Collision

Collisions are all about objects bumping into each other, causing changes in motion and energy. These encounters are a lot like the conflicts and interactions we face in real life, where different forces and impacts shape our paths. A collision occurs when two objects come into direct contact and exert forces on each other over a short period. There are two main types of collisions: elastic, where both momentum and kinetic energy are conserved, like billiard balls colliding; and inelastic, where momentum is conserved but some kinetic energy is lost, such as in car crashes. Understanding collisions helps us predict and analyze physical interactions in everyday life and various scientific fields.

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This Story also Contains

1. Collision
2. Momentum And Energy Conservation In Collision
3. Solved Examples Based on Collision
4. Summary

In this article, we are going to study about Collision stage of collision and more. This collision concept belongs to the chapter work, energy, and power, which is an important chapter in Class 11 physics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), National Eligibility Entrance Test (NEET), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), no questions have been asked on this concept. And for NEET twp questions were asked from this concept. Let's read this entire article to understand collision in depth.

Collision

The interaction between two or more objects is called a collision. And during this interaction strong force acts between two or more bodies for a short time as a result of which the energy and momentum of the interacting particle change.

Stages of Collision

There are three distinct identifiable stages in collision

1. Before the collision

The interaction forces are zero

2. During the collision

The interaction forces are very large and these forces act for a very short time. And because of these interaction forces the energy and momentum of the interacting particle change.

3. After the collision

The interaction forces are zero

Momentum And Energy Conservation In Collision

The magnitude of the interacting force is often unknown, therefore, Newton’s second law cannot be used. However, the law of conservation of momentum is useful in relating the initial and final velocities.

1. Momentum conservation

In a collision, the effect of external forces such as gravity or friction is not taken into account as due to the small duration of collision (t) average impulsive force responsible for the collision is much larger than the external force acting on the system and since this impulsive force is 'Internal' therefore the total momentum of the system always remains conserved.

2. Energy conservation

In a collision 'total energy' is also always conserved. Here total energy includes all forms of energy such as mechanical energy, internal energy, excitation energy, radiant energy, etc.

But in a collision Kinetic energy may or may not be conserved.

Coefficient of Restitution

The ratio of the relative velocity of separation to the relative velocity of approach.

$e=\frac{v_2-v_1}{u_1-u_2}=\frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}}$

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Solved Examples Based on Collision

Example 1: A ball of mass 'm' moving with velocity 'v ', collides inelastically with another identical ball. After the collision, the 1st ball moves with velocity $v/\sqrt{2}$ in a direction perpendicular to the initial direction of motion. Find the speed of the second ball after the collision?

1) $\frac{\sqrt{3}}{2}v$
2) $\frac{1}{2}v$
3) $\sqrt{\frac{3}{2}}v$
4) $\sqrt{2}v$

Solution:

Momentum conservation along X-axis
$\begin{array}{rl}& mv=m{v}_1\cos \theta \\ & v=v_1\cos \theta \dots .(1)\end{array}$

along $y$-axis

$\begin{array}{rl}& 0=mv/\sqrt{2}-m{v}_1\sin \theta \\ & \frac{v}{\sqrt{2}}=v_1\sin \theta \dots (2)\\ & v^2+\frac{v^2}{2}=v_1^2({\sin }^2\theta +{\cos }^2\theta )\\ & v_1^2=\frac{3v^2}{2}\Rightarrow v_1=\sqrt{\frac{3}{2}}v\end{array}$

Hence, the answer is the option (1).

Example 2: A large number (n) of identical beads, each of mass m and radius r are strung on a thin smooth rigid horizontal rod of length L (L>>r) and are at rest at random positions. The rod is mounted between two rigid supports (see figure). If one of the beads is now given a speed, the average force experienced by each support after a long time is (assume all collisions are elastic) :

1) $\frac{m{v}^2}{L-nr}$
2) $\frac{m{v}^2}{L-2nr}$
3) $\frac{m{v}^2}{2(L-nr)}$
4) Zero

Solution:

Perfectly Elastic Collision

The law of conservation of momentum and that of Kinetic Energy hold good.

wherein
$\begin{array}{rl}& \frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2\\ & m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2\\ & m_1,m_2:\text{masses}\end{array}$

$u_1,v_1$ : initial and final velocity of the mass $m_1$
$u_2,v_2$ : initial and final velocity of the mass $m_2$
Average time for one collision

$=\frac{2L-4nr}{v}$
Since the diameter of each base $=2\mathrm{r}$
Change in momentum per collision $=2\mathrm{mv}$

$\begin{array}{rl}& \therefore F_{av}=\frac{\mathrm{\Delta }{p}_{av}}{\mathrm{\Delta }{t}_{av}}=\frac{2mv}{(2L-4nr)/v}=\frac{m{v}^2}{L-2nr}\\ & F_{av}=\frac{m{v}^2}{L-2nr}\end{array}$

Hence, the answer is the option(2).

Example 3: A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss (in Joule) during the collision is :

1) 0.34 Joule

2) 0.16 Joule

3) 1 Joule

4) 0.67 Joule

Solution:
$\text{Initial kinetic energy}=\frac{1}{2}\times (0.50)\times 4=1J$
After collision momentum $=1.5\mathrm{v}$

From momentum conservation,

$1.5v=1\Rightarrow v=\frac{2}{3}\mathrm{m}/\mathrm{s}$
Final kinetic energy

$=\frac{1}{2}\times (1.5)\times {\left(\frac{2}{3}\right)}^2=\frac{3}{4}\times \frac{4}{9}=\frac{1}{3}J$
Loss in kinetic energy

$=K\cdot E_f-K\cdot E_i=-\frac{2}{3}J$
So, the loss will be 0.67 Joule

Hence, the answer is the option(4).

Summary

Collisions involve interactions where objects exert forces on each other, leading to changes in momentum and energy. There are two main types: elastic (both momentum and kinetic energy conserved) and inelastic (only momentum conserved). Understanding collisions is crucial for solving physics problems related to momentum and energy conservation, particularly in competitive exams like JEE and NEET.