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 :
![\[ \vec{p} = m \vec{v} \]](mastermathpng-1.png)
Note: this is a vector equation.
[B] Newton’s 2nd Law
This is identical to the previous definition (Key Point 2.2), provided the mass of the particle is constant
- This form of the 2nd law is useful when the mass of a body changes with time, for instance when a rocket takes off.
[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.
![\[ \int_0^t \vec{F}(t) dt = \vec{p}(t) - \vec{p}(0) = \vec{\Delta p} \]](mastermathpng-4.png)
Key Point 4.7
The average force acting on a body over time Δt
producing momentum change is defined as:
![\[ \vec{F}_{av} \Delta t = \vec{\Delta p} \]](mastermathpng-6.png)
Worked example
Reveal

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*}](mastermathpng-7.png)
The average force is

Learning Resources
![]() | HRW Chapter 9.4, 9.5 |
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