Understanding Elastic Collisions

Oct 13, 2025 by wbrum.org 96 views

Introduction

In physics, collisions are a fundamental concept that helps us understand the behavior of objects in motion. An elastic collision is a type of collision where the total kinetic energy of the system is conserved. In this article, we will explore the concept of elastic collisions and determine the velocity of mass 1 after an elastic collision between two masses.

Understanding Elastic Collisions

An elastic collision is a type of collision where the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies of the objects before the collision is equal to the sum of their kinetic energies after the collision. In an elastic collision, the objects do not lose any energy, and the total kinetic energy remains the same.

Conservation of Momentum

In an elastic collision, the total momentum of the system is also conserved. This means that the sum of the momenta of the objects before the collision is equal to the sum of their momenta after the collision. The momentum of an object is given by the product of its mass and velocity.

Mathematical Formulation

Let's consider two masses, mass 1 and mass 2, with initial velocities $v_{1i}$ and $v_{2i}$ , respectively. After an elastic collision, the final velocities of the masses are $v_{1f}$ and $v_{2f}$ , respectively. The conservation of momentum and kinetic energy can be expressed mathematically as follows:

Conservation of momentum: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
Conservation of kinetic energy: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Solving for the Final Velocity of Mass 1

We are given the following values:

Mass 1: $m_1 = 281 \, \text{kg}$
Mass 2: $m_2 = 209 \, \text{kg}$
Initial velocity of mass 1: $v_{1i} = +2.82 \, \text{m/s}$
Initial velocity of mass 2: $v_{2i} = -1.72 \, \text{m/s}$

We can use the conservation of momentum equation to solve for the final velocity of mass 1. Rearranging the equation, we get:

$m_1v_{1f} = m_1v_{1i} + m_2v_{2i} - m_2v_{2f}$

Substituting the given values, we get:

$281v_{1f} = 281(2.82) + 209(-1.72) - 209v_{2f}$

Simplifying the equation, we get:

$281v_{1f} = 790.22 - 358.28 - 209v_{2f}$

Combine like terms:

$281v_{1f} = 431.94 - 209v_{2f}$

Now, we need to find the value of $v_{2f}$ . We can use the conservation of kinetic energy equation to solve for $v_{2f}$ . Rearranging the equation, we get:

$\frac{1}{2}m_2v_{2f}^2 = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 - \frac{1}{2}m_1v_{1f}^2$

Substituting the given values, we get:

$\frac{1}{2}(209)v_{2f}^2 = \frac{1}{2}(281)(2.82)^2 + \frac{1}{2}(209)(-1.72)^2 - \frac{1}{2}(281)v_{1f}^2$

Simplifying the equation, we get:

$104.45v_{2f}^2 = 251.31 + 144.41 - 139.55v_{1f}^2$

Combine like terms:

$104.45v_{2f}^2 = 395.72 - 139.55v_{1f}^2$

Now, we have two equations with two unknowns. We can solve for $v_{1f}$ and $v_{2f}$ using these equations.

Solving the System of Equations

We can solve the system of equations using substitution or elimination. Let's use substitution. Rearranging the first equation, we get:

$v_{2f} = \frac{431.94 - 281v_{1f}}{209}$

Substituting this expression into the second equation, we get:

$104.45\left(\frac{431.94 - 281v_{1f}}{209}\right)^2 = 395.72 - 139.55v_{1f}^2$

Simplifying the equation, we get:

$104.45\left(\frac{431.94 - 281v_{1f}}{209}\right)^2 = 395.72 - 139.55v_{1f}^2$

Expanding and simplifying the equation, we get:

$104.45\left(\frac{431.94^2 - 2(431.94)(281)v_{1f} + (281)^2v_{1f}^2}{209^2}\right) = 395.72 - 139.55v_{1f}^2$

Simplifying the equation, we get:

$104.45\left(\frac{186,111.36 - 121,111.28v_{1f} + 78,841v_{1f}^2}{43,841}\right) = 395.72 - 139.55v_{1f}^2$

Combine like terms:

$104.45\left(\frac{186,111.36 - 121,111.28v_{1f} + 78,841v_{1f}^2}{43,841}\right) = 395.72 - 139.55v_{1f}^2$

Now, we can solve for $v_{1f}$ using this equation.

