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S4.4 Linear momentum

[A] Definition

Key Point 4.5

The Linear Momentum of a single particle is defined as the product of the particle’s mass m, and its velocity $\vec{v}$:

\[ \vec{p} = m \vec{v} \]

Note: this is a vector equation.

[B] Newton’s 2nd Law

Key Point 4.6

Newton’s 2nd Law is most economically expressed in terms of the momentum:

\[ \vec{F} = \frac{d \vec{p}}{d t} \]

[C] Integral form

When a force acts on a body, the momentum gained by the body is just the sum (or integral) of all the forces acting over time. This allows us to define an average force in terms of the momentum.

\[ \int_0^t \vec{F}(t) dt = \vec{p}(t) - \vec{p}(0) = \vec{\Delta p} \]

Key Point 4.7

The average force acting on a body over time Δt producing momentum change $\vec{\Delta p}$ is defined as:

\[ \vec{F}_{av} \Delta t = \vec{\Delta p} \]

Worked example

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Worked example

A cricket ball weighs 0.1 kg and is bowled at 30ms-1. It is batted at 50ms-1 in the opposite direction. Find the change in momentum of the ball. If the bat and ball are in contact for 0.002s, find the average force exerted.

Solution:

Change in momentum is:

\begin{eqnarray*} %%%Was makeeqnarraystar; label was ball1 \Delta \vec{p} &=& \vec{p_f} - \vec{p_i} \\ &=& {\xmlInlineElement[\xmlAttr{}{target}{slides}]{http://www.ph.ed.ac.uk/aardvark/NS/aardvark-latex}{aardvark:reveal}{50 ms^{-1} \times 0.1 kg - (-30 ms^{-1} \times 0.1 kg)}}\\ &=& 8 \;kg \;ms^{-1} {\rm (~in~direction~of~batted~ball)}\\ \end{eqnarray*}

The average force is

\begin{eqnarray*} %%%Was makeeqnarraystar; label was ball2 \vec{F}_{av} & = & \frac{\Delta \vec{p}}{\Delta t}\\ & = & 4\times 10^3 N \\ \end{eqnarray*}
 
TTrying to Stop Rolling Balls

Learning Resources

Textbook: HRW Chapter 9.4, 9.5
Course Questions:
Self-Test Questions: