# Collisions With VPython Student Guide

## Table of contents

In this exercise we will investigate elastic and inelastic collisions.

- There are questions asked of you throughout this assignment, labeled
,__Question 1__, etc. Write down the answers in the TERM booklet (to be handed in at the end of the Practical).__Question 2__ - At the end of the Practical, hand in your TERM booklet, along with a printed copy of your code for the final exercise (collisionexC.py). Please staple the code into the booklet.
- If you are working with a partner, the two of you must reach a consensus about an answer before moving on to the next part. Feel free to ask the TA for help if a consensus cannot be reached.
- All quantities are in SI units; distance in meters (m), time in seconds (s), velocity in m/s, acceleration in m/s
^{2}, and mass in kg.

If we consider a system of 2 particles with no external forces acting, then the momentum of the system must be conserved. If the particles collide, then the total momentum of the system after the collision must equal the total momentum of the system before the collision.

All collisions will conserve momentum. *Elastic* collisions also conserve kinetic energy. *Totally inelastic* collisions lose the maximum possible kinetic energy. Most collisions are *inelastic*, where some kinetic energy is lost, but not the maximum possible.

Which type of collision occurs depends on the material of the colliding objects and other conditions. For example, two billiard balls colliding will be very different from two balls of Velcro.

**Exercise A: Totally Inelastic Collisions**

In “Totally inelastic collisions” the particles completely stick together after the collision. This means their velocities after the collision are the same.

__Question 1__**:** For a totally inelastic collision, if the masses are *m _{1}* and

*m*and their velocities before the collision are \(\vec{v}_1\) and \(\vec{v}_2\), what is the velocity of both masses after the collision?

_{2}

Download the program collisionexA.py and save it on your computer. This program is not complete, so if you try to run it you will get a “syntax error” – see the steps below to fix it. The goal of the program is to create and display 2 balls with masses m1 and m2 and initial constant velocities v1 and v2. The positions of the balls are updated in the while loop. When the balls collide, the velocities need to be changed. This is done in the “if” conditional which is supposed to check that the balls have collided, and if so, changes the velocities of both balls.

** Question 2:** What condition would indicate that balls have collided? Write this conditional in place of the ???Q2??? in the if statement. Hint: the balls collide when the edge of each is ‘inside’ the other one. What is the relationship between the distance between the balls’ centres and their radii when this happens? Your condition should be an inequality involving the position and radius properties of both balls (ball1.pos, ball2.pos, ball1.radius and ball2.radius). You will want to use the mag(...) function, which calculates the magnitude of a vector. Keep in mind that the position is a vector, but the radius is a scalar.

** Question 3:** Fill in the ???Q3??? field with the value of ball1’s velocity after the collision (use your result from question 1). Notice we have set ball2’s velocity to equal ball1’s velocity after the collision. Now you should be able to save and run the program without getting a syntax error. The program prints out the velocities, momenta, and kinetic energies before and after the collision. Is momentum conserved? Is kinetic energy conserved or some of it lost? Is this what you expected?

** Question 4:** By keeping the initial velocities as they are, and changing the masses of the particles in the program, what can you say about the final velocity for the following five conditions?

- m1 = m2
- m1 > m2
- m1 < m2
- m1 >> m2
- m2 >> m1

**Exercise B: Elastic Collisions**

The other extreme in collisions is an “elastic collision”. In this case, kinetic energy is conserved. We can’t use the constraint that \(\vec{v}_2=\vec{v}_1\) after the collision so we need another piece of information to find the velocities of the 2 balls after the collision.

