Elastic Collisions Continued

Consider 2 marbles. Marble 1 has mass \(\text{50}\) \(\text{g}\) and marble 2 has mass \(\text{100}\) \(\text{g}\). Edward rolls marble 2 along the ground towards marble 1 in the positive x-direction. Marble 1 is initially at rest and marble 2 has a velocity of \(\text{3}\) \(\text{m·s$^{-1}$}\) in the positive x-direction. After they collide elastically, both marbles are moving. What is the final velocity of each marble?

Step 1: Decide how to approach the problem

We are given:

• mass of marble 1, \({m}_{1}=50~\text{g}\)

• mass of marble 2, \({m}_{2}=100~\text{g}\)

• initial velocity of marble 1, \(\vec{v}_{i1}=\text{0}\text{ m·s$^{-1}$}\)

• initial velocity of marble 2, \(\vec{v}_{i2}=\text{3}\text{ m·s$^{-1}$}\) to the right

• the collision is elastic

The masses need to be converted to SI units.

\begin{align*} {m}_{1}& = 0,05\text{kg}\\ {m}_{2}& = 0,1\text{kg} \end{align*}

We are required to determine the final velocities:

• final velocity of marble 1, \(\vec{v}_{f1}\)

• final velocity of marble 2, \(\vec{v}_{f2}\)

Since the collision is elastic, we know that

• momentum is conserved, \(\vec{p}_{Ti}=\vec{p}_{Tf}\).

• energy is conserved, \(K{E}_{Ti}=K{E}_{Tf}\).

We have two equations and two unknowns (\(\vec{v}_{1}\), \(\vec{v}_{2}\)) so it is a simple case of solving a set of simultaneous equations.

Step 2: Choose a frame of reference

Choose to the right as positive.

Step 3: Draw a rough sketch of the situation

Step 4: Solve the problem

Momentum is conserved. Therefore:

\begin{align*} \vec{p}_{i}& = \vec{p}_{f}\\ \vec{p}_{i1}+\vec{p}_{i2}& = \vec{p}_{f1}+\vec{p}_{f2}\\ {m}_{1}\vec{v}_{i1}+{m}_{2}\vec{v}_{i2}& = {m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2}\\ {m}_{2}\vec{v}_{i2}& = {m}_{1}\vec{v}_{f1}+{m}_{2}\vec{v}_{f2}\\ {m}_{1}\vec{v}_{f1} & = {m}_{2}\vec{v}_{i2}- {m}_{2}\vec{v}_{f2} \\ \vec{v}_{f1} & = \frac{{m}_{2}}{{m}_{1}}(\vec{v}_{i2}- \vec{v}_{f2}) \end{align*}

Energy is also conserved. Therefore:

\begin{align*} K{E}_{i}& = K{E}_{f}\\ K{E}_{i1}+K{E}_{i2}& = K{E}_{f1}+K{E}_{f2}\\ \frac{1}{2}{m}_{1}{v}_{i1}^{2}+\frac{1}{2}{m}_{2}{v}_{i2}^{2}& = \frac{1}{2}{m}_{1}{v}_{f1}^{2}+\frac{1}{2}{m}_{2}{v}_{f2}^{2}\\ \frac{1}{2}{m}_{2}{v}_{i2}^{2}& = \frac{1}{2}{m}_{1}{v}_{f1}^{2}+\frac{1}{2}{m}_{2}{v}_{f2}^{2}\\ {m}_{2}{v}_{i2}^{2}& = {m}_{1}{v}_{f1}^{2}+{m}_{2}{v}_{f2}^{2} \end{align*}

Substitute the expression we derived from conservation of momentum into the expression derived from conservation of kinetic energy and solve for \({v}_{f2}\).

\begin{align*} {m}_{2}{v}_{i2}^{2}& = {m}_{1}{v}_{f1}^{2}+{m}_{2}{v}_{f2}^{2}\\ &={m}_{1}{\left(\frac{{m}_{2}}{{m}_{1}}\left({v}_{i2}-{v}_{f2}\right)\right)}^{2}+{m}_{2}{v}_{f2}^{2}\\ &={m}_{1}\frac{{m}_{2}^{2}}{{m}_{1}^{2}}{\left({v}_{i2}-{v}_{f2}\right)}^{2}+{m}_{2}{v}_{f2}^{2}\\ &=\frac{{m}_{2}^{2}}{{m}_{1}}{\left({v}_{i2}-{v}_{f2}\right)}^{2}+{m}_{2}{v}_{f2}^{2}\\ {v}_{i2}^{2}& = \frac{{m}_{2}}{{m}_{1}}{\left({v}_{i2}-{v}_{f2}\right)}^{2}+{v}_{f2}^{2}\\ &=\frac{{m}_{2}}{{m}_{1}}\left({v}_{i2}^{2}-2·{v}_{i2}·{v}_{f2}+{v}_{f2}^{2}\right)+{v}_{f2}^{2}\\ 0& = \left(\frac{{m}_{2}}{{m}_{1}}-1\right){v}_{i2}^{2}-2\frac{{m}_{2}}{{m}_{1}}{v}_{i2}·{v}_{f2}+\left(\frac{{m}_{2}}{{m}_{1}}+1\right){v}_{f2}^{2}\\ &=\left(\frac{0.1}{0.05}-1\right){\left(3\right)}^{2}-2\frac{0.1}{0.05}\left(3\right)·{v}_{f2}+\left(\frac{0.1}{0.05}+1\right){v}_{f2}^{2}\\ &=\left(2-1\right){\left(3\right)}^{2}-2·2\left(3\right)·{v}_{f2}+\left(2+1\right){v}_{f2}^{2}\\ &=9-12{v}_{f2}+3{v}_{f2}^{2}\\ &=3-4{v}_{f2}+{v}_{f2}^{2}\\ &=\left({v}_{f2}-3\right)\left({v}_{f2}-1\right) \end{align*}

