During a collision between two objects there will be a sudden spike in the contact force between them, which can be difficult to measure and quantify. It is therefore not possible to use Newton's law $F=ma$ to predict the accelerations that occur during collisions. In order to predict the motion of the objects after the collision we must use a momentum calculation. The law of conservation of momentum states that the total amount of momentum before and after the collision is the same. Thus, if we know the momenta of the objects before the collision, it will be possible to calculate their momenta after the collision and from this figure out their subsequent motion.

To illustrate why the notion of momentum is important, consider the following situation. Say you have a 1[g] piece of paper and a 1000[kg] car moving at the same speed 100[km/h]. Which of the two objects would you rather get hit by? Momentum, denoted $\vec{p}$, is the precise physical concept which measures the “amount of moving stuff”. An object of mass $m$ moving with velocity $\vec{v}$ has momentum $\vec{p}\equiv m\vec{v}$. Momentum plays a key role in collisions, so your gut feeling about the piece of paper and the car is correct. The car weights $1000\times1000=10^{6}$ times more than the piece of paper, so it has $10^6$ times more momentum when moving at the same speed. A collision with the car will “hurt” a million times more than the collision with the piece of paper even though they were moving at the same speed.

In this section we will learn how to use the law of conservation of momentum to predict the outcomes of collisions.

• $m$: the mass of the moving object.
• $\vec{v}$: the velocity of the moving object.
• $\vec{p}=m\vec{v}$: the momentum of the moving object.
• $\sum \vec{p}_{in}$: the sum of the momenta of particles before a collision.
• $\sum \vec{p}_{out}$: the sum of the momenta after the collision.

The momentum of a moving object is equal to the velocity of the moving object multiplied by the object's mass: \[ \vec{p} = m\vec{v} \qquad [\text{kg}\:\text{m}/\text{s}]. \] If the velocity of the object is $\vec{v}=20\hat{\imath}=(20,0)$[m/s] and it has a mass of 100[kg] then its momentum is $\vec{p}=2000\hat{\imath}=(2000,0)$[kg$\:$m/s].

Momentum is a vector quantity, so we will often have to convert momentuma from the length-and-direction form to the components form: \[ \vec{p}= \|\vec{p}\| \angle \theta = (\|\vec{p}\|\cos\theta, \|\vec{p}\|\sin\theta) = (p_x, p_y). \] The component form makes it easy to add and subtract vectors: $\vec{p}_1 + \vec{p}_2 = (p_{1x}+p_{2x},p_{1y}+p_{2y})$. To express the final answer, we will have to convert from the component form back to the length-and-direction form using: \[ \|\vec{p}\| = \sqrt{ p_x^2 + p_y^2 }, \qquad \theta = \tan^{-1}\!\left( \frac{ p_{y} }{ p_{x} } \right). \]

Newton's first law states that in the absence of acceleration ($\vec{a}=0$), an object will maintain a constant velocity. This is kind of obvious if you know Calculus, since $\vec{a}$ is the derivative of $\vec{v}$. For example, if an object is stationary and there are no forces on it to cause it to accelerate, then it will remain stationary. If an object is moving with velocity $\vec{v}$ and there is no acceleration (or deceleration), then it will keep moving with velocity $\vec{v}$ forever. In the absence of acceleration, objects will conserve their velocity: \[ \vec{v}_{in}= \vec{v}_{out}. \] This is equivalent to saying that objects conserve their momentum (just multiply the velocity by the constant mass of the object).

More generally, if you have a situation involving multiple moving objects, you can say that the “overall momentum”, i.e., the sum of the momenta of all the interacting particles stays constant. This reasoning is particularly useful when analyzing collisions since it allows us to connect the sum of the momenta before the collision and after the collision: \[ \sum \vec{p}_{in} = \sum \vec{p}_{out}. \] Whatever momentum comes into a collision must come out. This equation is known as the law of conservation of momentum.

This conservation law is one of the furthest reaching laws of physics you will learn in Mechanics. We learned about the conservation of momentum in a simple context of two colliding particles, but the law applies much more generally: for multiple particles, for fluids, for fields, and even for collisions involving atomic particles described by quantum mechanics. The quantity of motion (momentum) cannot be created or destroyed, it can only be exchanged between systems.

You throw a piece of rolled up carton of mass $0.4$[g] from your balcony on a rainy day. You throw it horizontally with a speed of 10[m/s]. Shortly after it leaves your hand it collides with a rain drop of weight $2$[g] falling straight down at a speed of $30$[m/s]. What will be resulting velocity if the two objects stick together after the collision?

The conservation of momentum equation says that: \[ \vec{p}_{in,1} + \vec{p}_{in,2} = \vec{p}_{out}. \] Plugging in the values we get \[ 0.4\times (10,0) \ \ + \ \ 2\times (0,-30) \ \ = \ \ 2.4 \times \vec{v}_{out}, \] or solving for $\vec{v}_{out}$ we find: \[ \vec{v}_{out} = \ \frac{ 0.4(10,0) - 2 (0,30)} {2.4} = (1.666, - 25.0) = 1.666\hat{\imath} - 25.0\hat{\jmath}. \]

Two hipsters on single-speed bicycles are headed towards the same intersection. Say they are both speeding down Parc street at 50[km/h] and the first hipster is crossing the street at a diagonal of 30 degrees when they collide. I mean you saw this coming right? Well the second hipster didn't, because he was busy turning the pedals as fast as he can.

