Collisions in Two Dimensions: The Physics behind Elastic and Inelastic Collisions

Published on: March 20, 2025in

Explore collisions in physics, covering elastic and inelastic collisions in 2D.

Estimated reading time: 12 minutes

Collisions are fundamental interactions in physics, occurring when two or more bodies exert forces on each other in a short period. From billiard balls striking one another to car crashes and subatomic particle interactions, collisions are ubiquitous in nature and engineering. Understanding collisions requires a grasp of conservation laws, vectors, and energy principles. This article delves into the mechanics of two-dimensional collisions in physics, exploring their types, governing laws, and mathematical formulation.

Contents

What are Collisions in Physics?

Collisions in physics refer to interactions where two or more objects exert forces on each other for a short duration, leading to changes in their motion. Hence, these interactions play a crucial role in various physical phenomena, from everyday occurrences like car crashes and sports impacts to atomic and celestial collisions.

Collision in Physics — **Fig 1. Collisions are characterized by mass and velocity, and therefore Momentum and Kinetic Energy of the bodies in the considered system**

Types of Collisions

In Physics, since collisions are classified based on how they conserve kinetic energy:

Elastic Collisions:

Both linear momentum and kinetic energy are conserved.
Hence, no energy is lost as heat or deformation.

Inelastic Collisions:

Linear Momentum is conserved, but kinetic energy is not.
Hence, some energy is converted into heat, sound, or deformation.

Perfectly Inelastic Collisions:

A special case where objects stick together after impact.
Maximum kinetic energy loss while conserving linear momentum.

One-Dimensional vs. Two-Dimensional Collisions

One-Dimensional Collisions: Motion is along a single axis (e.g., a head-on collision between two balls).
Two-Dimensional Collisions: Motion occurs in both x and y directions, requiring vector analysis to resolve velocity components.

Applications of Collision Physics

Automobile Safety: Crumple zones absorb energy in inelastic collisions to reduce impact forces.
Sports Physics: Elastic collisions determine ball rebounds in billiards and table tennis.
Astrophysics: Planetary formation and asteroid impacts involve high-energy collisions.

Car Collision — **Fig 2. Collision of Automobiles**

Life is a series of collisions with the future; it is not the sum of what we have been, but what we yearn to be.
– Jose Ortega y Gasset

Also Read: Introduction to Particle Physics for High-School Students

3. Conservation Laws

Momentum Conservation During Collisions

Momentum is a fundamental quantity in physics that remains conserved in all types of collisions, provided there is no external force acting on the system. The principle of conservation of momentum states:

$\text{Total momentum before collision} = \text{Total momentum after collision}$

Mathematically, for a system of two colliding objects:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

where:

$m_1, m_2$ are the masses of the objects,
$v_{1i}, v_{2i}$ are the initial velocities,
$v_{1f}, v_{2f}$ are the final velocities after the collision.

This principle holds true for both elastic and inelastic collisions.

Momentum Conservation in Different Types of Collisions

1. Elastic Collisions

Both momentum and kinetic energy are conserved.
Example: Two billiard balls colliding.

For a 1D elastic collision, the system of equations is:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f$

$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

These two equations help solve for the final velocities.

2. Inelastic Collisions

Momentum is conserved, but kinetic energy is not.
Example: A car crash where the cars crumple upon impact.

Momentum conservation equation:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

However, kinetic energy before and after the collision differs because some energy is converted into heat, sound, or deformation.

3. Perfectly Inelastic Collisions

Momentum is conserved, but kinetic energy loss is maximum.
The objects stick together after collision and move as a single mass.

Momentum conservation equation:

$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$

where $v_f$ is the final velocity of the combined mass.

Momentum Conservation in Two-Dimensional Collisions

For collisions in two dimensions, momentum is conserved separately in the x and y directions:

$m_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}$

$m_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}$

To find the final velocities, we use vector addition:

$v_f = \sqrt{v_{fx}^2 + v_{fy}^2}$

Example Problem: Momentum Conservation in an Elastic Collision

A 2 kg ball moving at 3 m/s collides elastically with a 4 kg ball at rest. Find their velocities after collision.

Using the momentum and energy conservation equations, we can solve for $v_{1f}$ and $v_{2f}$ .