Solving for $v_{1f}$

We can solve for $v_{1f}$ using numerical methods or algebraic manipulation. Let's use algebraic manipulation. Rearranging the equation, we get:

$78,841v_{1f}^2 - 121,111.28v_{1f} + 186,111.36 - 395.72(43,841) = 0$

Simplifying the equation, we get:

$78,841v_{1f}^2 - 121,111.28v_{1f} + 186,111.36 - 17,255,111.96 = 0$

Combine like terms:

$78,841v_{1f}^2 - 121,111.28v_{1f} - 17,068,999.6 = 0$

Now, we can solve for $v_{1f}$ using the quadratic formula.

Quadratic Formula

The quadratic formula is given by:

$v_{1f} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation.

In this case, $a = 78,841$ , $b = -121,111.28$ , and $c = -17,068,999.6$ . Substituting these values into the quadratic formula, we get:

$v_{1f} = \frac{-(-121,111.28) \pm \sqrt{(-121,111.28)^2 - 4(78,841)(-17,068,999.6)}}{2(78,841)}$

Simplifying the equation, we get:

$v_{1f} = \frac{121,111.28 \pm \sqrt{14,573,111,974.24 + 5,384,111,111,936}}{157,682}$

Combine like terms:

$v_{1f} = \frac{121,111.28 \pm \sqrt{20,957,223,086.16}}{157,682}$

Simplifying the equation, we get:

$v_{1f} = \frac{121,111.28 \pm 145,011.28}{157,682}$

Now, we have two possible solutions for $v_{1f}$ .

Solving for $v_{1f}$ (continued)

Let's consider the two possible solutions for $v_{1f}$ :

$v_{1f} = \frac{121,111.28 + 145,011.28}{157,682}$

Simplifying the equation, we get:

$v_{1f} = \frac{266,122.56}{157,682}$

$v_{1f} = 1.69 \, \text{m/s}$

$$v_{1<br/> **Elastic Collision: Determining the Velocity of Mass 1**$

Q&A: Elastic Collision

Q: What is an elastic collision?

A: An elastic collision is a type of collision where the total kinetic energy of the system is conserved. This means that the sum of the kinetic energies of the objects before the collision is equal to the sum of their kinetic energies after the collision.

Q: What is the difference between an elastic and inelastic collision?

A: In an inelastic collision, the total kinetic energy of the system is not conserved. This means that some of the kinetic energy is converted into other forms of energy, such as heat or sound.

Q: How do you determine the velocity of an object after an elastic collision?

A: To determine the velocity of an object after an elastic collision, you need to use the conservation of momentum and kinetic energy equations. These equations can be used to solve for the final velocities of the objects involved in the collision.

Q: What is the formula for the conservation of momentum?

A: The formula for the conservation of momentum is:

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

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

Q: What is the formula for the conservation of kinetic energy?

A: The formula for the conservation of kinetic energy is:

$\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

Q: How do you solve for the final velocity of an object after an elastic collision?

A: To solve for the final velocity of an object after an elastic collision, you need to use the conservation of momentum and kinetic energy equations. These equations can be used to solve for the final velocities of the objects involved in the collision.

Q: What is the significance of the coefficient of restitution in an elastic collision?

A: The coefficient of restitution is a measure of the elasticity of a collision. It is defined as the ratio of the final velocity of an object to its initial velocity. A coefficient of restitution of 1 indicates a perfectly elastic collision, while a coefficient of restitution of 0 indicates a perfectly inelastic collision.

Q: How do you determine the coefficient of restitution in an elastic collision?

A: To determine the coefficient of restitution in an elastic collision, you need to use the conservation of momentum and kinetic energy equations. These equations can be used to solve for the final velocities of the objects involved in the collision, and then the coefficient of restitution can be calculated using the formula:

$e = \frac{v_{1f} - v_{2f}}{v_{1i} - v_{2i}}$

Q: What is the significance of the coefficient of restitution in real-world applications?

A: The coefficient of restitution is an important parameter in many real-world applications, including the design of crash test dummies, the development of safety features in vehicles, and the analysis of the impact of objects in various fields, such as engineering and physics.

Q: How do you apply the concept of elastic collision in real-world scenarios?

A: The concept of elastic collision can be applied in various real-world scenarios, including the design of safety features in vehicles, the development of crash test dummies, and the analysis of the impact of objects in various fields, such as engineering and physics.

Conclusion

In conclusion, the concept of elastic collision is an important one in physics and engineering. It can be used to analyze the impact of objects in various fields, including the design of safety features in vehicles and the development of crash test dummies. By understanding the principles of elastic collision, we can design safer and more efficient systems that can withstand various types of impacts.