Calculator™ © - Free Online Calculators

Online Calculators since 2009

- Conservation Of Momentum In 1 D Calculator
- Parallel Resonant Frequency Calculator
- Gravitational Field Strength Calculator
- Elastic Collision Calculator
- Final Temperature Of Mixture Calculator
- Biquad Filter Coefficient Calculator
- Torque Calculator
- Gravitational Force Calculator
- Uniform Circular Motion Calculator
- Capacitance Calculator
- Physics Tutorials, Physics Revision and Physics Calculators

In this Physics tutorial, you will learn:

- What is impulse?
- How impulse is related to momentum?
- When does the phenomenon of impulse take place?
- How can we represent the impulse graphically?
- What is the definition of collision?
- How many types of collision arte there?
- What occurs during each type of collision?

Suppose you throw two identical stones from the same height. Obviously, they hit the ground at the same velocity. The first stone falls on a hard ground while the other on a kind of sponge. Both stones stop moving after their respective collision.

Do you think both stones experience the same change in momentum? Why?

Which stone causes more harm on the surface they hit? Why?

Which stone do you think is more damaged after the collision? Why?

Let's try to answer the above questions. Thus, apparently, it looks like both objects have experienced the same change in momentum, as since both of them fall down from the same height, their velocity near the ground is equal. On the other hand, they have a zero momentum after the collision, as both of them stop moving after hitting the ground.

However, it is obvious the balls have experienced different processes during their contacts with the respective surfaces. The collision of the first ball (the one that falls on a hard ground) causes a greater impact on both the ball and the ground as they may be easily damaged. On the other hand, the collision with the sponge does not cause any harm in any of the objects in contact as it is "amortized" i.e. absorbed by sponge's softness. Hence, the collision is slower and more extended in time compared to the first case due to the elastic properties of sponge. Look at the figure:

But what can we say about the collision force? Is it the same in both cases?

To answer this question, let's transform the equation of the change in momentum we obtained in the previous tutorial, "Linear Momentum". We have:

= m × v

= m × ∆v

If we divide both sides of the above equation by the collision interval Δt, we obtain

From Kinematics, we know that

where a*⃗* is the acceleration of the object. Therefore, we obtain after substitutions

The right side of the equation (2) represents the right side of the Newton's Second Law of Motion,

where F*⃗* is the force of collision. Therefore, combining the left sides of equations (2) and (3), we obtain

The equation (4) therefore represents another way of writing the Newton's Second Law of Motion. If we transform the equation (4) into

we obtain a new concept known as "Impulse" which represents the relationship between the change in momentum, collision force and time interval during which the collision occurs. Impulse is denoted by J*⃗* and it is a vector quantity as it is obtained by subtracting two vectors. Therefore, an object has an impulse only during the collision while before and after the collision it has only momentums. The following table clarifies this point:

Impulse as a change in momentum has the same unit as momentum, i.e. kg × m / s. In many textbooks, the concept of force is not expressed in its traditional form, i.e. Force = Mass ×; Acceleration based on Newton's Second Law of Motion, but is the form of equation (4), i.e. in terms of impulse and the collision interval.

Now, let's return to the question regarding the collision force, i.e. whether it is the same in both cases mentioned above or not. When the object falls on hard ground, it jumps immediately after the impact, although this jump may be very small to be noticeable for our eyes. This means the impact interval is very short and therefore, the force is big. On the other hand, when the object falls on a soft surface such as a sponge, it takes a much longer to it pushing the surface and reach the bottom. Therefore, as force and collision interval are inversely proportional to each other, the impact force when the object falls on a soft ground is much smaller than when the same object falls on hard ground.

A 50 g steel ball hits a metal surface at 12 m/s and bounces back at 8 m/s. the ball is in contact with the metal surface for 0.02 s as shown in the figure.

What is the force of collision between the steel ball and the metal surface?

The ball changes direction after the collision, so we must take its final velocity as negative. As for the other clues, we must not forget to convert the mass of the ball from 50 g into 0.05 kg.

Given the relationship between the change in momentum and impulse, we have

F*⃗* × ∆t = m × v*⃗*_{1} - m × v*⃗*_{2}

Substituting the known values, we obtain

F*⃗* × 0.02 = 0.05 × 12-0.05 × (-8)

F*⃗* × 0.02 = 0.6 + 0.4

F*⃗* × 0.02 = 1

F*⃗* = *1**/**0.02* = 50 N

F

F

F

The direction of force depends on what we are concerned for. If we want to know the force by which the ball hits the metal surface, it is easy to find that in the specific case, it is due right. On the other hand, the force by which the metal reacts to the ball is 50 N as well but due left, based on the action-reaction principle.

