Momentum and collisions

Energy conservation is one scalar equation and a collision has more unknowns than that, which is why the subject needs a conserved quantity that carries direction.

Two results are carried in. Newton's second law in its original form says $\sum \vec{F} = d \vec{p} / d t$ with $\vec{p} = m \vec{v}$ , and the third law says that when two bodies interact the forces on them are equal and opposite at every instant.

Impulse: integrating over time

The previous two lessons integrated the second law over distance. Integrating it over time instead gives something simpler and just as useful. From $\vec{F} = d \vec{p} / d t$ ,

\vec{J} = \int_{t_{1}}^{t_{2}} \vec{F} d t = {\vec{p}}_{2} - {\vec{p}}_{1} = Δ \vec{p}

The integral is the impulse, measured in newton seconds, which are the same as kg m s⁻¹. The statement is that the impulse delivered equals the change in momentum, exactly, whatever shape the force has in between.

That last clause is the point. A collision involves a force that rises from nothing to an enormous peak and falls back within milliseconds, and nobody knows its detailed shape. The impulse does not need it: only the area under the curve matters, and the area is fixed by the momentum change, which is measurable at leisure before and after. Where a shape is wanted, the useful summary is the average force, $\overline{F} = Δ p / Δ t$ .

The engineering consequence is that a required momentum change can be bought either with a large force over a short time or a small force over a long one, and safety design is entirely the second option. A 70 kg driver at 15 m s⁻¹ has 1050 kg m s⁻¹ of momentum to lose. Stopped by a rigid steering column in 0.10 s, the average force is 10.5 kN, roughly fifteen times body weight and enough to break a sternum. Stopped by an airbag and a crumple zone in 0.50 s, it is 2.1 kN. Nothing about the momentum changed; the collision was merely made longer.

Example. A 0.145 kg baseball arrives at 40 m s⁻¹ and leaves at 50 m s⁻¹ in the opposite direction. What impulse did the bat deliver, and what average force acted if the contact lasted 0.70 ms?

Take the outgoing direction as positive, so the incoming momentum is $- (0.145) (40) = - 5.80$ kg m s⁻¹ and the outgoing is $+ (0.145) (50) = + 7.25$ . The impulse is the difference, $+ 13.05$ N s, and the sign convention has done real work: the speeds add rather than subtract because the direction reversed. The average force is $13.05 / 0.00070 = 18.6$ kN, about two and a half tonnes weight, applied for less than a thousandth of a second.

Now you. A 0.058 kg tennis ball arrives at 30 m s⁻¹ and is returned at 45 m s⁻¹ the other way, with 5.0 ms of contact. Find the impulse and the average force.

Answer

$J = 0.058 (45 + 30) = 4.35$ N s, and $\overline{F} = 4.35 / 0.0050 = 870$ N.

Conservation, and why it is the third law in disguise

Take two bodies that interact only with each other. By the third law, ${\vec{F}}_{12} = - {\vec{F}}_{21}$ at every instant, so

\frac{d {\vec{p}}_{1}}{d t} + \frac{d {\vec{p}}_{2}}{d t} = {\vec{F}}_{21} + {\vec{F}}_{12} = 0

and the total momentum ${\vec{p}}_{1} + {\vec{p}}_{2}$ has zero derivative. The total momentum of a system is constant whenever the net external force on it is zero. Internal forces, however violent, cancel in pairs and cannot shift the total.

Three features make this more useful than energy conservation in a collision. It is a vector statement, so it is really one conservation law per axis, and each can be applied separately. It holds regardless of what the internal forces are, so nothing need be known about how the bodies deform. And it holds even when mechanical energy does not, because the third law says nothing about energy.

Two qualifications keep it honest. External forces do have to be absent, or at least negligible: gravity acts throughout a collision, but over a millisecond it changes the momentum by an amount too small to matter against the collision forces, which is why momentum is treated as conserved during an impact and not afterwards. And momentum is conserved along an axis only if the external force along that axis vanishes: two cars colliding on a road conserve horizontal momentum while the ground supplies whatever vertical impulse it likes.

Collisions, and the two extreme kinds

A collision is any brief interaction in which the internal forces are much larger than the external ones. Momentum is conserved in all of them; kinetic energy is not.

