Centre of Mass, Momentum & Collisions

A spin bowler releases a ball that twists and dips, every point on its surface tracing a tangled path. Yet its centre glides along a perfect parabola. That centre is special.

One point that behaves#

Throw a spinning ball straight up. A mark on its seam corkscrews through space, but the centre of the ball goes straight up and down. Lob it to a fielder and the centre traces a parabola, exactly like a thrown pebble, however much the ball spins. (Air adds small deviations that bowlers exploit; we ignore them here.) The centre of a ball is an example of a centre of mass: the point that moves as if it were a single particle carrying all the mass.

Definition#

Take N particles; particle i has mass $m_i$ and position $\vec r_i = (x_i, y_i, z_i)$. Let $M = \sum m_i$ be the total mass.

Two particles#

Put $m_1$ at the origin and $m_2$ a distance d away on the x-axis. Then $X = m_2 d/(m_1 + m_2)$: the centre of mass lies on the line joining them, at distances

Combining parts#

Split a system into two groups with masses $M_1, M_2$ and centres of mass at $X_1, X_2$. The sum $\sum m_ix_i$ over the first group is just $M_1X_1$, and likewise for the second, so

Replace each part by a point mass at its own centre of mass, then combine. This trick saves enormous effort.

A real object isn’t a handful of dots. Treat it as continuous, slice it into infinitely many tiny pieces, and the sum becomes an integral.

From sums to integrals#

For a body with continuously spread matter, replace “particle i” with a tiny element of mass $dm$ at position (x, y, z):

The art is choosing the element: as the integration variable runs between its limits, the elements must cover the whole body exactly once, and every part of one element should have (almost) the same coordinate you are integrating.

(a) A uniform rod

Rod of mass M and length L along the x-axis from 0 to L. Mass per unit length is M/L, so an element between x and x + dx has $dm = (M/L)\,dx$:

The midpoint, as symmetry predicts.

(b) A uniform semicircular wire

Radius R, centre at the origin, ends on the x-axis. Choose the element between angles θ and θ + dθ: length $R\,d\theta$, mass $dm = \dfrac{M}{\pi R}R\,d\theta = \dfrac M\pi d\theta$, coordinates $(R\cos\theta, R\sin\theta)$. As θ goes from 0 to π the elements cover the wire.

(c) A uniform semicircular plate

No new integral over angles is needed. Slice the plate into thin semicircular strips of radius r and width dr. Each strip is a semicircular wire, whose centre of mass we just found: at height $2r/\pi$. Its mass is its area $\pi r\,dr$ times the mass per area $M/(\pi R^2/2)$, i.e. $dm = \dfrac{2M}{R^2}r\,dr$. Replace each strip by a point mass at its own centre of mass and add:

You cannot lift yourself by your own bootstraps. Physics puts it more precisely: internal forces can never move the centre of mass.

Two particles pulling on each other#

Two particles on a line attract each other and nothing else acts on them. Their centre of mass is at $X = (m_1x_1 + m_2x_2)/(m_1 + m_2)$, so its velocity and acceleration are mass-weighted averages too:

With mutual force F, $a_1 = F/m_1$ and $a_2 = -F/m_2$, so the numerator is $F - F = 0$. The centre of mass does not accelerate. If the particles started at rest, the centre of mass stays put while they rush together.

Any system of particles#

For particle i, Newton’s second law reads $m_i\vec a_i = \sum_{j\ne i}\vec F_{ij} + \vec F_i^{\,ext}$. Add these up over all particles. The internal forces come in third-law pairs ($\vec F_{ij} + \vec F_{ji} = 0$) and cancel, leaving only the external forces. Since $\sum m_i\vec a_i = M\vec a_{cm}$:

The exploding shell#

A projectile’s only external force is gravity, so its centre of mass follows the parabola $x = u_xt$, $y = u_yt - \tfrac12gt^2$ (9.10). Now let it burst in mid-air. The pieces fly off on different parabolas, but the burst is internal: the centre of mass of the fragments carries on along the original curve, exactly as if nothing had happened, until one piece hits the ground and an external force joins in.

Fire a rifle and it kicks your shoulder. Throw a ball off a skateboard and you roll backwards. Both are one law at work: the total momentum of an isolated system never changes.

Momentum#

Because $\sum m_i\vec r_i = M\vec R_{cm}$, differentiating gives a neat result: a system’s total momentum is its mass times the velocity of its centre of mass.

