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Physics Lesson 6.6.2 - Types of Collision
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Welcome to our Physics lesson on Types of Collision, this is the second lesson of our suite of physics lessons covering the topic of Collision and Impulse. Types of Collision, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Types of Collision
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.
Example 2
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?
Solution 2
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 v⃗1 before the impact, and the velocity of the ball immediately after leaving the ground, which we consider as the final velocity v⃗2.
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
m × g × h0 = m × v21/2
Simplifying the mass from both sides and substituting the known values, we obtain for the velocity of the ball just before striking the ground:
v21 = 10 × 6 × 2 = 120
v1 = √120 ≈ 11.0 m/s downwards
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
m × v22/2 = m × g × h3
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:
v22 = 2 × 10 × 3.6 = 72
v2 = √72 ≈ 8.5 m/s upwards
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
F⃗ × 0.05 = 0.4 × 11.0 - 0.4 × (-8.5)
F⃗ × 0.05 = 4.4 + 3.4 = 7.8
F⃗ = 7.8/0.05 = 156 N
You have reached the end of Physics lesson 6.6.2 Types of Collision. There are 4 lessons in this physics tutorial covering Collision and Impulse. Types of Collision, you can access all the lessons from this tutorial below.
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