** Question 5:** Using the facts that both momentum and kinetic energy are conserved, you will have two equations. Use these two equations to find expressions for the final velocities

*v*

_{1f}and

*v*

_{2f}in terms of the initial velocities

*v*

_{1i}and

*v*

_{2i}and the masses

*m*

_{1}and

*m*

_{2}. [Hint: It is helpful as intermediate step if you show that the two equations taken together imply that velocity of approach is equal to velocity of escape:

*v*

_{1i}–

*v*

_{2i}=

*v*

_{2f}–

*v*

_{1f}. This is equation 9.14 from page 157 of Wolfson. This elastic collision equation can be combined with the momentum conservation equation (9.12) to find the expressions we are looking for. ]

Save a new copy of the .py file you had at the end of Exercise A, call it “collisionexB.py”. Replace the two lines under the comment “#Update the post-collision velocities” with the following:

```
v1final= ??
v2final= ??
v1=v1final
v2=v2final
```

Note that we calculated the final velocities first before updating v1 and v2 values.

** Question 6:** Replace the two ?? in the lines you inserted in your code with the python equivalent for the expressions you found in Question 5. Note that if you are multiplying by an integer, add a .0 to make it a float rather than an integer (e.g. use 2.0 instead of 2).

Run your program.

** Question 7:** Are momentum and kinetic energy conserved? Is this what you expected?

** Question 8**: Now change the initial velocity of the second ball, v2, to zero. What can you say about the final velocities of both masses for the following five conditions? Run your program to confirm your predictions.

- m1=m2
- m1 > m2
- m1 < m2
- m1 >> m2
- m1 << m2

**Exercise C: Elastic Collisions in Two Dimensions**

For an elastic collision in two dimensions we have three conservation equations from the bottom of page 156 of Wolfson:

\(\begin{eqnarray} m_1 v_{1{\rm i}x} + m_2 v_{2{\rm i}x} = m_1 v_{1{\rm f}x} + m_2 v_{2{\rm f}x} \end{eqnarray}\) (C1. Conservation of *x*-component of momentum)

\(\begin{eqnarray} m_1 v_{1{\rm i}y} + m_2 v_{2{\rm i}y} = m_1 v_{1{\rm f}y} + m_2 v_{2{\rm f}y} \end{eqnarray}\) (C2. Conservation of *y*-componentof momentum)

\(\begin{eqnarray} \frac{1}{2}m_1 v_{1{\rm i}}^2 + \frac{1}{2}m_2 v_{2{\rm i}}^2 = \frac{1}{2}m_1 v_{1{\rm f}}^2 + \frac{1}{2}m_2 v_{2{\rm f}}^2 \end{eqnarray}\) (C3. Conservation of kinetic energy)

To solve problems, let's start in the rest-frame of ball 2, so that \(\vec{v}_{2{\rm i}} = 0\). Also, let's rotate our reference frame so that, initially, ball 1 travels in the +*x*-direction, so that \(v_{1{\rm i}y} = 0\). We can write the initial *x*-component of the velocity of ball 1 simply as *v*_{1i}. The conservation equations are simplified somewhat to be:

\(\begin{eqnarray} m_1 v_{1{\rm i}} = m_1 v_{1{\rm f}x} + m_2 v_{2{\rm f}x} \end{eqnarray}\) (C4)

\(\begin{eqnarray} m_1 v_{1{\rm f}y} = -m_2 v_{2{\rm f}y} \end{eqnarray}\) (C5)

\(\begin{eqnarray} \frac{1}{2}m_1 v_{1{\rm i}}^2 = \frac{1}{2}m_1 v_{1{\rm f}}^2 + \frac{1}{2}m_2 v_{2{\rm f}}^2 \end{eqnarray}\) (C6)

Let's place the centre of ball 2 at the origin. Define the impact parameter, *b*, as the initial *y*-position of the centre of ball 1.

