Vector Nature of Momentum

A soccer ball of mass \(\text{420}\) \(\text{g}\) is kicked at \(\text{20}\) \(\text{m·s$^{-1}$}\) towards the goal post. Calculate the momentum of the ball.

Step 1: Identify what information is given and what is asked for

The question explicitly gives:

• the mass of the ball, and

• the velocity of the ball.

The mass of the ball must be converted to SI units.

\(\text{420}\text{ g}=\text{0.42}\text{ kg}\)

We are asked to calculate the momentum of the ball. From the definition of momentum, \(\vec{p}=m\vec{v}\) we see that we need the mass and velocity of the ball, which we are given.

Step 2: Do the calculation

We calculate the magnitude of the momentum of the ball,

\begin{align*} \vec{p}& = m\vec{v} \\& = \left(\text{0.42}\right)\left(20\right) \\& = \text{8.40}\text{ kg·m·s$^{-1}$}\end{align*}

Step 3: Quote the final answer

We quote the answer with the direction of motion included,\(\vec{p}\) = \(\text{8.40}\) \(\text{kg·m·s$^{-1}$}\) in the direction of the goal post.

A cricket ball of mass \(\text{160}\) \(\text{g}\) is bowled at \(\text{40}\) \(\text{m·s$^{-1}$}\) towards a batsman. Calculate the momentum of the cricket ball.

Step 1: Identify what information is given and what is asked for

The question explicitly gives

• the mass of the ball (m = \(\text{160}\) \(\text{g}\) = \(\text{0.16}\) \(\text{kg}\)), and

• the velocity of the ball \((\vec{v}\) = \(\text{40}\) \(\text{m·s$^{-1}$}\) towards the batsman)

To calculate the momentum we will use

\(\vec{p}=m\vec{v}.\)

Step 2: Do the calculation

\begin{align*} \vec{p}& = m\vec{v} \\ & = \left(\text{0.16}\right)\left(40\right) \\ & = \text{6.4}\text{ kg·m·s$^{-1}$} \\ & = \text{6.4}\text{ kg·m·s$^{-1}$} \text{in the direction of the batsman}\end{align*}

Step 3: Quote the final answer

The momentum of the cricket ball is \(\text{6.4}\) \(\text{kg·m·s$^{-1}$}\) in the direction of the batsman.

The centre of the Moon is approximately \(\text{384 400}\) \(\text{km}\) away from the centre of the Earth and orbits the Earth in \(\text{27.3}\) days. If the Moon has a mass of \(\text{7.35} \times \text{10}^{\text{22}}\) \(\text{kg}\), what is the magnitude of its momentum (using the definition given in this tutorial) if we assume a circular orbit? The actual momentum of the Moon is more complex but we do not cover that in this tutorial.

Step 1: Identify what information is given and what is asked for

The question explicitly gives

• the mass of the Moon (\(m\) = \(\text{7.35} \times \text{10}^{\text{22}}\) \(\text{kg}\))

• the distance to the Moon (\(\text{384 400}\) \(\text{km}\) = \(\text{384 400 000}\) \(\text{m}\) = \(\text{3.844} \times \text{10}^{\text{8}}\) \(\text{m}\))

• the time for one orbit of the Moon (\(\text{27.3} \text{ days} = \text{27.3} \times 24 \times 60 \times 60 = \text{2.36} \times \text{10}^{\text{6}}\text{ s}\))

We are asked to calculate only the magnitude of the momentum of the Moon (i.e. we do not need to specify a direction). In order to do this we require the mass and the magnitude of the velocity of the Moon, since

\(\vec{p}=m\vec{v}.\)

Step 2: Find the magnitude of the velocity of the Moon

The magnitude of the average velocity is the same as the speed. Therefore:

\(v=\frac{\Delta x}{\Delta t}\)

We are given the time the Moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the Moon and the fact that the Moon has a circular orbit. Using the equation for the circumference, C, of a circle in terms of its radius, we can determine the distance travelled by the Moon in oneorbit:

\begin{align*} C& = 2\pi r \\ & = 2\pi \left(\text{3.844} \times \text{10}^{\text{8}}\right) \\& = \text{2.42} \times \text{10}^{\text{9}}\text{ m}\end{align*}

Combining the distance travelled by the Moon in an orbit and the timetaken by the Moon to complete one orbit, we can determine themagnitude of the Moon's velocity or speed,

\begin{align*} v& = \frac{\Delta x}{\Delta t} \\ & = \frac{C}{T} \\ & = \frac{\text{2.42} \times \text{10}^{\text{9}}\text{ m}}{\text{2.36} \times \text{10}^{\text{6}}\text{ s}} \\ & = \text{1.02} \times \text{10}^{\text{3}}\text{ m·s$^{-1}$}\end{align*}

Step 3: Finally calculate the momentum and quote the answer

The magnitude of the Moon's momentum is:

\begin{align*} \vec{p}& = m\vec{v} \\ {p}& = m{v} \\& = \left(\text{7.35} \times \text{10}^{\text{22}}\right)\left(\text{1.02} \times \text{10}^{\text{3}}\right) \\& = \text{7.50} \times \text{10}^{\text{25}}\text{ kg·m·s$^{-1}$}\end{align*}

The magnitude of the momentum of the Moon is \(\text{7.50} \times \text{10}^{\text{25}}\) \(\text{kg·m·s$^{-1}$}\).