Conservation of linear momentum#

From the last lesson, if the external forces add up to zero then $\vec a_{cm} = 0$, so $\vec V_{cm}$ and hence $\vec P$ are constant:

Note what is not required: kinetic energy may change wildly (explosions create it, crashes destroy it), and the internal forces may be of any kind (springs, friction, chemical, nuclear). As long as the outside world exerts no net force, momentum is fixed.

In space there is nothing to push against. A rocket pushes against its own exhaust: momentum conservation in its purest form.

Setting up#

Fuel burns and hot gas is expelled backwards through a nozzle. Let the rocket plus remaining fuel have mass $M_0$ at t = 0; gas leaves at a steady rate $r = -dM/dt$, at speed $u$ relative to the rocket. At time t the mass is $M = M_0 - rt$ and the speed is v. Ignore gravity and air.

A momentum balance#

In a short time Δt, a mass $\Delta M = r\Delta t$ of gas leaves. Relative to the ground it moves at $v - u$, while the rest of the rocket speeds up to $v + \Delta v$. No external force acts, so

Divide by Δt and let it shrink:

Integrate from rest at t = 0:

Two bodies meet, push hard on each other for a moment, and separate (or stick). We rarely know the forces during that moment, but momentum lets us connect “before” and “after” anyway.

What happens inside a collision#

Picture a block sliding toward an identical block at rest, with a light spring bumper on the front of the second one. When the spring touches, it compresses: it slows the first block and speeds up the second. While block 1 is still faster, the gap keeps closing and the spring squeezes harder. Compression is greatest at the instant both blocks move with the same velocity. After that the spring pushes them apart until it regains its natural length.

Throughout, the spring forces are internal, so momentum is conserved at every instant. Kinetic energy is not: at maximum compression some of it is stored in the spring. Whether all of it comes back decides the type of collision.

Elastic and inelastic#

For any perfectly inelastic collision on a line, a little algebra gives the kinetic energy lost:

Coefficient of restitution#

Real collisions fall between the two extremes. Measure where with the ratio of the speed of separation after the collision to the speed of approach before it:

Combine it with momentum conservation and the final velocities follow for any masses and any e:

When no kinetic energy is lost we get two conservation laws for two unknowns, and the outcome of a head-on collision is fixed completely. Billiard balls, Newton’s cradle and gas molecules all obey the same two formulas.

Head-on elastic collision#

Masses $m_1, m_2$ move along one line with velocities $u_1, u_2$ before and $v_1, v_2$ after. Write both conservation laws with each mass’s terms collected on one side:

Divide the second by the first (the factors of difference-of-squares cancel nicely):

This is linear, so combining it with momentum conservation is easy:

Three cases worth knowing

• Equal masses: $v_1 = u_2, \ v_2 = u_1$. The bodies simply swap velocities. A moving ball hitting an identical one at rest stops dead and the other carries on (Newton’s cradle).
• Very heavy target at rest ($m_2 \gg m_1$): $v_1 \approx -u_1, \ v_2 \approx 0$. A ball bounces back off a wall with its speed unchanged.
• Very heavy projectile ($m_1 \gg m_2$, target at rest): $v_1 \approx u_1, \ v_2 \approx 2u_1$. A bowling ball barely notices a ping-pong ball, which flies off at twice its speed.

Collisions in two dimensions#

If the balls don’t meet head-on, the push during contact acts along the line joining their centres. Momentum is conserved as a vector, so we get one equation for each component, plus energy if the collision is elastic. That is three equations for four unknowns (two speeds and two angles), so one more piece of information, such as one outgoing angle, is needed.

A cricketer catching a fast ball pulls the hands back. A high jumper lands on a thick mat. A car crumples at the front. All three are stretching out time to shrink a force.

Force over time#

During a collision the force rises sharply, peaks, and drops back to zero, all within milliseconds. We rarely know its shape. But Newton’s second law in the form $\vec F = d\vec p/dt$ tells us what its total effect must be. Integrate over the collision:

On a force–time graph the impulse is the area under the curve. A large force for a short time and a small force for a long time can have exactly the same area, and so produce exactly the same change in momentum.

Impulsive forces#

A force that is very large but acts for a very short time, such as a bat on a ball or a hammer on a nail, is called an impulsive force. During that brief moment, ordinary forces like gravity or friction deliver negligible impulse (a modest force × a tiny time). That is why we can apply momentum conservation across a collision even when gravity or friction acts on the bodies.