6.1 Conservation of momentum

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Conservation of momentum in two dimensions

We have seen in Grade 11 that the momentum of a system is conserved when there are no external forces acting on the system. Conversely, an external force causes a change in momentum $Δ p$ , with the impulse delivered by the force, $F$ acting for a time $Δ t$ given by:

Δ p = F \cdot Δ t

The same principles that were studied in applying the conservation of momentum to problems in one dimension, can be applied to solving problems in two dimensions.

The calculation of momentum is the same in two dimensions as in one dimension. The calculation of momentum in two dimensions is broken down into determining the $x$ and $y$ components of momentum and applying the conservation of momentum to each set of components.

Consider two objects moving towards each other as shown in [link] . We analyse this situation by calculating the $x$ and $y$ components of the momentum of each object.

Before the collision

Total momentum:

\begin{matrix} p_{i 1} & = & m_{1} v_{i 1} \\ p_{i 2} & = & m_{2} v_{i 2} \end{matrix}

$x$ -component of momentum:

\begin{matrix} p_{i 1 x} & = & m_{1} v_{i 1 x} = m_{1} v_{i 1} cos θ_{1} \\ p_{i 2 x} & = & m_{2} u_{i 2 x} = m_{2} v_{i 2} cos θ_{2} \end{matrix}

$y$ -component of momentum:

\begin{matrix} p_{i 1 y} & = & m_{1} v_{i 1 y} = m_{1} v_{i 1} sin θ_{1} \\ p_{i 2 y} & = & m_{2} v_{i 2 y} = m_{2} v_{i 2} sin θ_{2} \end{matrix}

After the collision

Total momentum:

\begin{matrix} p_{f 1} & = & m_{1} v_{f 1} \\ p_{f 2} & = & m_{2} v_{f 2} \end{matrix}

$x$ -component of momentum:

\begin{matrix} p_{f 1 x} & = & m_{1} v_{f 1 x} = m_{1} v_{f 1} cos φ_{1} \\ p_{f 2 x} & = & m_{2} v_{f 2 x} = m_{2} v_{f 2} cos φ_{2} \end{matrix}

$y$ -component of momentum:

\begin{matrix} p_{f 1 y} & = & m_{1} v_{f 1 y} = m_{1} v_{f 1} sin φ_{1} \\ p_{f 2 y} & = & m_{2} v_{f 2 y} = m_{2} v_{f 2} sin φ_{2} \end{matrix}

Conservation of momentum

The initial momentum is equal to the final momentum:

p_{i} = p_{f}

\begin{matrix} p_{i} & = & p_{i 1} + p_{i 2} \\ p_{f} & = & p_{f 1} + p_{f 2} \end{matrix}

This forms the basis of analysing momentum conservation problems in two dimensions.

Phet simulation for momentum

In a rugby game, Player 1 is running with the ball at 5 m $\cdot$ s $^{- 1}$ straight down the field parallel to the edge of the field. Player 2 runs at 6 m $\cdot$ s $^{- 1}$ an angle of $60^{\circ}$ to the edge of the field and tackles Player 1. In the tackle, Player 2 stops completely while Player 1 bounces off Player 2. Calculate the velocity (magnitude and direction) at which Player 1 bounces off Player 2. Both the players have a mass of 90 kg.

Identify what is required and what is given
The first step is to draw the picture to work out what the situation is. Mark the initial velocities of both players in the picture.

We also know that $m_{1} = m_{2} = 90$ kg and $v_{f 2}$ = 0 ms $^{- 1}$ .

We need to find the final velocity and angle at which Player 1 bounces off Player 2.
Use conservation of momentum to solve the problem. First find the initial total momentum:
Total initial momentum = Total final momentum. But we have a two dimensional problem, and we need to break up the initial momentum into $x$ and $y$ components.

