6.2 Types of collisions (Page 2/4)

Siyavula textbooks: grade 12 Page 2 / 4

Consider two 2 marbles. Marble 1 has mass 50 g and marble 2 has mass 100 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 3 m $\cdot$ s $^{- 1}$ in the positive $x$ -direction. After they collide elastically , both marbles are moving. What is the final velocity of each marble?

Decide how to approach the problem
We are given:
- mass of marble 1, $m_{1}$ =50 g
- mass of marble 2, $m_{2}$ =100 g
- initial velocity of marble 1, $v_{i 1}$ =0 m $\cdot$ s $^{- 1}$
- initial velocity of marble 2, $v_{i 2}$ =3 m $\cdot$ s $^{- 1}$
- the collision is elastic
The masses need to be converted to SI units.

$\begin{matrix} m_{1} & = & 0, 05 kg \\ m_{2} & = & 0, 1 kg \end{matrix}$

We are required to determine the final velocities:
- final velocity of marble 1, $v_{f 1}$
- final velocity of marble 2, $v_{f 2}$
Since the collision is elastic, we know that
- momentum is conserved, $p_{i} = p_{f}$ .
- energy is conserved, $K E_{i}$ = $K E_{f}$ .
We have two equations and two unknowns ( $v_{1}$ , $v_{2}$ ) so it is a simple case of solving a set of simultaneous equations.
Choose a frame of reference
Choose to the right as positive.
Draw a rough sketch of the situation
Solve the problem
Momentum is conserved. Therefore:

$\begin{matrix} p_{i} & = & p_{f} \\ p_{i 1} + p_{i 2} & = & p_{f 1} + p_{f 2} \\ m_{1} v_{i 1} + m_{2} v_{i 2} & = & m_{1} v_{f 1} + m_{2} v_{f 2} \\ (0, 05) (0) + (0, 1) (3) & = & (0, 05) v_{f 1} + (0, 1) v_{f 2} \\ 0, 3 & = & 0, 05 v_{f 1} + 0, 1 v_{f 2} \end{matrix}$

Energy is also conserved. Therefore:

$\begin{matrix} K E_{i} & = & K E_{f} \\ K E_{i 1} + K E_{i 2} & = & K E_{f 1} + K E_{f 2} \\ \frac{1}{2} m_{1} v_{i 1}^{2} + \frac{1}{2} m_{2} v_{i 2}^{2} & = & \frac{1}{2} m_{1} v_{f 1}^{2} + \frac{1}{2} m_{2} v_{f 2}^{2} \\ (\frac{1}{2}) (0, 05) {(0)}^{2} + (\frac{1}{2}) (0, 1) {(3)}^{2} & = & \frac{1}{2} (0, 05) {(v_{f 1})}^{2} + (\frac{1}{2}) (0, 1) {(v_{f 2})}^{2} \\ 0, 45 & = & 0, 025 v_{f 1}^{2} + 0, 05 v_{f 2}^{2} \end{matrix}$

Substitute Equation [link] into Equation [link] and solve for $v_{f 2}$ .

$\begin{matrix} m_{2} v_{i 2}^{2} & = & m_{1} v_{f 1}^{2} + m_{2} v_{f 2}^{2} \\ = & m_{1} {(\frac{m_{2}}{m_{1}}, (v_{i 2} - v_{f 2}))}^{2} + m_{2} v_{f 2}^{2} \\ = & m_{1} \frac{m_{2}^{2}}{m_{1}^{2}} {(v_{i 2} - v_{f 2})}^{2} + m_{2} v_{f 2}^{2} \\ = & \frac{m_{2}^{2}}{m_{1}} {(v_{i 2} - v_{f 2})}^{2} + m_{2} v_{f 2}^{2} \\ v_{i 2}^{2} & = & \frac{m_{2}}{m_{1}} {(v_{i 2} - v_{f 2})}^{2} + v_{f 2}^{2} \\ = & \frac{m_{2}}{m_{1}} (v_{i 2}^{2} - 2 \cdot v_{i 2} \cdot v_{f 2} + v_{f 2}^{2}) + v_{f 2}^{2} \\ 0 & = & (\frac{m_{2}}{m_{1}} - 1) v_{i 2}^{2} - 2 \frac{m_{2}}{m_{1}} v_{i 2} \cdot v_{f 2} + (\frac{m_{2}}{m_{1}} + 1) v_{f 2}^{2} \\ = & (\frac{0.1}{0.05} - 1) {(3)}^{2} - 2 \frac{0.1}{0.05} (3) \cdot v_{f 2} + (\frac{0.1}{0.05} + 1) v_{f 2}^{2} \\ = & (2 - 1) {(3)}^{2} - 2 \cdot 2 (3) \cdot v_{f 2} + (2 + 1) v_{f 2}^{2} \\ = & 9 - 12 v_{f 2} + 3 v_{f 2}^{2} \\ = & 3 - 4 v_{f 2} + v_{f 2}^{2} \\ = & (v_{f 2} - 3) (v_{f 2} - 1) \end{matrix}$