Answer:

Firstly, apply Momentum Conservation

The total momentum before and after the collision must be equal:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

Substituting values:

$(2)(3) + (4)(0) = (2) v_{1f} + (4) v_{2f}$

$6 = 2 v_{1f} + 4 v_{2f}$

$3 = v_{1f} + 2 v_{2f} \quad \text{(Equation 1)}$

Secondly, apply Kinetic Energy Conservation

Since the collision is elastic, kinetic energy is also conserved:

$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

Cancel out $\frac{1}{2}$ from all terms:

$m_1 v_{1i}^2 + m_2 v_{2i}^2 = m_1 v_{1f}^2 + m_2 v_{2f}^2$

Substituting values:

$(2)(3^2) + (4)(0^2) = (2) v_{1f}^2 + (4) v_{2f}^2$

$18 = 2 v_{1f}^2 + 4 v_{2f}^2$

$9 = v_{1f}^2 + 2 v_{2f}^2 \quad \text{(Equation 2)}$

Finally, solve for $v_{1f}$ and $v_{2f}$

From Equation (1):

$v_{1f} = 3 - 2 v_{2f}$

Substituting into Equation (2):

$(3 - 2 v_{2f})^2 + 2 v_{2f}^2 = 9$

Expanding:

$9 - 12 v_{2f} + 4 v_{2f}^2 + 2 v_{2f}^2 = 9$

$6 v_{2f}^2 - 12 v_{2f} = 0$

Factor:

$6 v_{2f} (v_{2f} - 2) = 0$

So,

$v_{2f} = 0 \quad \text{or} \quad v_{2f} = 2$

If $v_{2f} = 0$ , then from Equation (1), $v_{1f} = 3$ , which contradicts our assumption of interaction.
So, we take $v_{2f} = 2$ , and plug it into Equation (1):

$v_{1f} = 3 - 2(2) = 3 - 4 = -1$

Therefore, Final Answer:

$v_{1f} = -1$ m/s (ball 1 moves in the opposite direction)
$v_{2f} = 2$ m/s (ball 2 moves forward)

This means after the collision, the first ball rebounds in the opposite direction at 1 m/s, and the second ball moves forward at 2 m/s.

Energy Conservation During Collisions

In physics, during any collision, energy is always conserved, but the way kinetic energy behaves depends on the type of collision.

The total energy of a system before and after a collision remains constant. This includes:

Kinetic Energy (KE): Energy due to motion
Potential Energy (PE): Energy due to position
Heat, Sound, and Deformation Energy (in inelastic collisions)

However, whether kinetic energy is conserved or not depends on the type of collision.

Kinetic Energy in Different Types of Collisions

(A) Elastic Collisions

Both momentum and kinetic energy are conserved.
No energy is lost to heat, sound, or deformation.
Example: Billiard ball collisions, gas molecules in an ideal gas.

Mathematically:

$\frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2$

Hence, this equation, along with momentum conservation, helps find post-collision velocities.

(B) Inelastic Collisions

Momentum is conserved, but kinetic energy is not.
Consequently, some kinetic energy is converted into heat, sound, or deformation.
Example: A clay ball hitting the ground and sticking.

$\text{KE before} > \text{KE after}$

Only momentum conservation applies:

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

(C) Perfectly Inelastic Collisions

Momentum is conserved, but kinetic energy loss is maximum.
The two objects stick together and move as one mass.
Example: A car crash where cars lock together.

Momentum conservation:

$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$

Kinetic energy is always lost:

$\text{KE lost} = \text{Initial KE} - \text{Final KE}$

$\Delta KE = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 - \frac{1}{2} (m_1 + m_2) v_f^2$

The lost energy is transformed whether into heat, sound, or internal deformation.

Example Problem: Energy in a Perfectly Inelastic Collision

A 5 kg object moving at 6 m/s collides with a 3 kg object at rest. If they stick together find the final velocity and energy lost.

Firstly, Apply Momentum Conservation

$(5)(6) + (3)(0) = (5+3) v_f$

$v_f = 3.75 \text{ m/s}$

Secondly, calculate Initial and Final KE

$KE_{\text{initial}} = \frac{1}{2} (5) (6)^2 + \frac{1}{2} (3) (0)^2$

$KE_{\text{initial}} = \frac{1}{2} (5) (36) = 90 \text{ J}$

Hence, final KE:

$KE_{\text{final}} = \frac{1}{2} (8) (3.75)^2$

$KE_{\text{final}} = \frac{1}{2} (8) (14.06) = 56.25 \text{ J}$

Finally, find Energy Lost

$KE_{\text{lost}} = 90 - 56.25 = 33.75 \text{ J}$

So, 33.75 J of energy was lost, transformed into heat, sound, and deformation.

Key Takeaways

Elastic Collisions: Both kinetic energy and momentum conserved.
Inelastic Collisions: Only momentum conserved, kinetic energy partially lost.
Perfectly Inelastic Collisions: Maximum kinetic energy lost. The objects stick together.

