Collision: Class 11 Physics Notes, Meaning, Types & Equations

Collision or Impact is the interaction between two bodies during a minimal duration in which they exert relatively large forces on each other. Interaction forces during an impact are created either due to direct contact, strong repulsive force fields, or some connecting links.

The duration of the interaction is short enough to permit us only to consider the states of motion just before and after the event, and not during the impact. The duration of an impact ranges from ${10}^{-23}\mathrm{s}$ for impacts between elementary particles to millions of years for impacts between galaxies. The impacts we observe in our everyday life, such as those between two balls that last from ${10}^{-3}\mathrm{s}$ to few seconds.

Let's look into some examples of collision to understand it better.

Direct collision: - When a rubber ball strikes a floor, it remains in contact with the floor for a very short time during which it changes its velocity. This is an example of a collision where physical contact takes place between the colliding bodies.

Indirect collision: - When an $\alpha$ -particle passes by the nucleus of a gold atom in Rutherford's experiment, it gets deflected in a very short time. Deflection means a change in the direction of motion- a change in velocity. In this process, the particles do not touch each other.

Effects of collision, the momentum and kinetic energy of the interacting bodies change.

Interacting forces during collision are so large compared to other external forces acting on either of the bodies that the effects of the latter can be neglected. Then, forces involved in a collision are action-reaction pairs, i.e., the internal forces of the system and thus the total momentum remains conserved in any type of collision.

Understanding this concept of collision will be quite helpful for your JEE Mains. 

We can categorise different types of collisions based on the following criteria. 
(a) On the basis of direction

On the basis of direction there are two types of collision.

(i) Head-on Collision

A Collision in which the particles move along the same straight line before and after the collision is defined as Head-on Collision or One-dimensional Collision.

(ii) Oblique Collision

A Collision in which the particles move along the same plane at different angles before and after the collision is defined as an Oblique Collision or a Two-dimensional Collision.

(a) On the basis of kinetic energy: -

On the basis of kinetic energy, there are three types of collision.

(i) Elastic Collision

A Collision is said to be elastic if the total kinetic energy before and after the collision remains the same.

(ii) Inelastic Collision

A Collision is said to be Inelastic if the total kinetic energy before and after the collision does not remain same.

(iii) Perfectly Inelastic Collision

A Collision in which interacting particles get stuck together after the collision is called a Perfectly Inelastic Collision. In this type of collision, the loss of energy is maximum.

To understand mechanism of a collision, let us consider two balls $A$ and $B$ of masses $m_A$ and $m_B$ moving with velocities $u_A$ and $u_B$ in the same direction as shown. Velocity $u_A$ is larger than $u_B$ so that the ball $A$ hits the ball B. During the impact, both the bodies push each other and first they get deformed till the deformation reaches a maximum value and then they try to regain their original shapes due to elastic behaviour of the materials the balls formed of.

Deformation period: - The time interval during which deformation takes place.

During the period of deformation, due to push applied by the balls on each other during period of deformation speed of ball A decreases and that of ball B increases and at the end of the deformation period, when the deformation is maximum both the balls move with the same velocity say it is $u$ .

Thereafter, the balls will either move together with this velocity or follow the period of restitution.

Restitution period: - The time interval in which the bodies try to regain their original shapes.

During the period of restitution due to push applied by the balls on each other, speed of the ball A decreases further and that of ball B increases further till they separate from each other.
Let us denote the velocities of the balls $A$ and $B$ after the impact by $v_A$ and $v_B$ respectively.

(a) Deformation period

Applying impulse-momentum principle on ball A during deformation period

$\vec{J}=\mathrm{\Delta }\vec{\mathrm{P}}$

$\Rightarrow -\int Ddt=m_{A}u-m_A{u}_A\Rightarrow m_A{u}_A-\int Ddt=m_{A}u$        …..(i)

Applying impulse-momentum principle on ball $B$ during deformation period

$\int \mathrm{D}\mathrm{d}\mathrm{t}={\mathrm{m}}_{\mathrm{B}}\mathrm{u}-{\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}\mathrm{}\Rightarrow {\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}+\int \mathrm{D}\mathrm{d}\mathrm{t}={\mathrm{m}}_{\mathrm{B}}\mathrm{u}$         ….(ii)

(b) Restitution period

Applying impulse-momentum principle on ball A during restitution period

$-\int Rdt=m_A{v}_A-m_{A}u\Rightarrow m_{A}u-\int Rdt=m_A{v}_A$  …..(iii)

