Collision
The term “collision” often brings to mind images of objects crashing into each other. Common examples include cars crashing on a highway, a hammer hitting a nail, or billiard balls striking each other on a pool table.
Collisions occur across a vast range of scales—from tiny particles like atoms colliding at microscopic levels to massive celestial bodies like stars colliding in space. Although these events may last only a brief moment and are sometimes too fast to be seen, they usually involve significant alterations or deformations of the involved bodies.
Typically, a collision is characterized as an isolated incident where two or more entities exert considerable forces upon one another within a brief period. It’s important to note that collisions do not necessarily require the physical touching of objects involved; the impact forces are the critical component of the interaction.
For instance, an alpha particle heading towards an atom’s nucleus is repelled by electrostatic forces, causing what is known as a collision with the nucleus.
Studying collisions primarily helps us understand the forces at play during these events, informed by the conditions of particles before and after the collision. Our knowledge of the subatomic realm, including electrons, protons, neutrons, and quarks, is largely derived from collision experiments.
Two fundamental principles govern these interactions:
Types of collision
Collisions are generally categorized into:
- Elastic collisions
- Inelastic collisions
1. Elastic Collision
In an elastic collision, there is no loss of kinetic energy. Such collisions are often observed between atomic and subatomic particles and can practically be seen in collisions between ivory balls.
Characteristics of an elastic collision include:
- Conservation of linear momentum
- Conservation of total system energy
- Conservation of kinetic energy.
It is noted that in such collisions, mechanical energy, such as kinetic energy, is not transformed into other energy forms like heat or sound. - Forces involved in these collisions are typically conservative.
2. Inelastic Collision
An inelastic collision is defined by some loss of kinetic energy during the event.
Most everyday collisions result in some kinetic energy loss, making them inelastic by nature.
If colliding bodies remain joined post-collision, such as mud sticking to a wall after being thrown, the collision is termed perfectly inelastic. This is also true for arrows that embed in targets and meteorites that strike the Earth, integrating into its mass.
Basic characteristics of inelastic collisions
- Linear momentum remains conserved,
- Total energy of the system is conserved,
- There is a loss of kinetic energy,
Often, energy from motion (kinetic energy) is converted into other forms such as heat or sound, and not all forces involved are conservative.
For example, dropping a superball onto a hard floor might seem like an elastic collision because it does not lose kinetic energy due to the impact and can bounce back to its original height. However, practical observations show that some kinetic energy is lost due to factors like air resistance and imperfect elastic recovery, hinting at inelastic characteristics. If a golf ball is dropped and rebounds to a fraction of its original height, or a ball of wet putty sticks where it lands without rebounding, these scenarios illustrate perfectly inelastic collisions.
It is essential to recognize that both total energy and linear momentum are preserved in all types of collisions, whether elastic or inelastic. However, in the case of inelastic collisions, kinetic energy is not maintained.
Coefficient of Restitution Explained
Typically, real-world collisions between objects don’t conform to the ideal models of completely elastic or inelastic interactions. These encounters are often categorized as imperfect or semi-elastic. The degree to which a collision approximates an elastic encounter is quantified by the coefficient of restitution, a measure of how the relative speed after the collision compares to the relative speed before the collision. This coefficient is denoted as ‘e’.
$$e=\frac{{\text{relative velocity after collision}}}{{\text{relative velocity before collision}}}$$
or
$$\boxed{e=\frac{{{{v}_{2}}-{{v}_{1}}}}{{{{u}_{1}}-{{u}_{2}}}}}$$
Here, \( u_1, u_2 \) represent the velocities of the two bodies before the collision, and \( v_1, v_2 \) their velocities after the collision.