First, let's explain what is collision, because we have mentioned this term many times in the last two articles without giving its definition.

In physics, **Collision, also called impact, represents the sudden, forceful coming together in direct contact of two bodies, such as, for example, two balls, a hammer and a nail head, or a falling object and a floor.**

There are two main types of collision. They are:

**Elastic collision**in which objects move apart after the contact, and**Inelastic collision**in which objects move together after the collision.

In all examples discussed in the last two articles, we considered only elastic collisions in which one of the objects in contact was unmoveable (wall, metal surface, thin sheet, etc.) while the other (usually a ball or a stone) was moveable and approached the unmoveable object and eventually hit it. Then, the objects moved apart.

The highest stage of elastic collision is when an object moves at the same speed (not velocity) after the impact. This means it conserves the kinetic energy and therefore, no energy turns into other forms such as heat, sound, light (in sparks), etc. Such collision is called **"absolutely (completely) elastic collision"** and obviously, this is an idealization because it is clear that it is impossible not hearing any noise or not producing any heat during an impact. For example, when you hit a nail using a hammer, the nail head will become hot after a number of hits. This means some of hammer's initial kinetic energy converts into heat during the process. You hear a loud noise as well, so some kinetic energy of the hammer converts into sound. Finally, you can see some sparks at the nail's head. This means some of the initial kinetic energy of hammer converts into light energy, and so on.

However, for convenience we will consider perfectly elastic collisions in some of the future exercise. This means we will neglect the other energy conversions that take place during the process.

A 400 g ball falls from 6 m above the ground and rises up to 3.6 m after the impact with the ground. The collision lasts for 0.05 s. Ignore air resistance and take g = 10 m/s2 for convenience.

- What kind of collision is this?
- What is the force of impact with the ground?

a. It is obvious this collision is elastic as the ball moves away from the ground after the impact. Thus, we have to determine whether the collision is absolutely (completely) elastic or not. We can find this by working out the velocities before and after the impact to the ground. If these velocities (despite they may have opposite direction) have the same magnitude, the collision is absolutely (completely) elastic, otherwise it is not. At this point, we can use the energetic approach to determine the velocity of the ball just before touching the ground, which we will consider as the initial velocity v*⃗*_{1} before the impact, and the velocity of the ball immediately after leaving the ground, which we consider as the final velocity v*⃗*_{2}.

When the object is initially at 6 m above the ground (at the instant zero), it possesses only gravitational potential energy, which is converted into kinetic energy at the instant just before the object touches the ground (instant 1). Thus, we have

GPE_{0} = KE_{1}

m × g × h_{0} = *m × v*^{2}_{1}*/**2*

m × g × h

Simplifying the mass from both sides and substituting the known values, we obtain for the velocity of the ball just before striking the ground:

10 × 6 = *v*^{2}_{1}*/**2*

v^{2}_{1} = 10 × 6 × 2 = 120

v_{1} = √**120** ≈ 11.0 *m**/**s* downwards

v

v

To determine the value of the ball's velocity just after leaving the ground, we must keep in mind that the kinetic energy of the ball just when it leaves the ground is equal to the final gravitational potential energy at h3 = 3.6 m. Therefore, we have

KE_{2} = GPE_{3}

*m × v*^{2}_{2}*/**2* = m × g × h_{3}

Again, simplifying mass from both sides and substituting the known values, we obtain for the velocity of the ball just immediately after leaving the ground:

v

v

Since the magnitudes of velocity are not equal, this is an elastic collision but not absolutely (completely) elastic.

b. The force of impact to the ground is determined by calculating the impulse. Thus, taking the velocity before the impact as positive and that after the impact as negative (and given that m = 400 g = 0.4 kg), we write

F*⃗* × ∆t = m × v*⃗*_{1} - m × v*⃗*_{2}

F*⃗* × 0.05 = 0.4 × 11.0 - 0.4 × (-8.5)

F*⃗* × 0.05 = 4.4 + 3.4 = 7.8

F*⃗* = 7.8/0.05 = 156 N

F

F

F

Since in impulse the impact interval usually is very short, we take the impact force as uniform. Therefore, in the simplified version, impulse is represented by the area under the Force vs Time graph involved. Look at the figure below.

