec r_C$.
• Centre: an $F$–$t$ graph with the shaded area marked $\vec J = \int \vec F\,dt$ and a dashed rectangle of the same area whose height is the average force.
• Right: an exploding shell drawn at three successive instants: the fragments fly apart while the dashed parabola of the centre of mass continues unchanged.

A phone of mass $m = 0.20\ \text{kg}$ slides off a table and falls from a height $h = 1.0\ \text{m}$. Just before impact its speed is $v = \sqrt{2gh} \approx 4.4\ \text{m/s}$ and its momentum is $p = mv \approx 0.89\ \text{kg}\cdot\text{m/s}$. Hitting bare tiles, the phone stops in about $0.002\ \text{s}$; landing on a $2\ \text{cm}$ carpet it stops in about $0.04\ \text{s}$.

The impulse–momentum theorem $\vec{F}\Delta t = \Delta\vec{p}$ is what acts here: the loss of momentum is the same in both cases, $0.89\ \text{kg}\cdot\text{m/s}$. On tiles $\Delta t = 0.002\ \text{s}$, so the average force is $F \approx 0.89/0.002 \approx 450\ \text{N}$; on the carpet $\Delta t = 0.04\ \text{s}$ and $F \approx 22\ \text{N}$ — twenty times smaller, which is why the screen survives.

A driver of mass $m = 70\ \text{kg}$ travels at $v = 15\ \text{m/s}$ (that is $54\ \text{km/h}$), so his momentum is $p = mv = 1050\ \text{kg}\cdot\text{m/s}$. With belt and airbag he loses this speed in $\Delta t = 0.15\ \text{s}$; without them he is stopped by the steering column in $\Delta t = 0.02\ \text{s}$.

The change in momentum $\Delta p = 1050\ \text{kg}\cdot\text{m/s}$ is fixed by the mass and the speed and does not depend on what the driver hits. The average force is $F = \Delta p/\Delta t$: with the airbag $F \approx 7000\ \text{N}$, without it $F \approx 52500\ \text{N}$. The crumple zone and the airbag lengthen the stopping time $\Delta t$ and therefore lower the force — a direct consequence of the link between the impulse of a force and the change in momentum of a body.

The bar is set at $2.00\ \text{m}$. At the top of the jump the athlete's centre of mass is at about $1.85\ \text{m}$, below the bar, because the arched body curves around the bar while the centre of mass itself passes underneath. To lift the centre of mass from $1.0\ \text{m}$ to $1.85\ \text{m}$, that is by $0.85\ \text{m}$, the take-off speed must be $v = \sqrt{2 \cdot 9.8 \cdot 0.85} \approx 4.1\ \text{m/s}$.

The centre of mass of a system is the point whose motion is set by external forces alone: $M\vec{a}_{CM} = \sum\vec{F}_{ext}$. While the athlete is in the air the muscles act only as internal forces and cannot change the trajectory of the centre of mass — it moves along a parabola like a thrown point. The example also shows that the centre of mass can lie outside the body itself, and it is its height, not the height of any part of the body, that decides what can be cleared.