Applying impulse-momentum principle on ball  during restitution period

$\int Rdt=m_B{v}_B-m_{B}u\mathrm{}\Rightarrow m_{B}u+\int Rdt=m_B{v}_B$         ……(iv)

Conservation of Momentum during impact

From equations, (i) and (ii) we have $m_A{u}_A+m_B{u}_B=\left(m_A+m_B\right)u$

From equations, (iii) and (iv) we have $\left(m_A+m_B\right)u=m_A{v}_A+m_B{v}_B$

From equations, (v) and (vi) we obtain the following equation.

${\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}$

Hence, Total Initial Momentum = Total Final Momentum

The above equation elucidates the principle of conservation of momentum.

Usually the force D applied by the bodies A and B on each other during the period of deformation differs from the force R applied by the bodies on each other during the period of restitution.

Therefore, it is not necessary that the magnitude of impulse $\int Ddt$ due to deformation equals to that of impulse $\int Rdt$ due to restitution.

The ratio of magnitudes of impulse of restitution to that of deformation is called the coefficient of restitution and is denoted by e.

$\begin{array}{rr}& e=\frac{\text{impulse of recovery}}{\text{impulse of deformation}}=\frac{\int Rdt}{\int Ddt}\\ & e=\frac{\text{velocity of separation along line of impact}}{\text{velocity of approach along line of impact}}=\frac{\left|{\stackrel{\dot{}}{v}}_B-{\stackrel{\dot{}}{v}}_A\right|}{\left|{\stackrel{\dot{}}{u}}_A-{\stackrel{\dot{}}{u}}_B\right|}\end{array}$

Coefficient of restitution depends on various factors as elastic properties of materials forming the bodies, velocities of the contact points before impact, state of rotation of the bodies and temperature of the bodies. In general, its value ranges from zero to one but in collisions where additional kinetic energy is generated, its value may exceed one.

Depending on the values of coefficient of restitution, two particular cases are of special interest.

Perfectly Plastic or Inelastic Impact For these impacts e $=0$ , and bodies undergoing impact stick to each other after the impact.

Perfectly Elastic Impact For these impacts e $=1$ .

Strategy to solve problems of head-on impact:

Write the momentum conservation equation;

${\mathrm{m}}_{\mathrm{A}}{\mathrm{v}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{v}}_{\mathrm{B}}={\mathrm{m}}_{\mathrm{A}}{\mathrm{u}}_{\mathrm{A}}+{\mathrm{m}}_{\mathrm{B}}{\mathrm{u}}_{\mathrm{B}}$

Write the equation involving coefficient of restitution

${\mathrm{v}}_{\mathrm{B}}-{\mathrm{v}}_{\mathrm{A}}=\mathrm{e}\left({\mathrm{u}}_{\mathrm{A}}-{\mathrm{u}}_{\mathrm{B}}\right)$

If velocity vectors of the colliding bodies are directed along the line of impact, the impact is called a direct or Head-on impact.

Head on Elastic Collision

The head on elastic collision is one in which the colliding bodies move along the same straight-line path before and after the collision.

 

Assuming initial direction of motion to be positive and $u_1>u_2$ (so that collision may take place) and applying law of conservation of linear momentum     

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\Rightarrow m_1\left(u_1-v_1\right)=m_2\left(v_2-u_2\right)$   …..(i)

For elastic collision, kinetic energy before collision must be equal to kinetic energy after collision, i.e.,

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\Rightarrow m_1\left(u_1^2-v_1^2\right)=m_2\left(v_2^2-u_2^2\right)$ …..(ii)

Dividing equation (ii) by (i)

$u_1+v_1=v_2+u_2\Rightarrow \left(u_1-u_2\right)=\left(v_2-v_1\right)$ …..(iii)

In 1-D elastic collision 'velocity of approach' before collision is equal to the 'velocity of separation after collision, no matter what the masses of the colliding particles are.
This law is called Newton's law for elastic collision.
Thus, value of coefficient of restitution for elastic collision; $e=\frac{v_2-v_1}{u_1-u_2}=1$
If we multiply equation (iii) by ${\mathrm{m}}_2$ and subtract it from (i)