In this case, we have F*⃗* = 20 N and Δt = 0.2 s - 0.1 s = 0.1 s. Hence,

J*⃗* = F*⃗* × ∆t = 20 N × 0.1 s = 2 *kg × m**/**s*

If the collision time interval is prolonged, we cannot speak anymore for impulse, as impulse takes place only when a force is exerted in a very short time interval on an object. Therefore, we cannot discuss about impulse in inelastic collision because in such collisions, the objects stick together after the impact and as a result, the impact time practically extends at infinity if no other force in the opposite direction is exerted to detach them.

In physics, Collision, also called impact, represents the sudden, forceful coming together in direct contact of two bodies, such as, for example, two balls, a hammer and a nail head, or a falling object and a floor.

There are two main types of collision. They are:

- Elastic collision in which objects move apart after the contact, and
- 2- Inelastic collision in which objects move together after the collision.

The highest stage of elastic collision is when an object moves at the same speed (not velocity) after the impact. This means it conserves the kinetic energy and therefore, no energy turns into other forms such as heat, sound, light (in sparks), etc. Such collision is called "absolutely (completely) elastic collision" and obviously, this is an idealization because it is clear that it is impossible not hearing any noise or not producing any heat during an impact.

"Impulse" represents the relationship between the change in momentum, collision force and time interval during which the collision occurs. Impulse is a vector quantity. It is denoted by J*⃗* and it is measured by the same unit as momentum [kg × m / s], as it is obtained by subtracting the initial and final momentum of an object.

Mathematically, we have:

J*⃗* = ∆p*⃗* = F*⃗* × ∆t

= m × ∆v*⃗*

= m × v*⃗*_{1} - m × v*⃗*_{2}

= m × ∆v

= m × v

We can speak for impulse only during the collision; neither before, nor after it.

Newton's Second Law of Motion is often expressed in terms of impulse and time interval, i.e. as

F*⃗* = *J**⃗**/**∆t* = *∆**⃗*p*/**∆t*

Since in impulse the impact interval usually is very short, we take the impact force as uniform. Therefore, in the simplified version, impulse is represented by the area under the Force vs Time graph involved.

If the collision time interval is prolonged, we cannot speak anymore for impulse, as impulse takes place only when a force is exerted in a very short time interval on an object. Therefore, we cannot discuss about impulse in inelastic collision because in such collisions, the objects stick together after the impact and as a result, the impact time practically extends at infinity if no other force in the opposite direction is exerted to detach them.

**1)** A 40 g bullet hits a 20 cm thick wooden trunk at 200 m/s. The bullet decelerates due to the friction and leaves the trunk in the opposite side at 100 m/s. Calculate the average force exerted by the bullet in the trunk.

- 300 N
- 3000 N
- 30 000 N
- 300 000 N

**Correct Answer: B**

**2)** A 3 kg object hits a hard surface at 6 m/s and then turns back. The collision lasts for 0.1 s and the force of impact is 360 N. What type of collision is it?

- Completely elastic
- Elastic but not completely
- Inelastic but nut completely
- Completely inelastic

**Correct Answer: A**

**3)** A 2 kg object hits a hard surface at 20 m/s due North. The Force vs Time graph for this object is shown in the figure below.

- 35 m/s due North
- 15 m/s due South
- 70 m/s due South
- 35 m/s due South

**Correct Answer: B**

We hope you found this Physics tutorial "Collision and Impulse. Types of Collision" 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 "Collision And Impulse" with our Physics tutorial on Law of Conservation of Momentum and Kinetic Energy.

The following Physics Calculators are provided in support of the Centre of Mass and Linear Momentum tutorials.

- Centre Of Mass Calculator
- Conservation Of Momentum In 1 D Calculator
- Conservation Of Momentum In 2 D Calculator
- Equilibrium Using Moments Calculator
- Impulse Calculator
- Newtons Second Law For A System Of Particles Calculator
- Torque Calculator

You may also find the following Physics calculators useful.

- Horsepower To Amps Unit Calculator
- Partially Filled Horizontal Tank Volume Calculator
- Parallel Plate Capacitor Calculator
- Dc Motor Horsepower Calculator
- Velocity Of Alpha Particle Nuclear Decay Calculator
- Density Of Sand Calculator
- Convective Heat Transfer Calculator
- Plasma Frequency Calculator
- Newtons Second Law For A System Of Particles Calculator
- Elastic Collision Calculator
- Solid Pressure Calculator
- Electromagnetic Field Energy Density Calculator
- Capacitance Calculator
- Hollow Shaft Maximum Torque Calculator
- Effective Aperture Calculator
- 3 D Work Calculator
- Torque Calculator
- Acceptance Angle In Optical Fibre Calculator
- Rectangular Weir Flow Rate Calculator
- Helical Spring Rate Calculator