A perfectly inelastic collision is one in which the bodies move off together, and it is the case that loses the most energy consistent with conserving momentum. With $m_{1}$ at $u_{1}$ striking a stationary $m_{2}$ :

v = \frac{m_{1} u_{1}}{m_{1} + m_{2}}

Example. A 1200 kg car at 20 m s⁻¹ runs into a stationary 900 kg car and the two lock together. Find the common speed and the fraction of kinetic energy lost.

$v = (1200) (20) / 2100 = 11.43$ m s⁻¹. The kinetic energy before is $\frac{1}{2} (1200) (400) = 240$ kJ and after is $\frac{1}{2} (2100) ({11.43}^{2}) = 137$ kJ, so 103 kJ, or 42.9 per cent, has gone into crushing metal, heat and noise. That lost energy is not a defect of the calculation: it is the design intent, since a car that bounced off elastically would deliver a far larger impulse to its occupants.

Now you. A 0.40 kg lump of putty at 6.0 m s⁻¹ hits a stationary 1.6 kg block and sticks. Find the common speed and the energy lost.

Answer

$v = (0.40) (6.0) / 2.0 = 1.20$ m s⁻¹. Before, $K = \frac{1}{2} (0.40) (36) = 7.20$ J; after, $K = \frac{1}{2} (2.0) (1.44) = 1.44$ J. The loss is 5.76 J, which is 80 per cent.

An elastic collision is one in which kinetic energy is also conserved. No macroscopic collision is exactly elastic, but hard steel balls, billiard balls and gas molecules come close, and collisions between subatomic particles can be exactly so.

Solving the elastic collision

Take $m_{1}$ moving at $u_{1}$ into $m_{2}$ at rest, in one dimension. Two conservation laws give two equations:

m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2} \frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}

Solving these directly involves a quadratic and a wrong root. The elegant route is to rearrange each as a difference: $m_{1} (u_{1} - v_{1}) = m_{2} v_{2}$ from the first, and $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} v_{2}^{2}$ from the second. Dividing the second by the first and using the difference of two squares gives $u_{1} + v_{1} = v_{2}$ , which rearranges to

u_{1} - u_{2} = v_{2} - v_{1}

once a general $u_{2}$ is carried through. In an elastic collision the relative speed of separation equals the relative speed of approach. That is one linear equation replacing the quadratic, and with the momentum equation it solves in two lines:

v_{1} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}} u_{1} v_{2} = \frac{2 m_{1}}{m_{1} + m_{2}} u_{1}

The limits are worth reading off. Equal masses give $v_{1} = 0$ and $v_{2} = u_{1}$ : the incoming body stops dead and the target leaves with the whole velocity, which is the shot every snooker player learns first. A light body hitting a much heavier one gives $v_{1} \to - u_{1}$ and $v_{2} \to 0$ : it bounces back at nearly its original speed, which is why a ball rebounds from a wall. A heavy body hitting a light one gives $v_{1} \to u_{1}$ and $v_{2} \to 2 u_{1}$ : the projectile is barely slowed and the target leaves at twice the incoming speed, which is how a golf club moving at 50 m s⁻¹ sends a ball off at nearly 70.

That middle case is also why nuclear reactors are moderated with light nuclei. A neutron loses the largest fraction of its energy to a target of its own mass, which is hydrogen, so water and graphite slow neutrons in a few dozen collisions while lead would take thousands.

Example. A 0.50 kg ball moving at 4.0 m s⁻¹ collides elastically with a stationary 0.30 kg ball. Find both final velocities and check the energy.

$v_{1} = (0.50 - 0.30) / (0.80) \times 4.0 = 1.00$ m s⁻¹ and $v_{2} = 2 (0.50) / (0.80) \times 4.0 = 5.00$ m s⁻¹, both forward. Momentum: before $2.00$ , after $(0.50) (1.00) + (0.30) (5.00) = 2.00$ kg m s⁻¹. Energy: before $\frac{1}{2} (0.50) (16) = 4.00$ J, after $0.25 + 3.75 = 4.00$ J. And the separation speed, $5.00 - 1.00 = 4.00$ m s⁻¹, equals the approach speed, as it must.