In a perfectly elastic collision where no kinetic energy is lost, relative velocity of separation after collision is equal to relative velocity of approach before collision. Therefore, the value of \( e \) is 1:
$$ \boxed{e = 1} $$
In a perfectly inelastic collision, where bodies coalesce and stop moving relative to each other, \( e \) is 0:
$$ \boxed{ e = 0} $$
In all other types of collisions, the coefficient of restitution falls between 0 and 1:
$$ \bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\color{#000000}{0 < e < 1}}} $$
Remember: Discussion here pertains to collisions in closed and isolated systems, where no mass enters or exits the system, and external forces like gravity or friction are negligible compared to the internal impulsive forces experienced during collisions.
Example 1: A ball is dropped from a height of \(12 \, \text{m}\) and loses 25% of its kinetic energy upon striking the ground. What height will it rebound to, accounting for the kinetic energy loss?
Solution:
Given: \( h = 12 \, \text{m} \), \( g = 9.8 \, \text{m/s}^2 \)
To find: Velocity on reaching the ground and rebound height.
From the equation:
$$ v^2 – u^2 = 2as $$
$$ v^2 = 2 \times 9.8 \times 12 $$
$$ v = \sqrt{2 \times 9.8 \times 12} = 15.34 \, \text{m/s} $$
Kinetic energy on reaching the ground:
$$ \color{green} KE = \frac{1}{2} M v^2 = \frac{1}{2} M (15.34)^2 $$
After losing 25% of kinetic energy:
$$ KE_{\text{after}} = 0.75 \times KE $$
Rebound velocity \( v’ \) using the remaining kinetic energy:
$$ \frac{1}{2} M v’^2 = KE_{\text{after}} $$
$$ {v}’=\sqrt{{2\times 88.2}}=13.2$$
Using the rebound velocity to calculate the rebound height \( h’ \):
$ \boxed{\begin{array}{l}\because {h}’=\dfrac{{{{{{v}’}}^{2}}}}{{2g}}\\\therefore {h}’=\dfrac{{{{{\left( {13.2} \right)}}^{2}}}}{{2\times 9.8}}\\\Rightarrow {h}’=9\end{array}}$
Example 2: A bullet of mass $0.012 \, \text{kg}$ and horizontal speed $70 \, \text{m/s}$ strikes a block of wood of mass $0.4 \, \text{kg}$ and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also estimate the amount of heat produced in the block.
Solution:
As soon as the bullet strikes the wooden target, the bullet comes to rest with respect to the block of wood, making the collision inelastic. Therefore, the principle of conservation of energy does not hold for this collision. Only the conservation of momentum will hold.
Let:
Mass of bullet, $M_1 = 0.012 \, \text{kg}$
Mass of block, $M_2 = 0.4 \, \text{kg}$
The initial velocity of the bullet, $u_1 = 70 \, \text{m/s}$; initial velocity of the block, $u_2 = 0$.
Let $v$ be the final velocity of the block and bullet combined. Then, according to the principle of conservation of momentum:
\[
M_1 u_1 + M_2 u_2 = (M_1 + M_2) v
\]
\[
v = \frac{M_1 u_1 + M_2 u_2}{M_1 + M_2}
\]
\[
v = \frac{0.012 \times 70 + 0.4 \times 0}{0.012 + 0.4} = 2.04 \, \text{m/s}
\]
Suppose that the block and the bullet rise to a height $h$. Then,
\[
\frac{1}{2} (M_1 + M_2) v^2 = (M_1 + M_2) gh
\]
Or,
\[
h = \frac{v^2}{2g} = \frac{(2.04)^2}{2 \times 9.8} = 0.212 \, \text{m} = 21.2 \, \text{cm}
\]
The loss in kinetic energy of the system will appear as heat. Therefore, heat produced, $H$:
\[
H = \frac{1}{2} M_1 u_1^2 – \frac{1}{2} (M_1 + M_2) v^2
\]
$ \displaystyle H=\frac{1}{2}\times 0.012\times {{70}^{2}}$ $ \displaystyle -\frac{1}{2}\times (0.012+0.4)\times {{(2.04)}^{2}}$
$ H=29.4-(0.85)$
$ \boxed{H=28.55\text{J}}$
