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S4.5 Linear momentum and its conservation

[A] Linear momentum of a system of particles

Using the velocity of the centre of mass (Key Point 4.2), and the definition of linear momentum for a single particle (Key Point 4.5), we can identify the total linear momentum of the system $\vec{P}_{tot}$ as follows:

Key Point 4.8

Linear momentum of centre of mass is defined as
\[ \vec{P}_{tot} = M \vec{v}_{cm} = \vec{p}_1 + \vec{p}_2 + ...\, \vec{p}_n \]

[B] Newton’s 2nd law

We can now express Newton’s 2nd law in terms of the momentum.

Key Point 4.9

Newton’s 2nd law for a system of particles is:

\[ \vec{F}_{ext} = \frac{d \vec{P}_{tot}}{dt} \]

Worked example

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Worked example

A 2kg body and a 3kg body are moving along the x-axis. At a particular instant the 2kg body is 1m from the origin and has a velocity of 3ms-1 and the 3kg body is 2m from the origin and has a velocity of -1ms-1. Find the position and velocity of the center of mass and also find the total momentum.

Solution:

The CoM is found from KP 5.1.

\begin{eqnarray*} %%%Was makeeqnarraystar; label was weg1 \vec{x}_{cm} & = &{\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(2 kg \times 1 m + 3 kg \times 2 m)/5 kg}} \\ & = & 1.6 m \end{eqnarray*}

The velocity of the CoM is found from KP 5.2

\begin{eqnarray*} %%%Was makeeqnarraystar; label was weg2 \vec{v}_{cm} & = &{\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(2 kg \times 3 ms^{-1} + 3 kg \times -1 ms^{-1})/5 kg}} \\ & = & 0.6 ms^{-1} \end{eqnarray*}

The total momentum is found from KP 5.8

\begin{eqnarray*} %%%Was makeeqnarraystar; label was weg3 \vec{P}_{tot} & = &{\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{2 kg \times 3 ms^{-1} + 3 kg \times -1 ms^{-1}}} \\ & = & 3 \;kg\; ms^{-1} \end{eqnarray*}
 

[C] Conservation of Linear Momentum

If the total external force acting on a system is zero then we can see that the total linear momentum of the system must be a constant. This is the law of conservation of linear momentum.

Key Point 4.10

Conservation of Linear Momentum: if a system of particles is isolated from its surroundings then linear momentum is conserved:

\[ \frac{d \vec{P}_{tot}}{d t} = 0 \Rightarrow \vec{P}_{tot} = {\rm constant} \]

Note: Since linear momentum is a vector quantity, each component of linear momentum is conserved separately. If a component of the net external force on a system is zero along an axis, then the component of linear momentum of the system along that axis cannot change.

[D] Isolated systems

Saying a system is isolated is really another way of saying that the total external force is zero; hence that linear momentum is conserved within the system.

Examples of isolated (or approximately isolated) systems:

When trying to apply the law of conservation of linear momentum, you should make sure the system you choose is isolated (at least in the direction you want). Additionally, if you want to apply the law of conservation of energy, the system must be closed (no particles enter or leave the system).

Worked example

Worked example

A 50kg child is riding on a 40kg go-cart moving at a speed of 3m/s. He jumps off with zero horizontal speed. What is the resulting change in the speed of the go-cart?

Solution:

Taking our system to be the child plus the cart and assuming negligible friction on the wheels of the cart, then linear momentum will be conserved i.e. $\vec{p}^{tot}_f=\vec{p}^{tot}_i$.

\[ \vec{p}^{tot}_i={\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(50 kg \times 3 ms^{-1} + 40 kg\times 3 ms^{-1})=270}} kg ms^{-1} \]

and

\[ \vec{p}^{tot}_f={\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{(50 kg \times 0+40kg\times \vec{v}^c_f)}} \]
where $\vec{v}^c_f$ is the final velocity of the cart.

Hence

\[ \vec{v}_f^c=6.75 ms^{-1} \]

and the change in speed is 3.75ms-1.

 

Learning Resources

Textbook: HRW Chapter 9.7
Course Questions:
Self-Test Questions: