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- Physics Tutorial: Linear Momentum
- Physics Tutorial: Moment of Force. Conditions of Equilibrium
- Physics Tutorial: Centre of Mass. Types of Equilibrium
- Physics Tutorial: Momentum and Impulse in Two Dimensions. Explosions.
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In this Physics tutorial, you will learn:

- The definition of linear momentum
- How to find linear momentum of a moving object
- What does linear momentum represent?
- Where does the linear momentum differ from the moment of force?

Suppose you throw horizontally a tennis ball towards a wall. It is obvious the tennis ball will turn back after hitting the wall. We don't expect it turn back at the same speed because it loses some kinetic energy during the collision - energy which turns into other forms such as heat, sound energy etc. Therefore, the speed after the collision may be slightly smaller than before hitting the wall.

Now think about the same tennis ball hitting a very thin sheet of paper. It is obvious that the tennis ball will pierce the sheet and move at the same direction as before. Sure, the ball may encounter some small resistance from the sheet and as a result, its speed may be slightly smaller than before. Hence, we may obtain the same value for the speed as in the case when the tennis ball hits the hard wall and turns back.

If we use the energetic approach, we get the same value for the initial and final kinetic energy in both cases, as energy is a scalar and as such, it doesn't involve any direction. Let's take some numbers only for illustration purpose. Thus, if we take the mass of the ball equal to 200 g (0.2 kg) and if the ball's initial speed in both cases is equal to 4 m/s and its final speed (in both cases again) is 3 m/s, we obtain for the initial kinetic energy in both cases,

KE_{initial} = *m × v*^{2}_{initial}*/**2* = *0.2 × 4*^{2}*/**2* = 1.6 J

and for the final kinetic energy (in both cases),

KE_{initial} = *m × v*^{2}_{final}*/**2* = *0.2 × 3*^{2}*/**2* = 0.9 J

Therefore, it would not make any difference whether the ball hits a wall or a sheet when dealing with mass and speed. We are only able to calculate the decrease in kinetic energy (1.6 J - 0.9 J = 0.7 J) which represents the amount of kinetic energy that has been converted into other forms during the collision.

Even if we used the respective velocities instead of speeds, the result would not change, as velocity is raised at power two in the formula of kinetic energy and therefore, we would obtain always a positive result, regardless the direction of velocity.

Thus, the energetic approach would be not so useful in making a thorough investigation of such events. If no figure or written information is provided, we are not able to detect whether the ball encounters a hard or a soft obstacle.

Hence, we need another approach, which besides the values of mass and velocity, consider the direction of motion as well. Such an approach, will be discussed and explained in the following paragraph.

By definition, Linear Momentum (p*⃗*) is the product of an object's mass (m) and its velocity (v*⃗*).

Mathematically, we have

p*⃗* = m × v*⃗*

Since mass is measured in kg and velocity in m/s, the unit of momentum is [kg × m/s].

It is obvious that momentum is a vector quantity as it is obtained through the multiplication of a vector (velocity) by a scalar (mass). As stated in the tutorial "Multiplication of a Vector by a Scalar", the result is a new vector, which is collinear with the original one.

A 300 t asteroid is moving in space at 5 km/s towards the Earth. What is its momentum in kg × m/s?

First, we must convert the units into the basic ones. Thus, since 1 t = 1000 kg and 1 km = 1000 m, we have

m = 300 t = 300 000 kg

v*⃗* = 5 *km**/**s* = 5000 *m**/**s*

v

Therefore, we obtain for the momentum of the asteroid:

p*⃗* = m × v*⃗*

= 300 000 kg × 5000*m**/**s*

= 1 500 000 000 kg ×*m**/**s*

= 1.5 × 10^{9} kg × *m**/**s*

= 300 000 kg × 5000

= 1 500 000 000 kg ×

= 1.5 × 10

Thus, we say the momentum of the asteroid is 1.5 × 109 kg × m/s towards the Earth.

In the past, momentum has been known as the "amount of motion." This term is logical because a very large momentum as the one obtained in the example above, means the object's motion is very big. As a result, the object may cause a lot of harm when hitting another body, such as the Earth. On the other hand, a very light object (such as the tennis ball in the Introduction paragraph) which moves slowly has a small amount of motion, as its momentum is small.

In the next paragraph, it will be clearer why momentum has been called "amount of motion" in the past.

Like in Kinematics or Dynamics - in which more than in objects actual velocity we are interested in the change in velocity as this gives a better idea on the resultant force and therefore in the acceleration of the object itself - here we are more interested in the change in momentum than in the actual momentum itself. This point is explained by considering again the example provided in the Introduction paragraph.

If we take the original direction of the ball as positive, we have the following figure.

