(*Strictly, we should note that, at very high speeds, a relativistic factor γ must be included: p = γmv. See Einstein Light for a brief introduction to relativistic mechanics.)
Remember that, while momentum is proportional to the speed, kinetic energy (= ½mv2) is proportional to the square of the speed. So velocity is (proportionally) more important in kinetic energy, and mass is (proportionally) more important in momentum. A 100 kg man running at 10 m.s−1 has a kinetic energy of 5,000 J. A (large) bullet with mass 40 g and speed 500 m.s−1 also has a kinetic energy of 5,000 J. However, the man's momentum has a magnitude 1,000 kgm.s−1 while the magnitude of bullet's momentum is only 20 kgm.s−1.
Let's go to the source: Newton didn't use kinetic energy, but he did use momentum, which he called 'quantity of motion'. He wrote 'Quantitas motus est mensura ejusdem orta ex velocitate et quantite materię conjunctim.' or 'The quantity of motion is the measure of the same, arising from the velocity and the quantity of matter conjunctly.'
He then used that to write the most general version of the 1st and 2nd laws: 'Mutationem motus proportionalem esse vi motrici impressę, & fieri secundum lineaum rectam qua vis illa imprimitur.' or 'The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.' Let's now do that in algebra and modern notation:
F = dp/dt Newton's laws of motion
Substituting p = mv and differentiating (revise calculus) gives:
Two important conclusions follow from F = dp/dt. The first is the law of conservation of momentum:
The conditional clause is extremely important. If you are sitting in a chair, your momentum is probably very close to zero. When you get up and walk away, your momentum is not zero. Momentum, in general, is not conserved. When you start to walk, you push against the Earth and it pushes you in the opposite direction. So there is an external force acting on you. Momentum is only conserved if the total external force is zero.
If the system you consider is large enough, then any forces will be internal, not external. This is shown in the following cartoon, which is of course not to scale. A car accelerates from rest. The momentum of the car is not conserved. However, when the road pushes the car forwards, the car wheels also push the road Earth in the opposite direction. So the car gains momentum to the right, while the Earth gains a momentum to the left that is equal in magnitude. So, due to this interaction*, the momentum of the car-Earth system is not changed. Of course the Earth has such a large mass that we do not notice the (extra) acceleration of the Earth due to the force exerted by the car. (The Earth's mass is roughly 1022 times greater than that of a car, so the change in velocity of the Earth is smaller by that factor.) Note that the car and the road each exert a torque about the centre of the earth, so that both the Earth and the car change their angular momentum about that point. More about torques in the section on rotation. (While the car is accelerating with respect to the Earth, other external forces such as the gravitational forces exerted by the sun and the galaxy both act on the Earth and the car.)
Remembering that we have used F for the total external force, let's write our equation as Fexternal = dp/dt. The second conclusion we draw from it is Newton's third law. If we apply this equation to every particle in a system, and add all the equations, we obtain Ftotal = dp/dt. The total force is the sum of the internal and external forces, so it follows that the total internal force in any system is zero. This is perhaps the most elegant statement of Newton's third law, and can be revised here or in that section of Physclips. Thus we have seen that
Let's look at collisions involving cars because we can see from the deformations that rather large forces are involved. From examination of the film clip below, we estimate that this collision produces forces between car and barrier of a few hundred kN. The weight of a car (urban assault vehicles excluded) is usually around 10 kN.
Suppose two cars, each with mass m of one tonne, each travelling at 60 kph, collide head-on and remain in contact after the collision. From symmetry, both are stationary afterwards. Suppose that the crumple zone of each car is shortened by x = 0.5 m, which is the distance Δs that each travels during the collision. Let the magnitude of the average acceleration during the collision be a. The magnitude of the average force exerted by each car on the other during this collision is ma. From kinematics, we can write a = Δ(v2)/(2Δs). Substitution gives an average force of 300 kN.
Compare these large internal forces with the external forces: The weight of each car is 10 kN. The frictional force acting on each car (assuming braking hard with good road conditions) is about the same size, as indicated by the arrows in our animation (for friction, see Weight and Contact Forces).
It is often possible to neglect the effect of external forces during a collision. When you make an approximation such as this, you should always state it explicitly. An appropriate statement might be: "During the collision, external forces are negligible so the momentum of the system is conserved".
Notice the importance of x in the calculation above: a large crumple zone gives a smaller force on the car. What about the forces on its occupants? If you are not wearing a seatbelt then, when the car starts decelerating during the collision, you continue travelling at the initial speed until some part of you, such as your head, has its own collision with some other object, such as the window, steering wheel etc. In this case, the x for your head may depend on how much your head deforms. If you are wearing a seatbelt, then the distance over which your head decelerates will be usually rather more than the distance over which the car decelerates. A larger value of x gives a smaller deceleration and so smaller forces on your head.