Substituting back into Equation [link] , we get:

$\begin{matrix} v_{f 1} & = & \frac{m_{2}}{m_{1}} (v_{i 2} - v_{f 2}) \\ = & \frac{0.1}{0.05} (3 - 3) \\ = & 0 m \cdot s^{- 1} \\ or \\ v_{f 1} & = & \frac{m_{2}}{m_{1}} (v_{i 2} - v_{f 2}) \\ = & \frac{0.1}{0.05} (3 - 1) \\ = & 4 m \cdot s^{- 1} \end{matrix}$

But according to the question, marble 1 is moving after the collision, therefore marble 1 moves to the right at 4 m $\cdot$ s $^{- 1}$ . Substituting this value for $v_{f 1}$ into Equation [link] , we can calculate:

$\begin{matrix} v_{f 2} & = & \frac{0, 3 - 0, 05 v_{f 1}}{0, 1} \\ = & \frac{0, 3 - 0, 05 \times 4}{0, 1} \\ = & 1 m \cdot s^{- 1} \end{matrix}$

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

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Two billiard balls each with a mass of $150 g$ collide head-on in an elastic collision. Ball 1 was travelling at a speed of $2 m \cdot s^{- 1}$ and ball 2 at a speed of $1, 5 m \cdot s^{- 1}$ . After the collision, ball 1 travels away from ball 2 at a velocity of $1, 5 m \cdot s^{- 1}$ .

Calculate the velocity of ball 2 after the collision.
Prove that the collision was elastic. Show calculations.

1. Draw a rough sketch of the situation
2. 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.
3. Solve problem
  $\begin{matrix} p_{b e f o r e} & = & p_{a f t e r} \\ m_{1} v_{i 1} + m_{2} v_{i 2} & = & m_{1} v_{f 1} + m_{2} v_{f 2} \\ (\frac{150}{1000}) (2) + (\frac{150}{1000}) (- 1, 5) & = & (\frac{150}{1000}) (- 1, 5) + (\frac{150}{1000}) (v_{f 2}) \\ 0, 3 - 0, 225 & = & - 0, 225 + 0, 15 v_{f 2} \\ v_{f 2} & = & 3 m \cdot s^{- 1} \end{matrix}$
  So after the collision, ball 2 moves with a velocity of $3 m \cdot s^{- 1}$ .
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. Calculating the initial total kinetic energy:
$\begin{matrix} E K_{b e f o r e} & = & \frac{1}{2} m_{1} v_{i 1}^{2} + \frac{1}{2} m_{2} v_{i 2}^{2} \\ = & (\frac{1}{2}) (0, 15) {(2)}^{2} + (\frac{1}{2}) (0, 15) {(- 1, 5)}^{2} \\ = & 0.469 . . . . J \end{matrix}$
Calculating the final total kinetic energy:
$\begin{matrix} E K_{a f t e r} & = & \frac{1}{2} m_{1} v_{f 1}^{2} + \frac{1}{2} m_{2} v_{f 2}^{2} \\ = & (\frac{1}{2}) (0, 15) {(- 1, 5)}^{2} + (\frac{1}{2}) (0, 15) {(2)}^{2} \\ = & 0.469 . . . . J \end{matrix}$
So $E K_{b e f o r e} = E K_{a f t e r}$ and hence the collision is elastic.

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Source: OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2

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