$\begin{matrix} p_{i x} & = & p_{f x} \\ p_{i y} & = & p_{f y} \end{matrix}$

For Player 1:

$\begin{matrix} p_{i x 1} & = & m_{1} v_{i 1 x} = 90 \times 0 = 0 \\ p_{i y 1} & = & m_{1} v_{i 1 y} = 90 \times 5 \end{matrix}$

For Player 2:

$\begin{matrix} p_{i x 2} & = & m_{2} v_{i 2 x} = 90 \times 8 \times sin 60^{\circ} \\ p_{i y 2} & = & m_{2} v_{i 2 y} = 90 \times 8 \times cos 60^{\circ} \end{matrix}$
Now write down what we know about the final momentum:
For Player 1:

$\begin{matrix} p_{f x 1} & = & m_{1} v_{f x 1} = 90 \times v_{f x 1} \\ p_{f y 1} & = & m_{1} v_{f y 1} = 90 \times v_{f y 1} \end{matrix}$

For Player 2:

$\begin{matrix} p_{f x 2} & = & m_{2} v_{f x 2} = 90 \times 0 = 0 \\ p_{f y 2} & = & m_{2} v_{f y 2} = 90 \times 0 = 0 \end{matrix}$
Use conservation of momentum:
The initial total momentum in the $x$ direction is equal to the final total momentum in the $x$ direction.

The initial total momentum in the $y$ direction is equal to the final total momentum in the $y$ direction.

If we find the final $x$ and $y$ components, then we can find the final $t o t a l$ momentum.

$\begin{matrix} p_{i x 1} + p_{i x 2} & = & p_{f x 1} + p_{f x 2} \\ 0 + 90 \times 8 \times sin 60^{\circ} & = & 90 \times v_{f x 1} + 0 \\ v_{f x 1} & = & \frac{90 \times 8 \times sin 60^{\circ}}{90} \\ v_{f x 1} & = & 6.928 {ms}^{- 1} \end{matrix}$

$\begin{matrix} p_{i y 1} + p_{i y 2} & = & p_{f y 1} + p_{f y 2} \\ 90 \times 5 + 90 \times 8 \times cos 60^{\circ} & = & 90 \times v_{f y 1} + 0 \\ v_{f y 1} & = & \frac{90 \times 5 + 90 \times 8 \times cos 60^{\circ}}{90} \\ v_{f y 1} & = & 9.0 {ms}^{- 1} \end{matrix}$
Using the $x$ and $y$ components, calculate the final total $v$
Use Pythagoras's theorem to find the total final velocity:

$\begin{matrix} v_{f t o t} & = & \sqrt{v_{f x 1}^{2} + v_{f y 1}^{2}} \\ = & \sqrt{6 . 928^{2} + 9^{2}} \\ = & 11.36 \end{matrix}$

Calculate the angle $θ$ to find the direction of Player 1's final velocity:

$\begin{matrix} sin θ & = & \frac{v_{f y 1}}{v_{f t o t}} \\ θ & = & 52 . 4^{\circ} \end{matrix}$

Therefore Player 1 bounces off Player 2 with a final velocity of 11.36 m $\cdot s^{- 1}$ at an angle of 52.4 $^{\circ}$ from the horizontal.

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In a soccer game, Player 1 is running with the ball at 5 m $\cdot$ s $^{- 1}$ across the pitch at an angle of $75^{\circ}$ from the horizontal. Player 2 runs towards Player 1 at 6 m $\cdot$ s $^{- 1}$ an angle of $60^{\circ}$ to the horizontal and tackles Player 1. In the tackle, the two players bounce off each other. Player 2 moves off with a velocity in the opposite $x$ -direction of 0.3 m $\cdot$ s $^{- 1}$ and a velocity in the $y$ -direction of 6 m $\cdot$ s $^{- 1}$ . Both the players have a mass of 80 kg.What is the final total velocity of Player 1?

Identify what is required and what is given
The first step is to draw the picture to work out what the situation is. Mark the initial velocities of both players in the picture.

We need to define a reference frame: For y, choose the direction they are both running in as positive. For x, the direction player 2 is running in is positive.