People outside the car often fare worse in collisions: their mass is smaller than that of a car, so they often accelerate more, and their crumple zone is usually small. Most modern cars have a low, sloping and relatively flexible bonnet (or hood) at the front. In a collision, a pedestrian or cyclist recoils from this surface and, if speeds are not too great, survives. In some of the richer suburbs of big cities in Australia and elsewhere, there is a fashion for putting large metal bars on the front of the tall, heavy private vehicles used for commuting and delivering children to school. These reduce the value of x in a collision, and thus lead to larger forces on the occupants. However, their effect on pedestrians and cyclists is much greater. The bars on the front deform much less than the relatively flexible panel in the bonnet or hood. Further, the bars do not usually deflect the victim upwards. Indeed, there is often a bar at the level of a child's head. It's sad that these unnecessary and dangerous accessories have become fashionable.
The following example is included to remind us that momentum conservation can apply in only one or two dimensions, and therefore to only some vector components.
Here, the external forces acting on the hammer-skateboard system are gravity, normal force and friction. During the collision, the force between them is much greater than their weight, so weight may be neglected. Notice that skateboard has hardly any vertical acceleration, so the total vertical force on it is close to zero. However, during the collision, there are obviously large vertical forces between skateboard and hammer because the hammer has a large vertical acceleration. So the external normal force acting during this collision cannot be neglected. So momentum is not conserved in the vertical direction. However, this doesn't necessarily prohibit momentum conservation in the horizontal direction. If the mass of the wheels of the skateboard is negligible, then momentum is conserved in the x direction. (In fact, the friction between the wheels and the bench must increase suddenly during the collision, because the wheels are rolling with different angular velocities before and after (see Wheels and rolling), and this change requires a torque that is supplied by the friction on the bench. However, provided that the mass wheels is small, this force will be small compared to that between hammer and skateboard.)
You can check how well Σ px,initial = Σ px,final applies here: the mass of the hammer is 2.0 kg, that of the skateboard is 3.5 kg, so conservation of momentum in the x direction predicts that the velocity of the board after collision will be 2.0/(2.0+3.5) = 0.36 times the x component of the hammer's velocity between when it leaves my hand and when it hits the skateboard. The speed is proportional to the number of pixels travelled per frame.
The red line shows the horizontal position of the centre of mass. Because the two cars have equal mass, the centre of mass is halfway between their centres. In the example above, the centre of mass continues moving, at constant velocity, throughout the collision. The kinetic associated with the relative motion of the cars (the kinetic energy measured with respect to the centre of mass) is briefly turned into potential energy in the spring at the moment of maximum compression, and then converted back to kinetic energy.
Conservation of momentum requires:
pinitial = pfinal (1).
Because the collision is elastic, we can apply conservation of mechanical energy. The height doesn't change, so gravitational potential energy is constant throughout. The potential energy stored briefly in the spring is converted back to kinetic energy, so
which simplifies to
So, combining (2) (conservation of momentum) and (3) (conservation of energy), we have
This equation can only be true if v1 = 0 or if v2 = 0. So there are two possible solutions. One is v1 = 0 and v2 = v, which is what we see in the clip above: the left car comes to a complete stop and the right car leaves the collision with v. The other solution is v1 = v and v2 = 0, which is what happens if they don't collide at all. (Strictly, the second solution is possible only in two or three dimensions, not one.)
The example below shows a more general case: in the frame of the camera, the initial velocities of the two cars are in opposite directions and have different magnitudes.
Initially, the car on the left is travelling faster, so the centre of mass is moving to the right, which it does throughout. In a frame of reference in which the car on the right were stationary before the collision, however, this collision would look just like the previous example: the left hand car would appear stationary after the collision.
The collision is elastic, but conservation of momentum still applies so, as in the first example above, we have:
Here, of course, we cannot apply conservation of mechanical energy. Instead, we know that the two cars stick together after the collision, so
Combining (5) and (6) gives
Because the two masses are equal, the centre of mass is halfway between them. Before the collision, one car had velocity v and the other zero, so the centre of mass of the system was also v/2 before the collision. The total momentum is the total mass times the velocity of the centre of mass, so the total momentum, before and after, is (2m)(v/2) = mv.
In the next case we'll show, the intitial velocities are equal in magnitude and opposite in direction. The centre of mass, the point midway between the two cars, is therefore stationary before the collision. Again it's a completely inelastic collision, so again the two masses will have the same velocity after the collision. It is therefore easy to predict the final states.
Let's finish this set of examples with a collision in which the two initial velocities have different magnitudes, but opposite directions. So the initial state is rather like that of the second elastic collision above. However, the collision here is completely inelastic.