The initial momentum of the ball before the collision is

p*⃗*_{1} = m × v*⃗*_{1}

= 0.2 kg × 4*m**/**s*

= 0.8 kg ×*m**/**s*

= 0.2 kg × 4

= 0.8 kg ×

and the final momentum is

p*⃗*_{2} = m × v*⃗*_{2}

= 0.2 kg × (-3)*m**/**s*

= -0.6 kg ×*m**/**s*

= 0.2 kg × (-3)

= -0.6 kg ×

If we calculate the change in momentum, i.e. p*⃗*_{1} - p*⃗*_{2} (or ∆p*⃗*), we obtain

∆p*⃗* = p*⃗*_{1} - p*⃗*_{2}

= 0.8 kg ×*m**/**s* - (-0.6 kg × *m**/**s*)

= 1.4 kg ×*m**/**s*

= 0.8 kg ×

= 1.4 kg ×

On the other hand, when the ball hits the paper sheet, both velocities are positive as they are in the same direction. Therefore, we have the following figure

The initial momentum of the ball before the collision was

p*⃗*_{1} = m × v*⃗*_{1}

= 0.2 kg × 4*m**/**s*

= 0.8 kg ×*m**/**s*

= 0.2 kg × 4

= 0.8 kg ×

and the final momentum after the collision is

p*⃗*_{2} = m × v*⃗*_{2}

= 0.2 kg × 3*m**/**s*

= 0.6 kg ×*m**/**s*

= 0.2 kg × 3

= 0.6 kg ×

If we calculate the change in momentum, i.e. p*⃗*_{1} - p*⃗*_{2} (or ∆p*⃗*), we obtain

∆p*⃗* = p*⃗*_{1} - p*⃗*_{2}

= 0.8 kg ×*m**/**s* - 0.6 kg × *m**/**s*

= 0.2 kg ×*m**/**s*

= 0.8 kg ×

= 0.2 kg ×

The interpretation of the above results is as follows:

"When the ball hits a hard surface such as the wall, its motion changes drastically. Not only does the magnitude of velocity change, but also its direction as well. Therefore, a greater change in momentum (or change in the amount of motion) is produced compared to the case when it hits a thin sheet. This is because the wall offers a higher resistance than the sheet and as a result, it affects much more the motion of the ball."

**Remarks!**

- Do not confuse the terms "moment" and "momentum". They express two very different concepts. While the first refers to the turning effect of a force, the later expresses the amount of motion of an object.
- If we want to calculate the change in momentum of an object or system, we must take as initial velocity the velocity of the object just before hitting the obstacle, not the very initial velocity the object may have had at the beginning of process. Likewise, the final velocity is taken just immediately after the object leaves the obstacle, not the final velocity of object's motion.
- It is called "Linear Momentum" because there also exists a rotational momentum, which we will discuss later, in the Rotational Dynamics section. In linear momentum, we take the object as moving linearly or according such a path that is a combination of multiple linear segments.

We cannot use the energetic approach to analyse the motion of objects as kinetic energy is obtained by raising the velocity at power two and we may get lost the traces of objects direction because kinetic energy is always positive, regardless the moving direction. Therefore, it would not make any difference whether an object hits a hard wall or a thin sheet as long as the magnitudes of velocities are the same.

To avoid such problems, a new concept known as "linear momentum" or simply "momentum" is introduced. By definition,

**Linear Momentum (p ⃗) is the product of an object's mass (m) and its velocity (v⃗). **

Mathematically, we have

p*⃗* = m × v*⃗*

Since mass is measured in kg and velocity in m/s, the unit of momentum is [kg × m/s]. Momentum is a vector quantity. In the past, momentum has been known as the "amount of motion."

Usually, we are more interested in the change in momentum than in the actual momentum itself. We write for the change in momentum

p*⃗* = m × ∆v*⃗*

= m × (v*⃗*_{1} - v*⃗*_{2})

= m × v*⃗*_{1} - m × v*⃗*_{2}

= m × (v

= m × v

where m is the mass of the object, v*⃗*_{1} and v*⃗*_{2} are its initial and final velocities respectively.

When the a moving object hits a hard surface such as a wall, its motion changes drastically. Not only does the magnitude of velocity change, but also its direction as well. Therefore, a greater change in momentum (or change in the amount of motion) is produced compared to the case when it hits a thin sheet. This is because the wall offers a higher resistance than the sheet and as a result, it affects much more the motion of the object itself.

Do not confuse the terms "moment" and "momentum". They express two very different concepts. While the first refers to the turning effect of a force, the later expresses the amount of motion of an object.

If we want to calculate the change in momentum of an object or system, we must take as initial velocity the velocity of the object just before hitting the obstacle, not the very initial velocity the object may have had at the beginning of process. Likewise, the final velocity is taken just immediately after the object leaves the obstacle, not the final velocity of object's motion.

It is called "Linear Momentum" because there also exists a rotational momentum, which we will discuss later, in the Rotational Dynamics section. In linear momentum, we take the object as moving linearly or according such a path that is a combination of multiple linear segments.

**1)** A 3 kg object has the same linear momentum as a 12 kg object moving at 4 m/s due north. What is the velocity of the first object?

- 16 m/s
- 9 m/s
- 16 m/s due north
- 9 m/s due north

**Correct Answer: C**

**2)** A 300 g ball moving at 8 m/s hits a hard surface and turns back at 6 m/s. What is the change in momentum the object experiences during this process?

- 0.6 kg × m / s
- 4.2 kg × m / s
- 15.0 kg × m / s
- 14.0 kg × m / s

**Correct Answer: B**

**3)** What change does occur in the amount of motion of a 20 g bullet moving initially at 400 m/s if it stops after plunging on tree trunk?

- It decreases by 1600 kg × m / s
- It increases by 1600 kg × m / s
- It decreases by 8 kg × m / s
- It increases by 8 kg × m / s

**Correct Answer: C**

We hope you found this Physics tutorial "Linear Momentum" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of "Linear Momentum" with our Physics tutorial on Collision and Impulse. Types of Collision.

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