We also know that $m_{1} = m_{2} = 80$ kg. And $v_{f x 2}$ =-0.3 ms $^{- 1}$ and $v_{f y 2}$ =6 ms $^{- 1}$ .

We need to find the final velocity and angle at which Player 1 bounces off Player 2.
Use conservation of momentum to solve the problem. First find the initial total momentum:
Total initial momentum = Total final momentum. But we have a two dimensional problem, and we need to break up the initial momentum into $x$ and $y$ components. Remember that momentum is a vector and has direction which we will indicate with a '+' or '-' sign.

$\begin{matrix} p_{i x} & = & p_{f x} \\ p_{i y} & = & p_{f y} \end{matrix}$

For Player 1:

$\begin{matrix} p_{i x 1} & = & m_{1} v_{i 1 x} = 80 \times (- 5) \times cos 75^{\circ} \\ p_{i y 1} & = & m_{1} v_{i 1 y} = 80 \times 5 \times sin 75^{\circ} \end{matrix}$

For Player 2:

$\begin{matrix} p_{i x 2} & = & m_{2} v_{i 2 x} = 80 \times 6 \times cos 60^{\circ} \\ p_{i y 2} & = & m_{2} v_{i 2 y} = 80 \times 6 \times sin 60^{\circ} \end{matrix}$
Now write down what we know about the final momentum:
For Player 1:

$\begin{matrix} p_{f x 1} & = & m_{1} v_{f x 1} = 80 \times v_{f x 1} \\ p_{f y 1} & = & m_{1} v_{f y 1} = 80 \times v_{f y 1} \end{matrix}$

For Player 2:

$\begin{matrix} p_{f x 2} & = & m_{2} v_{f x 2} = 80 \times (- 0.3) \\ p_{f y 2} & = & m_{2} v_{f y 2} = 80 \times 6 \end{matrix}$
Use conservation of momentum:
The initial total momentum in the $x$ direction is equal to the final total momentum in the $x$ direction.

The initial total momentum in the $y$ direction is equal to the final total momentum in the $y$ direction.

If we find the final $x$ and $y$ components, then we can find the final $t o t a l$ momentum.

$\begin{matrix} p_{i x 1} + p_{i x 2} & = & p_{f x 1} + p_{f x 2} \\ - 80 \times 5 cos 75^{\circ} + 80 \times 6 \times cos 60^{\circ} & = & 80 \times v_{f x 1} + 80 \times (- 0.3) \\ v_{f x 1} & = & \frac{- 80 \times 5 cos 75^{\circ} + 80 \times 6 \times cos 60^{\circ} - 80 \times (- 0.3)}{80} \\ v_{f x 1} & = & 2.0 {ms}^{- 1} \end{matrix}$

$\begin{matrix} p_{i y 1} + p_{i y 2} & = & p_{f y 1} + p_{f y 2} \\ 80 \times 5 \times sin 75^{\circ} + 80 \times 6 \times sin 60^{\circ} & = & 80 \times v_{f y 1} + 80 \times 6 \\ v_{f y 1} & = & \frac{80 \times 5 sin 75^{\circ} + 80 \times 6 \times sin 60^{\circ} - 80 \times 6}{80} \\ v_{f y 1} & = & 4.0 {ms}^{- 1} \end{matrix}$
Using the $x$ and $y$ components, calculate the final total $v$
Use Pythagoras's theorem to find the total final velocity:

$\begin{matrix} v_{f t o t} & = & \sqrt{v_{f x 1}^{2} + v_{f y 1}^{2}} \\ = & \sqrt{2^{2} + 4^{2}} \\ = & 4.5 \end{matrix}$

Calculate the angle $θ$ to find the direction of Player 1's final velocity:

$\begin{matrix} tan θ & = & \frac{v_{f y 1}}{v_{f x 1}} \\ θ & = & 63 . 4^{\circ} \end{matrix}$

Therefore Player 1 bounces off Player 2 with a final velocity of 4.5 m $\cdot s^{- 1}$ at an angle of 63.4 $^{\circ}$ from the horizontal.

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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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