# FHSST Physics/Momentum/Definition

Momentum The Free High School Science Texts: A Textbook for High School Students Studying Physics. Main Page - << Previous Chapter (Rectilinear Motion) - Next Chapter (Work and Energy) >> Definition - Momentum of a System - Change in Momentum - Properties - Impulse - Important Quantities, Equations, and Concepts

# What is Momentum?Edit

Momentum is a physical quantity which is closely related to forces. We will learn about this connection a little later. Remarkably momentum is a conserved quantity. This makes momentum extremely useful in solving a great variety of real-world problems. Firstly we must consider the definition of momentum.

 Definition: The momentum of an object is defined as its mass multiplied by its velocity.

Mathematically,

 ${\displaystyle {\overrightarrow {p}}=m{\overrightarrow {v}}}$ ${\displaystyle {\overrightarrow {p}}}$ : momentum (${\displaystyle kg.m.s^{-1}}$  + direction) m : mass (kg) ${\displaystyle {\overrightarrow {v}}}$ : velocity (m.s-1 + direction)

Thus, momentum is a property of a moving object and is determined by its velocity and mass. A large truck travelling slowly can have the same momentum as a much smaller car travelling relatively fast.

Note the arrows in the equation defining momentum- momentum is a vector with the same direction as the velocity of the object.

Since the direction of an object's momentum is given by the direction of its motion, one can calculate an object's momentum in two steps:

• include in the final answer the direction of the object's motion

## Worked Example 31 Calculating Momentum 1Edit

Question: A ball of mass 3kg moves at 2m.s−1 to the right. Calculate the ball's momentum.

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

• the ball's mass, and
• the ball's velocity

in the correct units!

Step 2 :

What is being asked? We are asked to calculate the ball's momentum. From the definition of momentum,

${\displaystyle {\begin{matrix}{\overrightarrow {p}}=m{\overrightarrow {v}},\end{matrix}}}$

we see that we need the mass and velocity of the ball, which we are given.

Step 3 :

Firstly we calculate the magnitude of the ball's momentum,

${\displaystyle {\begin{matrix}p&=&mv\\&=&(3kg)(2m.s^{-1})=6\ kg.m.s^{-1}.\end{matrix}}}$

Finally we quote the answer with the direction of the ball's motion included,

${\displaystyle {\begin{matrix}{\overrightarrow {p}}=6\ kg.m.s^{-1}{\textbf {\ to\ the\ right}}\end{matrix}}}$

## Worked Example 32 Calculating Momentum 2Edit

Question: A ball of mass 500g is thrown at 2m.s−1. Calculate the ball's momentum.

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

• the ball's mass, and
• the magnitude of the ball's velocity

but with the ball's mass in the incorrect units!

Step 2 :

What is being asked? We are asked to calculate the momentum which is defined as

${\displaystyle {\begin{matrix}{\overrightarrow {p}}=m{\overrightarrow {v}}.\end{matrix}}}$

Thus, we need the mass and velocity of the ball but we have only its mass and the magnitude of its velocity.

Step 3 :

In order to determine the velocity of the ball we need the direction of the ball's motion. If the problem does not give an explicit direction we are forced to be general. In a case like this we could say that the direction of the velocity is in the direction of motion of the ball. This might sound silly but the lack of information in the question has forced us to be, and we are certainly not wrong! The ball's velocity is then m.s−1 in the direction of motion.

Step 4 :

Next we convert the mass to the correct units,

${\displaystyle {\begin{matrix}1000g&=&1kg\\1&=&{\frac {1kg}{1000g}}\\500g\times 1&=&500g\times {\frac {1kg}{1000g}}\\&=&0.500kg\end{matrix}}}$

Step 5 :

Now, let us find the magnitude of the ball's momentum,

${\displaystyle {\begin{matrix}p&=&mv\\&=&(0.500kg)(2m.s^{-1})=1\ kg.m.s^{-1}\end{matrix}}}$

Step 6 :

Finally, we quote the answer with the direction of the momentum included,

${\displaystyle {\begin{matrix}{\overrightarrow {p}}&=&1\ kg.m.s^{-1}{\textbf {\ in\ the\ direction\ of\ motion\ of\ the\ ball}}\end{matrix}}}$

## Worked Example 33 Calculating the Momentum of the MoonEdit

Question: The moon is ${\displaystyle 384\ 400km}$  away from the earth and orbits the earth in 27.3 days. If the moon has a mass of ${\displaystyle 7.35\times 10^{22}kg}$  [footnode.html#foot11743 6.1] what is the magnitude of its momentum if we assume a circular orbit?

Step 1 :

Analyse the question to determine what is given. The question explicitly gives

• the moon's mass,
• the distance to the moon, and
• the time for one orbit of the moon

with mass in the correct units but all other quantities in the incorrect units.

The units we require are

• seconds (s) for time, and
• metres (m) for distance pesteng physics→→→→→→

Step 2 :

What is being asked? We are asked to calculate only the magnitude of the moon's momentum (i.e. we do not need to specify a direction). In order to do this we require the moon's mass and the magnitude of its velocity, since ${\displaystyle {\begin{matrix}p=mv.\end{matrix}}}$

Step 3 :

How do we find the speed or magnitude of the moon's velocity? Speed is defined as,

${\displaystyle {\begin{matrix}speed&=&{\frac {Distance}{time}}\end{matrix}}}$

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's orbit is circular. Firstly let us convert the distance to the moon to the correct units,

${\displaystyle {\begin{matrix}1km&=&1000m\\1&=&{\frac {1000m}{1km}}\\384\ 400km\times 1&=&384\ 400km\times {\frac {1000m}{1km}}\\&=&384\ 400\ 000m\\&=&3.844\times 10^{8}\ m\end{matrix}}}$

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 one orbit:

${\displaystyle {\begin{matrix}C&=&2\pi r\\&=&2\pi (3.844\times 10^{8}\ m)\\&=&2.42\times 10^{9}\ m.\\\end{matrix}}}$

Next we must convert the orbit time, T, into the correct units. Using the fact that a day contains 24 hours, an hour consists of 60 minutes, and a minute is 60 seconds long,

${\displaystyle {\begin{matrix}1day&=&(24)(60)(60)seconds\\1&=&{\frac {(24)(60)(60)s}{1day}}\\27.3days\times 1&=&27.3days{\frac {(24)(60)(60)s}{1day}}\\&=&2.36\times 10^{6}s\end{matrix}}}$

Therefore, ${\displaystyle {\begin{matrix}T=2.36\times 10^{6}s.\end{matrix}}}$

Combining the distance travelled by the moon in an orbit and the time taken by the moon to complete one orbit, we can determine the magnitude of the moon's velocity or speed,

${\displaystyle {\begin{matrix}v&=&{\frac {Distance}{time}}\\&=&{\frac {C}{T}}\\&=&1.02\times 10^{3}\ m.s^{-1}.\end{matrix}}}$

Step 4 :

Finally we can calculate the magnitude of the moon's momentum,

${\displaystyle {\begin{matrix}p&=&mv\\&=&(7.35\times 10^{22}kg)(1.02\times 10^{3}\ m.s^{-1})\\&=&7.50\times 10^{25}\ kg.m.s^{-1}.\end{matrix}}}$
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