Collisions
In the next several lessons, we'll use the principle of conservation of momentum to analyze various different kinds of collisions. We shall investigate two broad classes of collisions: perfectly elastic and perfectly inelastic collisions. The former (elastic collisions) is just whenever object bounce right off one another(i.e. when two protons are smashed into one another, they bounce off of each other). The latter (inelastic collisions) is when objects collide and stick together (for example, an arrow striking a massive body of rubber.
We'll begin our studies with the simpler of the two: inelastic collisions. And, furthermore, we'll only thjink about the collision between two different objects. Suppose that an object of mass \(m_i\) smashes into another object of mass \(m_2\). If the collision is perfectly inelastic, then the two objects will "stick" together and movwe as a single mass \(m_1+m_2\) at a velocity \(\vec{v}\). (That statement is best understood by experiance. For example, if an arrow strikes a firm rubber body, both of those objecs and the entire mass \(m_{arrow}+m_{rubber}\) will move off with a common velocity \(\vec{v}\). We'll assume that both objects can, in general, be initially moving with velocities \(\vec{v}_{1,i}\) and \(\vec{v}_{2,i}\). Thus, te total intial momentum of the system is \(m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}\). We'll assume that those two masses are completely isolated where there are no external forces acting upon them. Therefore, as we shjowed previously using Newton's laws, the total momentum of the system must be conserved and
$$\vec{p}_i=\vec{p}_j$$
$$m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}=m_1\vec{v}_{1,f}+m_2\vec{v}_{2,f}.\tag{4}$$
Figure 1: An inelastic collision is when two objects collide and stick together. Each object becomes conjoined and moves away with equal velocities.
As I said earlier, in a perfectly inelastic collision (i.e. an arrow striking a rubber body can be assumed, to a good degree of approximation, to be a perfectly inelastic collision), both objects will become conjoined and move away at the same velocity \(\vec{v}\). Thus, \(\vec{v}_{1,f}=\vec{v}_{2,f}=\vec{v}\) and we can simplify Equation (4) to
$$m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}=(m_1+m_2)\vec{v}_f.\tag{5}$$
Let's divide both sides of Equation (5) by \(m_1+m_2\) to get
$$\vec{v}_f=\frac{m_1\vec{v}_{1,i}+m_2\vec{v}_{2,i}}{m_1+m_2}.\tag{6}$$
Equation (6) is very usefull. If two objects inelastically collide, if we know there masses and how they'reinitially moving, we can use Equation (6) to determine how fast they'll be going when those two objects collide and stick together.
Example
Problem: An archer at rest fires an arrow of mass \(m_1=0.1kg\) at \(60m/s\) at a stationary rubber block of mass \(m_2=10kg\). The arrow inelastically collides with the block and both move off to the right at a common velocity. How fast does the arrow-black system move away?
This is a simple substitution problem. Since the archer (including the arrow) and the block are initially not moving, \(\vec{v}_{1,i}=\vec{v}_{2,i}=0\). Pluggging these values and \(m_1=0.1kg\) and \(m_2=10kg\) in Equation (6), we find that
$$\vec{v}_f=\frac{(0.1kg)(60m/s)+(10kg)(0m/s)}{10kg+0.1kg}=0.59\frac{m}{s}.$$
Thus, the arrow-block system moves off to the right at a speed of \(0.59m/s\).