Therefore \({v}_{f2}=1\) or \({v}_{f2}=3\)

Substituting back into the expression from conservation of momentum, we get:

\begin{align*} {v}_{f1}& = \frac{{m}_{2}}{{m}_{1}}\left({v}_{i2}-{v}_{f2}\right)\\ &=\frac{0.1}{0.05}\left(3-3\right)\\ &=0\\ & \text{or} \\ {v}_{f1}&=\frac{{m}_{2}}{{m}_{1}}\left({v}_{i2}-{v}_{f2}\right)\\ &=\frac{0.1}{0.05}\left(3-1\right)\\ &=\text{4}\text{ m·s$^{-1}$} \end{align*}

But according to the question, marble 1 is moving after the collision, therefore marble 1 moves to the right at \(\text{4}\) \(\text{m·s$^{-1}$}\).

Therefore marble 2 moves with a velocity of \(\text{1}\) \(\text{m·s$^{-1}$}\) to the right.

Two billiard balls each with a mass of \(\text{150}\) \(\text{g}\) collide head-on in an elastic collision. Ball 1 was travelling at a speed of \(\text{2}\) \(\text{m·s$^{-1}$}\) and ball 2 at a speed of \(\text{1.5}\) \(\text{m·s$^{-1}$}\). After the collision, ball 1 travels away from ball 2 at a velocity of \(\text{1.5}\) \(\text{m·s$^{-1}$}\).

1. Calculate the velocity of ball 2 after the collision.

2. Prove that the collision was elastic. Show calculations.

Step 1: Choose a frame of reference

Choose to the right as positive.

Step 2: Draw a rough sketch of the situation

Step 3: Decide how to approach the problem

Since momentum is conserved in all kinds of collisions, we can use conservation of momentum to solve for the velocity of ball 2 after the collision.

Step 4: Solve problem

\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(\frac{150}{1000}\right)\left(2\right)+\left(\frac{150}{1000}\right)\left(-1,5\right)& = \left(\frac{150}{1000}\right)\left(-1,5\right)+\left(\frac{150}{1000}\right)\left(\vec{v}_{f2}\right)\\ 0,3-0,225& = -0,225+0,15\vec{v}_{f2}\\ \vec{v}_{f2}& = \text{3}\text{ m·s$^{-1}$} \end{align*}

So after the collision, ball 2 moves with a velocity of \(\text{3}\) \(\text{m·s$^{-1}$}\) to the right.

Step 5: Elastic collisions

The fact that characterises an elastic collision is that the total kinetic energy of the particles before the collision is the same as the total kinetic energy of the particles after the collision. This means that if we can show that the initial kinetic energy is equal to the final kinetic energy, we have shown that the collision is elastic.

Step 6: Calculating the initial total kinetic energy

\begin{align*} E{K}_{before}& = \frac{1}{2}{m}_{1}{v}_{i1}^{2}+\frac{1}{2}{m}_{2}{v}_{i2}^{2}\\ & = \left(\frac{1}{2}\right)\left(0,15\right){\left(2\right)}^{2}+\left(\frac{1}{2}\right)\left(\text{0.15}\right){\left(-\text{1.5}\right)}^{2}\\ & = \text{0.469...}\text{ J} \end{align*}

Step 7: Calculating the final total kinetic energy

\begin{align*} E{K}_{after}& = \frac{1}{2}{m}_{1}{v}_{f1}^{2}+\frac{1}{2}{m}_{2}{v}_{f2}^{2}\\ & = \left(\frac{1}{2}\right)\left(\text{0.15}\right){\left(-\text{1.5}\right)}^{2}+\left(\frac{1}{2}\right)\left(\text{0.15}\right){\left(2\right)}^{2}\\ & = \text{0.469...}\text{ J} \end{align*}

So \(E{K}_{Ti}=E{K}_{Tf}\) and hence the collision is elastic.