Now you. A 2.0 kg body at 3.0 m s⁻¹ collides elastically with a stationary 6.0 kg body. Find both final velocities.

Answer

$v_{1} = (2.0 - 6.0) / 8.0 \times 3.0 = - 1.50$ m s⁻¹, so the light body rebounds, and $v_{2} = 2 (2.0) / 8.0 \times 3.0 = + 1.50$ m s⁻¹. Momentum checks: $- 3.0 + 9.0 = 6.0$ kg m s⁻¹, as before. Energy checks: $2.25 + 6.75 = 9.00$ J.

Two dimensions, and the missing equation

In a plane, momentum conservation supplies two equations, one per axis, and the unknowns are two outgoing speeds and two outgoing directions: four unknowns, two equations. Even adding energy conservation for an elastic collision leaves one short. The missing information is physical, not mathematical: it is the impact parameter, how squarely the bodies hit, and no conservation law can supply it. In practice one outgoing direction is measured and the rest follows.

One result does survive without that measurement. For an elastic collision between equal masses with one initially at rest, the two bodies always separate at 90° to each other. The proof is short: momentum gives $\vec{u} = {\vec{v}}_{1} + {\vec{v}}_{2}$ , and squaring gives $u^{2} = v_{1}^{2} + v_{2}^{2} + 2 {\vec{v}}_{1} \cdot {\vec{v}}_{2}$ , while energy for equal masses gives $u^{2} = v_{1}^{2} + v_{2}^{2}$ . So ${\vec{v}}_{1} \cdot {\vec{v}}_{2} = 0$ , which for two non-zero velocities means perpendicular. Snooker players rely on it constantly, and it visibly fails when the balls have spin or the collision is not quite elastic.

The ballistic pendulum

The instrument that made this lesson practical was Benjamin Robins' ballistic pendulum of 1742, the first device that could measure the speed of a musket ball. A bullet is fired into a heavy block hanging on a cord; the block swings up by a measured height; the bullet's speed is inferred.

The critical point, and the reason the device is a teaching classic, is that the problem has two stages governed by different laws, and using the wrong one in either stage gives an answer wrong by a factor of hundreds.

Stage one is the embedding, which takes about a millisecond. It is perfectly inelastic, so momentum is conserved and kinetic energy is emphatically not: $m u = (m + M) v$ .

Stage two is the swing, which takes about a second. Now the only forces doing work are gravity and the cord tension, both conservative or workless, so mechanical energy is conserved: $\frac{1}{2} (m + M) v^{2} = (m + M) g h$ , giving $v = \sqrt{2 g h}$ .

Combining, $u = \frac{m + M}{m} \sqrt{2 g h}$ .

Example. A 10 g bullet is fired into a 2.00 kg block and the block rises 0.202 m. What was the bullet's speed, and what fraction of the kinetic energy survived the impact?

From the swing, $v = \sqrt{2 (9.81) (0.202)} = 1.99$ m s⁻¹. From the embedding, $u = (2.010 / 0.010) (1.99) = 400$ m s⁻¹, a plausible musket velocity. The energy before is $\frac{1}{2} (0.010) (400^{2}) = 800$ J and after is $\frac{1}{2} (2.010) ({1.99}^{2}) = 3.98$ J, so 99.5 per cent was lost in the first millisecond. Anyone who had used energy conservation across the impact would have obtained a bullet speed of about 28 m s⁻¹, low by a factor of fourteen.

Now you. An 8.0 g bullet is fired into a 1.50 kg block, which rises 0.176 m. What was the bullet's speed?

Answer

$v = \sqrt{2 (9.81) (0.176)} = 1.86$ m s⁻¹, and $u = (1.508 / 0.008) (1.86) = 350$ m s⁻¹.

What is still assumed

Every body in this lesson has been treated as a point. A car has been a point, a block has been a point, and the collision has been an event with no spatial extent. That was never justified, and it is plainly false: real bodies spin, deform, and have their mass spread over metres.

The justification exists and is worth having, because it is what licenses eight lessons of point particles retroactively. For any system of particles, however complicated, there is one special point that moves exactly as a single particle would under the external forces alone. Finding it, proving that claim and then using it on a body that throws its own mass away is the next lesson.