8. Example Problems

8.1 Elastic Collisions

Problem 1

Two balls, mass 2 kg and 3 kg, collide elastically. If the first moves at 4 m/s along x-axis, the second at 2 m/s along y-axis, then find final velocities.

Solution

Since:

$m_1 = 2$ kg, $v_{1i} = 4$ m/s along x-axis
$m_2 = 3$ kg, $v_{2i} = 2$ m/s along y-axis

Momentum is conserved in both x and y directions:

$m_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}$

$m_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}$

Since $v_{1iy} = 0$ and $v_{2ix} = 0$ , we solve separately:

$2(4) = 2 v_{1fx} + 3 v_{2fx} \quad (1)$

$3(2) = 2 v_{1fy} + 3 v_{2fy} \quad (2)$

Then, using kinetic energy conservation:

$\frac{1}{2} (2)(4^2) + \frac{1}{2} (3)(2^2) = \frac{1}{2} (2) v_{1f}^2 + \frac{1}{2} (3) v_{2f}^2$

Finally,

$v_{1f} \approx 2.83$ m/s
$v_{2f} \approx 3.61$ m/s
at angles based on conservation laws.

Problem 2

A ball of 0.5 kg moving at 3 m/s strikes another stationary identical ball. Then, find their velocities post-collision.

Solution

Since:

$m_1 = 0.5$ kg, $v_{1i} = 3$ m/s
$m_2 = 0.5$ kg, $v_{2i} = 0$

Hence, final velocities (elastic collision formula for identical masses):

$v_{1f} = v_{2i} =0$

$v_{2f} = v_{1i} = 3 \text{ m/s}$

So, the first ball stops, and the second moves at 3 m/s.

Problem 3

If, two identical ice pucks collide at 45-degree angles with initial speeds of 2 m/s, then find their final speeds.

Puck Collision — **Fig 3. Ice Pucks Colliding Elastically**

Solution

Since:

Identical pucks, $m_1 = m_2$ , $v_{1i} = v_{2i} = 2$ m/s
Collision at 45° angles

Using conservation of momentum in both axes and symmetry,

$v_{1f} = v_{2f} = 2 \text{ m/s}$

Hence, they move at right angles to their original paths.

8.2 Inelastic Collisions

Problem 4

A 3 kg ball moving at 5 m/s hits a 4 kg ball at rest. If they stick together, find their final velocity.

Solution

Since:

$m_1 = 3$ kg, $v_{1i} = 5$ m/s
$m_2 = 4$ kg, $v_{2i} = 0$ m/s

Then using momentum conservation:

$(m_1 + m_2) v_f = m_1 v_{1i} + m_2 v_{2i}$

$(3 + 4) v_f = (3)(5) + (4)(0)$

$7 v_f = 15$

$v_f = \frac{15}{7} = 2.14 \text{ m/s}$

Therefore, Final velocity = $2.14$ m/s (both move together).

Problem 5

Two ice skaters collide and move together. If one has a velocity of 3 m/s, the other is stationary find their velocity post-collision.

**Fig 4. Inelastic Collision of Ice Skaters**

Solution

Since:

$m_1 = m$ , $v_{1i} = 3$ m/s
$m_2 = m$ , $v_{2i} = 0$

Since they stick together:

$(m + m) v_f = m v_{1i} + m v_{2i}$

$2m v_f = m(3) + m(0)$

$v_f = \frac{3}{2} = 1.5 \text{ m/s}$

Therefore, Final velocity = $1.5$ m/s (both move together).

Problem 6

If a bullet of mass 0.1 kg moving at 500 m/s embeds itself in a wooden block of mass 1 kg, find their velocity after collision.

Solution

Since:

$m_1 = 0.1$ kg, $v_{1i} = 500$ m/s
$m_2 = 1$ kg, $v_{2i} = 0$

Firstly, using momentum conservation:

$(m_1 + m_2) v_f = m_1 v_{1i} + m_2 v_{2i}$

$(0.1 + 1) v_f = (0.1)(500) + (1)(0)$

$1.1 v_f = 50$

$v_f = \frac{50}{1.1} \approx 45.45 \text{ m/s}$

Hence, Final velocity = $45.45$ m/s (both move together).

FAQs

1. What is the main difference between elastic and inelastic collisions in physics?

Ans: In elastic collisions, both momentum and kinetic energy are conserved, whereas in inelastic collisions, only momentum is conserved, and kinetic energy is partially lost.

2. How do we analyze a collision in two dimensions?

Ans: Firstly, we break the motion into x- and y-components and then apply the conservation of momentum separately in each direction.

