Collisions

Collisions

We have shown that the net change in momentum is zero for an isolated system. The momenta of the individual objects can change but the total momentum of the system remains constant.

This means that it makes sense to define the total momentum at the begining of the problem as the initial total momentum, \(\vec{p}_{Ti}\), and the final total momentum \(\vec{p}_{Tf}\). Momentum conservation implies that, no matter what happens inside an isolated system:

\[\vec{p}_{Ti} = \vec{p}_{Tf}\]

This means that in an isolated system the total momentum before a collision or explosion is equal to the total momentum after the collision or explosion.

Consider a simple collision of two billiard balls. The balls are rolling on a frictionless, horizontal surface and the system is isolated. We can apply conservation of momentum. The first ball has a mass \({m}_{1}\) and an initial velocity \(\vec{v}_{i1}\). The second ball has a mass \({m}_{2}\) and moves first ball with an initial velocity \(\vec{v}_{i2}\). This situation is shown in the figure below.

Before the collision.

The total momentum of the system before the collision, \(\vec{p}_{Ti}\) is:

\(\vec{p}_{Ti}={m}_{1}\vec{v}_{i1}+{m}_{2}\vec{v}_{i2}\)

After the two balls collide and move away they each have a different momentum. If the first ball has a final velocity of \(\vec{v}_{f1}\) and the second ball has a final velocity of \(\vec{v}_{f2}\) then we have the situation shown in the figure below.

After the collision.

The total momentum of the system after the collision, \(\vec{p}_{Tf}\) is:

\(\vec{p}_{Tf}={m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2}\)

This system of two balls is isolated since there are no external forces acting on the balls. Therefore, by the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision. This gives the equation for the conservation of momentum in a collision of two objects,

This equation is always true - momentum is always conserved in collisions.

Example: Conservation of Momentum

Question

A toy car of mass \(\text{1}\) \(\text{kg}\) moves westwards with a speed of \(\text{2}\) \(\text{m·s$^{-1}$}\). It collides head-on with a toy train. The train has a mass of \(\text{1.5}\) \(\text{kg}\) and is moving at a speed of \(\text{1.5}\) \(\text{m·s$^{-1}$}\) eastwards. If the car rebounds at \(\text{2.05}\) \(\text{m·s$^{-1}$}\), calculate the final velocity of the train.

Solution

Step 1: Analyse what you are given and draw a sketch

The question explicitly gives a number of values which we identify and convert into SI units if necessary:

• the train's mass (\({m}_{1}= \text{1.50}\text{ kg}\)),

• the train's initial velocity (\(\vec{v}_{1i}=\text{1.5}\text{ m·s$^{-1}$}\) eastward),

• the car's mass (\({m}_{2}= \text{1.00}\text{ kg}\)),

• the car's initial velocity (\(\vec{v}_{2i}=\text{2.00}\text{ m·s$^{-1}$}\) westward), and

• the car's final velocity (\(\vec{v}_{2f}=\text{2.05}\text{ m·s$^{-1}$}\) eastward).

We are asked to find the final velocity of the train.

Step 2: Choose a frame of reference

We will choose to the East as positive.

Step 3: Apply the Law of Conservation of momentum

\begin{align*} \vec{p}_{Ti}& = \vec{p}_{Tf} \\ {m}_{1}\vec{v}_{i1}+{m}_{2}\vec{v}_{i2}& = {m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2} \\ \left(\text{1.5}\right)\left(\text{+1.5} \right)+\left(2\right)\left(-2 \right)& = \left(\text{1.5}\right)\left(\vec{v}_{f1}\right)+\left(2\right)\left(\text{2.05} \right) \\ \text{2.25} -4 -\text{4.1}& = \left(\text{1.5}\right)\vec{v}_{f1} \\ \text{5.85} & = \left(\text{1.5}\right)\vec{v}_{f1} \\ \vec{v}_{f1}& = \text{3.9} \text{m}·{\text{s}}^{-1} \text{eastwards}\end{align*}