$2m_1{\stackrel{\rightharpoonup }{u}}_1+\left(m_2-m_1\right){\stackrel{\rightharpoonup }{u}}_2=\left(m_2+m_1\right){\stackrel{\rightharpoonup }{v}}_2\Rightarrow {\bar{v}}_2=\frac{m_2-m_1}{m_1+m_2}{\stackrel{\rightharpoonup }{u}}_2+\frac{2m_1{\stackrel{\rightharpoonup }{u}}_1}{m_1+m_2}$ …..(iv)

Similarly, multiplying equation (iii) by  and adding it to equation (i)

$2m_1{\stackrel{\rightharpoonup }{u}}_1+\left(m_2-m_1\right){\stackrel{\rightharpoonup }{u}}_2=\left(m_2+m_1\right){\stackrel{\rightharpoonup }{v}}_2\Rightarrow {\bar{v}}_2=\frac{m_2-m_1}{m_1+m_2}{\stackrel{\rightharpoonup }{u}}_2+\frac{2m_1{\stackrel{\rightharpoonup }{u}}_1}{m_1+m_2}$ …..(v)

Case-1: If the two bodies are of equal masses: $m_1=m_2=m$ then $v_1=u_2$ and $v_2=u_1$

Thus, if two bodies of equal masses undergo elastic collision in one dimension, then the bodies exchange their velocities after the collision.

Case-2: If the two bodies are of equal masses and second body is at rest.
$m_1=m_2$ and initial velocity of second body $u_2=0,v_1=0,v_2=u_1$

When body $A$ collides against body $B$ of equal mass at rest, then body $A$ comes to rest and body $B$ moves on with the velocity of body $A$ . In this case transfer of energy is hundred percent e.g. Billiard's Ball, Nuclear moderation.

Case-3: If the mass of one body is negligible as compared to the other.
If $m_1\gg m_2$ and $u_2=0$ then $v_2=0,v_1=-u_1$
When a heavy body A collides against a light body B at rest, then body A should keep on moving with same velocity whereas body $B$ moves with velocity double that of $A$ .
If $m_2\gg m_1$ and $u_2=0$ then $v_2=0,v_1=-u_1$
When a light body A collides against a heavy body B at rest, the body A starts moving with same speed just in opposite direction while the body B practically remains at rest.
Head on Inelastic Collision

${\mathrm{m}}_1{\vec{\mathrm{u}}}_1+{\mathrm{m}}_2{\vec{\mathrm{u}}}_2={\mathrm{m}}_1{\vec{\mathrm{v}}}_1+{\mathrm{m}}_2{\vec{\mathrm{v}}}_2$

By definition of coefficient of restitution

$\begin{array}{rr}& {\vec{\mathrm{v}}}_2-{\vec{\mathrm{v}}}_1=\mathrm{e}\left({\vec{\mathrm{u}}}_1-{\vec{\mathrm{u}}}_2\right)\\ & {\vec{\mathrm{v}}}_1=\left(\frac{{\mathrm{m}}_1-{\mathrm{e}\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\vec{\mathrm{u}}}_1+\left(\frac{(1+\mathrm{e}){\mathrm{m}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}\right){\vec{\mathrm{u}}}_2\mathrm{}\text{and}\end{array}$

${\vec{v}}_2=\left(\frac{m_2-e{m}_1}{m_1+m_2}\right){\vec{u}}_2+\left(\frac{(1+e)m_1}{m_1+m_2}\right){\vec{u}}_1$

Loss in Kinetic Energy of particles $=\frac{1}{2}\frac{m_1{m}_2}{\left(m_1+m_2\right)}\left(1-e^2\right){\left({\vec{u}}_1-{\vec{u}}_2\right)}^2$
Value of coefficient of restitution for Inelastic collision is $0<\mathrm{e}<1$

Head on Perfectly Inelastic Collision

${\mathrm{m}}_1{\vec{\mathrm{u}}}_1+{\mathrm{m}}_2{\vec{\mathrm{u}}}_2=\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)\vec{\mathrm{v}}$

Where final velocity $\vec{v}=\frac{m_1{\vec{u}}_1+m_2{\vec{u}}_2}{m_1+m_2}$
Loss in Kinetic Energy of particles $=\frac{1}{2}\frac{{\mathrm{m}}_1{\mathrm{m}}_2}{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}{\left({\vec{\mathrm{u}}}_1-{\vec{\mathrm{u}}}_2\right)}^2$
Value of coefficient of restitution for Perfectly Inelastic collision e $=0$ .