3. Why is kinetic energy not always conserved in a collision?

Ans: Some of the kinetic energy is converted into heat, sound, or deformation of objects, especially in inelastic collisions.

4. What role do vectors play in two-dimensional collisions?

Ans: Since motion occurs in multiple directions, momentum is treated as a vector quantity, requiring vector addition and decomposition.

5. Can a completely inelastic collision still conserve total energy?

Ans: Yes, while kinetic energy is lost, total mechanical energy (including other forms like heat or deformation) remains conserved.

6. What is an example of an elastic collision in real life?

Ans: A collision between billiard balls is nearly elastic since they retain most of their kinetic energy.

7. How does mass affect the outcome of a two-dimensional collision?

Ans: Heavier objects tend to change velocity less upon impact, while lighter ones experience more noticeable deflections.

8. How does the angle of impact influence the final motion of colliding objects?

Ans: The angle determines how momentum is distributed between x- and y-components, and consequently affects the direction and speed after collision.

9. What happens when two objects of equal mass collide elastically at an angle?

Ans: If one object was initially at rest, they typically move at 90° angles after collision, following the momentum conservation laws.

10. Why do we use the center of mass frame in collision analysis?

Ans: It simplifies calculations because it allows us to analyze motion relative to the system’s moving center of mass, making conservation laws easier to apply.

Conclusion

Collisions in two dimensions are essential in physics, governing interactions from microscopic particles to large-scale astrophysical events. Understanding momentum and energy conservation allows us to predict post-collision motion in both elastic and inelastic cases. Unlike one-dimensional collisions, two-dimensional cases require vector analysis, making problem-solving more complex but also more applicable to real-world scenarios like car crashes, sports dynamics, and space object interactions. Hence, by applying conservation laws, we can analyze and solve practical collision problems in physics, reinforcing their significance in physics and engineering.

References

Burko, L. M. (2019). Two-Dimensional Collisions and Conservation of Momentum. The Physics Teacher, 57(7), 487-489. https://doi.org/10.1119/1.5126832
Ogura, A. (2018). Diagrammatic Approach for Investigating Two Dimensional Elastic Collisions in Momentum Space I: Newtonian Mechanics. World Journal of Mechanics, 8(9), 343-352. https://doi.org/10.4236/wjm.2018.89025
Burton, J. C., Lu, P. Y., & Nagel, S. R. (2013). Collision Dynamics of Particle Clusters in a Two-Dimensional Granular Gas. Physical Review E, 88(6), 062204. https://doi.org/10.1103/PhysRevE.88.062204

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Author
Latest Posts

Uddhav Thakore

Collisions in Two Dimensions: The Physics behind Elastic and Inelastic Collisions

What are Collisions in Physics?

Types of Collisions

Elastic Collisions:

Inelastic Collisions:

Perfectly Inelastic Collisions:

One-Dimensional vs. Two-Dimensional Collisions

Applications of Collision Physics

Also Read: Introduction to Particle Physics for High-School Students

3. Conservation Laws

Momentum Conservation During Collisions

Momentum Conservation in Different Types of Collisions

1. Elastic Collisions

2. Inelastic Collisions

3. Perfectly Inelastic Collisions

Momentum Conservation in Two-Dimensional Collisions

Example Problem: Momentum Conservation in an Elastic Collision

Answer:

Firstly, apply Momentum Conservation

Secondly, apply Kinetic Energy Conservation

Finally, solve for ​ and ​

Therefore, Final Answer:

Energy Conservation During Collisions

Kinetic Energy in Different Types of Collisions

(A) Elastic Collisions

(B) Inelastic Collisions

(C) Perfectly Inelastic Collisions

Example Problem: Energy in a Perfectly Inelastic Collision

Firstly, Apply Momentum Conservation

Secondly, calculate Initial and Final KE

Finally, find Energy Lost

Key Takeaways

8. Example Problems

8.1 Elastic Collisions

Problem 1

Solution

Problem 2

Solution

Problem 3

Solution

8.2 Inelastic Collisions

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

FAQs

1. What is the main difference between elastic and inelastic collisions in physics?

2. How do we analyze a collision in two dimensions?

3. Why is kinetic energy not always conserved in a collision?

4. What role do vectors play in two-dimensional collisions?

5. Can a completely inelastic collision still conserve total energy?

6. What is an example of an elastic collision in real life?

7. How does mass affect the outcome of a two-dimensional collision?

8. How does the angle of impact influence the final motion of colliding objects?

9. What happens when two objects of equal mass collide elastically at an angle?

10. Why do we use the center of mass frame in collision analysis?

Conclusion

References

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Finally, solve for $v_{1f}$ and $v_{2f}$