**Figure C1:** Before Collision

Next, let's make the assumption for this part that both balls have equal mass, *m*_{1} = *m*_{2} = *m*. Then the conservation equations become:

\(\begin{eqnarray} v_{1{\rm i}} = v_{1{\rm f}x} + v_{2{\rm f}x} \end{eqnarray}\) (C7)

\(\begin{eqnarray} v_{1{\rm f}y} = -v_{2{\rm f}y} \end{eqnarray}\) (C8)

\(\begin{eqnarray} v_{1{\rm i}}^2 = v_{1{\rm f}}^2 + v_{2{\rm f}}^2 \end{eqnarray}\) (C9)

Lastly, let's define \(\theta_1\) to be the angle of the final velocity of ball 1 with respect to the *x*-axis, and \(\theta_2\) to be the angle of the final velocity of ball 2 with respect to the *x*-axis. In both cases, we use the convention that positive angles are measured counterclockwise with respect to the x-axis. Therefore \(\theta_2\) is a negative number.

**Figure C2: **After Collision

The conservation equations then become:

\(\begin{eqnarray} v_{1{\rm i}} = v_{1{\rm f}}\cos\theta_1 + v_{2{\rm f}}\cos\theta_2 \end{eqnarray}\) (C10)

\(\begin{eqnarray} v_{1{\rm f}}\sin\theta_1 = -v_{2{\rm f}}\sin\theta_2 \end{eqnarray}\) (C11)

\(\begin{eqnarray} v_{1{\rm i}}^2 = v_{1{\rm f}}^2 + v_{2{\rm f}}^2 \end{eqnarray}\) (C12)

It can be shown that for balls of equal mass, \(\theta_1 - \theta_2 = 90^\circ\). (See Wolfson Example 9.11, and Problem 9.72.). If the balls are spheres and are quite rigid, then the collision is very short, and during the collision, the normal force of ball 1 on ball 2 is the only force which accelerates ball 2 from rest. Therefore, the final velocity of ball 2 can be found from the geometry of the collision itself, shown in Figure C3.

**Figure C3:** During Collision. Note that the balls only will collide if *b* < 2*r*.

**To Do: **Save a new copy of the .py file you had at the end of Exercise B, call it “collisionexC.py”. For purposes of checking the initial and final momentum and energy, set the masses of the two balls equal to 1 kg. Under the lines to set the radius of the balls, Set the impact parameter [in m] to be b = 0.2.

** Question 9:** Find an expression for \(\theta_2\) in terms of

*b*and

*r*. Note that the arcsin() function in Python returns an angle in units of radians.

** Question 10:** Knowing that \(\theta_1 - \theta_2 = 90^\circ\), find an expression for \(\theta_1\) in terms of \(\theta_2\) . Once again, you will be using python, so come up with an expression in radians, not degrees. Python uses the variable math.pi as being equal to \(\pi\) .

** Question 11:** Combine equations C10 and C11 to find

*v*

_{1f}in terms of

*v*

_{1i}, \(\theta_1\) and \(\theta_2\). Note that, in your code,

*v*

_{1i}wll be represented by mag(v1). You should set a variable v1fmag for

*v*

_{1f}.

** Question 12:** Combine equations C10 and C11 to find

*v*

_{2f}in terms of

*v*

_{1i}, \(\theta_1\) and \(\theta_2\). You should set a variable v2fmag for

*v*

_{2f}.

** Question 13:** Now it's time to type all of these equations from Questions 9-12 into your Python code to compute theta2, theta1, v1fmag and v2fmag, respectively. After that, you will want to specify the vector velocities \(\vec{v}_1\) and \(\vec{v}_2\) with the following lines:

```
v1final=vector(v1fmag*cos(theta1),v1fmag*sin(theta1),0)
v2final=vector(v2fmag*cos(theta2),v2fmag*sin(theta2),0)
```

Check your code, and see if it does what you expect for various values of *b*. Write down the final velocities for the following impact parameters:

*b = r*/2*b = r**b*= 3*r*/2.

** Bonus Question :** (If You Have Time)

What if the balls are not the same mass? How would you modify the code to account for this?

[Note: This Module was written by Omar Gamel in September 2014, based on activities written by Sabine Stanley 2010-2014.]