Question

drag and match each equation to the step that produces it.

step resulting form of newtons second law
multiply both sides by ( m ) ( sum vec { f } = \frac { delta m vec { v } } { delta t } )
substitute using ( vec { a } = \frac { delta vec { v } } { delta t } ) ( sum vec { f } = m vec { a } )
move the scalar ( m ) inside ( delta vec { v } ) ( sum vec { f } = dot { m } \frac { delta vec { v } } { delta t } )
substitute using ( vec { p } = m vec { v } ) ( sum vec { f } = \frac { delta vec { p } } { delta t } )

Explanation:

Step1: Multiply both sides by \(m\)

When we start with \( \sum\vec{F}=\frac{\Delta\vec{v}}{\Delta t}\) (original form of Newton's second law if we consider \(a = \frac{\Delta\vec{v}}{\Delta t}\) and assume \(m = 1\) initially for the sake of showing the multiplication step conceptually), multiplying both sides by \(m\) gives \( \sum\vec{F}=m\frac{\Delta\vec{v}}{\Delta t}\)

Step2: Substitute using \( \vec{a}=\frac{\Delta\vec{v}}{\Delta t}\)

If we substitute \( \vec{a}=\frac{\Delta\vec{v}}{\Delta t}\) into \( \sum\vec{F}=m\frac{\Delta\vec{v}}{\Delta t}\), we get \( \sum\vec{F}=m\vec{a}\)

Step3: Move the scalar \(m\) inside \( \Delta\vec{v}\)

Since \(m\) is a scalar and \( \Delta(m\vec{v})=m\Delta\vec{v}\) (by the property of scalar - vector multiplication for finite differences), \( \sum\vec{F}=m\frac{\Delta\vec{v}}{\Delta t}\) can be written as \( \sum\vec{F}=\frac{\Delta(m\vec{v})}{\Delta t}\)

Step4: Substitute using \( \vec{p}=m\vec{v}\)

Substituting the momentum definition \( \vec{p} = m\vec{v}\) into \( \sum\vec{F}=\frac{\Delta(m\vec{v})}{\Delta t}\), we get \( \sum\vec{F}=\frac{\Delta\vec{p}}{\Delta t}\)

Answer:

• Multiply both sides by \(m\): \( \sum\vec{F}=m\frac{\Delta\vec{v}}{\Delta t}\)
• Substitute using \( \vec{a}=\frac{\Delta\vec{v}}{\Delta t}\): \( \sum\vec{F}=m\vec{a}\)
• Move the scalar \(m\) inside \( \Delta\vec{v}\): \( \sum\vec{F}=\frac{\Delta(m\vec{v})}{\Delta t}\)
• Substitute using \( \vec{p}=m\vec{v}\): \( \sum\vec{F}=\frac{\Delta\vec{p}}{\Delta t}\)