Let us assume that the combined weight of the straight-going hipster and his bike is 100[kg], whereas the street-crossing-at-30-degrees hipster has a lighter, more expensive bicycle frame. We put his weight at 90[kg].

(I am going to continue with the story, but I want to point out that we have been given, the following information so far: \[ \begin{align*} \vec{p}_{in,1} &= 90\times50 \angle 30=90(50\cos30,50\sin30), \nl \vec{p}_{in,2} &= 100\times50 \angle 0=(5000,0), \end{align*} \] where the $x$ coordinate points down Park street, and the $y$ coordinate is perpendicular to the street.)

Surprisingly, nobody gets hurt in this collision. They bump shoulder-to-shoulder and the one that was trying to cross the street gets redirected straight down the street, while the one going straight down gets deflected to the side and right onto the bike path. I know what you are thinking: couldn't they get hurt at least a little bit? OK, let's say that the whiplash from their shoulder-to-shoulder collision sends their heads flying towards each other and their glasses get smashed. There you have it.

Suppose the velocity of the first hipster after the collision is 60 [km/h], what is the velocity and the deflected direction of the second hipster? (I have just told you that the outgoing momentum of the first hipster is $\vec{p}_{out,1}=(90\times60,0)$, and asked you to find $\vec{p}_{out,2}$.)

We can solve this problem using the conservation of momentum formula, which tells us that: \[ \vec{p}_{in,1} + \vec{p}_{in,2} = \vec{p}_{out,1} + \vec{p}_{out,2}. \] We know three of the above quantities so we can solve for the one (vector) unknown by isolating it on one side of the equation: \[ \vec{p}_{out,2} = \vec{p}_{in,1} + \vec{p}_{in,2} - \vec{p}_{out,1}, \] \[ \vec{p}_{out,2} = 90(50\cos30,50\sin30)\ +\ (5000,0)\ - \ (90\times60,0). \] The $x$ component of the momentum $\vec{p}_{out,2}$ is: \[ p_{out,2,x} = 90\times50\cos30 + 5000 - 90\times 60 = 3497.11, \] and the $y$ component is $p_{out,2,y} = 90\times 50\sin30 = 2250$.

The magnitude of the momentum of hipster 2 is given by: \[ \|\vec{p}_{out,2}\| = \sqrt{ p_{out,2,x}^2 + p_{out,2,y}^2 } = 4158.39, \quad \textrm{[kgkm/h]}. \] Note the units we use for the momentum is not the standard choice [kgm/s]. That is fine. So long as you keep in mind which units you are using, you don't have to always convert to SI units.

The final velocity of hipster two is $v_{out,2} = 4158.39/100= 41.58$[km/h]. The deflection angle is obtained by \[ \phi_{def} = \tan^{-1}\!\!\left( \frac{ p_{out,2,y} }{ p_{out,2,x} } \right)= 32.76^\circ. \]

We defined the concept of momentum in terms of the velocity of the object, but in fact, momentum is a more fundamental concept than velocity. If you go on to take more advanced physics classes, you will learn that the natural variables to describe the state of a particle are their positions and momenta $(\vec{x}, \vec{p})$. You will also learn that the real form of Newton's second law is written in terms of the momentum: \[ \vec{F} = \frac{d \vec{p} }{dt} \quad \text{for } m \text{ constant } \Rightarrow \quad \vec{F}=\frac{d (m\vec{v}) }{dt}=m\frac{d \vec{v} }{dt} =m\vec{a}. \] In most physics problems the mass of objects will stay constant so using $\vec{F}=m\vec{a}$ is perfectly fine.

The law of conservation of momentum follows from Newton's third law: for each force $\vec{F}_{12}$ exerted by Object 1 on Object 2, there exists a counter force $\vec{F}_{21}$ of equal magnitude and opposite direction, which is the force of Object 2 pushing back on Object 1. Earlier I said that it is difficult to quantify the magnitude of the exact forces $\vec{F}_{12}$ and $\vec{F}_{21}$ that occur during a collision. Indeed, the amount of force suddenly shoots up as the two objects collide and then suddenly drops. Complicated as these forces may be, we know that during the entire collision they obey Newton's third law. Assuming there are no other forces acting on the objects we have: \[ \vec{F}_{12} = -\vec{F}_{21} \quad \text{using the above} \Rightarrow \quad \frac{d \vec{p}_1 }{dt} = -\frac{d \vec{p}_2 }{dt}. \] If now move both terms to the left-hand side we obtain the equation: \[ \frac{d \vec{p}_1 }{dt} + \frac{d \vec{p}_2 }{dt} = \frac{d}{dt}\left( \vec{p}_1 + \vec{p}_2 \right) = 0, \] which implies that quantity $\vec{p}_1 + \vec{p}_2$ is constant over time.

In this section we saw how to use a momentum calculation to predict the motion of the particles after a collision. In the next section, we will learn about the concept of energy which is another useful concept for understanding and predicting the motion of objects.

[ Animations of simple collisions between objects. ]
http://en.wikipedia.org/wiki/Conservation_of_linear_momentum