Elastic Collisions: Videos & Practice Problems

In elastic collisions, both momentum and kinetic energy are conserved. The conservation of momentum is expressed as (m1+m2)(v1+v2)=(m1+m2)(v1+v2). For a moving object colliding with a stationary one, simplified equations for final velocities are (m_1 - m_2) / (m_1 + m_2) × ⁢v1 initial and (2⁢m_1 /(m1+m2))⁢ × v1initial. Understanding these principles is crucial for analyzing collisions in physics.

Intro To Elastic Collisions

In physics, collisions between objects can be categorized into two main types: elastic and inelastic collisions. In elastic collisions, both momentum and kinetic energy are conserved. This means that the total momentum before the collision equals the total momentum after the collision, and the total kinetic energy before the collision equals the total kinetic energy after the collision.

To analyze elastic collisions, we utilize the principle of conservation of momentum, which can be expressed with the equation:

$$m_1 v_{1,\text{initial}} + m_2 v_{2,\text{initial}} = m_1 v_{1,\text{final}} + m_2 v_{2,\text{final}}$$

In this equation, \(m_1\) and \(m_2\) represent the masses of the two colliding objects, while \(v_{1,\text{initial}}\) and \(v_{2,\text{initial}}\) are their initial velocities. The final velocities are denoted as \(v_{1,\text{final}}\) and \(v_{2,\text{final}}\).

For elastic collisions, an additional equation is necessary to account for the conservation of kinetic energy, which can be represented as:

$$v_{1,\text{initial}} + v_{1,\text{final}} = v_{2,\text{initial}} + v_{2,\text{final}}$$

This equation highlights the relationship between the initial and final velocities of the two objects involved in the collision. It is important to note that the order of the variables differs from the momentum equation, which can help in memorization: the momentum equation follows the pattern \(m_1, m_2\) and \(v_{1,\text{initial}}, v_{1,\text{final}}, v_{2,\text{initial}}, v_{2,\text{final}}\), while the elastic collision equation follows the pattern \(v_{1,\text{initial}}, v_{1,\text{final}}, v_{2,\text{initial}}, v_{2,\text{final}}\).

When solving problems involving elastic collisions, you often end up with a system of equations due to the presence of two unknowns (the final velocities). A common method to solve this system is through equation addition, where you align the equations to eliminate one of the variables. For instance, if you have:

$$-2 = 5 v_{1,\text{final}} + 3 v_{2,\text{final}}$$

and

$$2 + v_{1,\text{final}} = -4 + v_{2,\text{final}}$$

you can manipulate these equations to isolate and solve for one of the unknowns. After finding one variable, you can substitute it back into either equation to find the other variable.

In summary, understanding the principles of conservation of momentum and kinetic energy is crucial for solving elastic collision problems. By applying the appropriate equations and methods, you can determine the final velocities of colliding objects effectively.

Head-On Elastic Collision

In this scenario, we analyze a head-on elastic collision between two blocks of equal mass. The initial velocities are given as \( v_{A, \text{initial}} = 5 \, \text{m/s} \) for block A and \( v_{B, \text{initial}} = -3 \, \text{m/s} \) for block B, indicating that block B is moving to the left. The goal is to determine the final velocities of both blocks after the collision.

To solve this problem, we apply the principles of conservation of momentum and the characteristics of elastic collisions. The conservation of momentum equation can be expressed as:

\[ m_A v_{A, \text{initial}} + m_B v_{B, \text{initial}} = m_A v_{A, \text{final}} + m_B v_{B, \text{final}} \]

Since the masses of the blocks are equal (\( m_A = m_B \)), we can simplify the equation by canceling out the mass terms:

\[ v_{A, \text{initial}} + v_{B, \text{initial}} = v_{A, \text{final}} + v_{B, \text{final}} \]

Substituting the initial velocities into the equation gives:

\[ 5 + (-3) = v_{A, \text{final}} + v_{B, \text{final}} \]

This simplifies to:

\[ 2 = v_{A, \text{final}} + v_{B, \text{final}} \quad \text{(1)} \]

Next, we use the elastic collision equation, which states that the sum of the initial velocities equals the sum of the final velocities:

\[ v_{A, \text{initial}} - v_{B, \text{initial}} = v_{B, \text{final}} - v_{A, \text{final}} \]

Substituting the known values results in:

\[ 5 - (-3) = v_{B, \text{final}} - v_{A, \text{final}} \]

This simplifies to:

\[ 8 = v_{B, \text{final}} - v_{A, \text{final}} \quad \text{(2)} \]

Now, we have a system of equations to solve. From equation (1), we can express \( v_{B, \text{final}} \) in terms of \( v_{A, \text{final}} \):

\[ v_{B, \text{final}} = 2 - v_{A, \text{final}} \]

Substituting this into equation (2) gives:

\[ 8 = (2 - v_{A, \text{final}}) - v_{A, \text{final}} \]

Solving this results in:

\[ 8 = 2 - 2v_{A, \text{final}} \]

Rearranging yields:

\[ 2v_{A, \text{final}} = 2 - 8 \]

\[ 2v_{A, \text{final}} = -6 \]

\[ v_{A, \text{final}} = -3 \, \text{m/s} \]

Now substituting \( v_{A, \text{final}} \) back into equation (1) to find \( v_{B, \text{final}} \):

\[ 2 = -3 + v_{B, \text{final}} \]

Thus, we find:

\[ v_{B, \text{final}} = 5 \, \text{m/s} \]

In conclusion, after the collision, block A moves to the left with a velocity of \( -3 \, \text{m/s} \), while block B moves to the right with a velocity of \( 5 \, \text{m/s} \). An interesting observation is that in elastic collisions involving equal masses, the blocks effectively exchange their velocities. This principle can be a useful shortcut in similar problems.

Special Equations in Elastic Collisions

In the study of elastic collisions, a common scenario involves a moving object colliding with a stationary object. This setup allows for simplified calculations of the final velocities of both objects using specific equations derived from the principles of momentum conservation and elastic collisions. In these problems, the stationary object's initial velocity is zero, which streamlines the equations used to determine the final velocities.

The final velocities can be calculated using the following formulas:

For the moving object (object 1):

$$ v_{1 \text{ final}} = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{1 \text{ initial}} $$

For the stationary object (object 2):

$$ v_{2 \text{ final}} = \frac{2 m_1}{m_1 + m_2} \cdot v_{1 \text{ initial}} $$

In these equations, \( m_1 \) is the mass of the moving object, \( m_2 \) is the mass of the stationary object, and \( v_{1 \text{ initial}} \) is the initial velocity of the moving object. Both equations share a common denominator of the total mass, which highlights the relationship between the masses and their effect on the final velocities.

To illustrate these concepts, consider a scenario with a boulder (mass = 40 kg) and a golf ball (mass = 0.1 kg) colliding in three different cases. In the first case, when two boulders of equal mass collide, the first boulder stops, and the second boulder moves forward with the same speed as the first boulder had initially. This outcome aligns with the principle that equal masses exchange velocities during an elastic collision.

In the second case, when the golf ball strikes the boulder, the golf ball rebounds backward at nearly the same speed it approached with, while the boulder gains a small amount of speed. This occurs because the boulder, being significantly more massive, absorbs only a tiny fraction of the golf ball's momentum.

In the final case, when the boulder hits the golf ball, the boulder retains most of its speed, while the golf ball accelerates to a much higher speed. This demonstrates that a massive object can impart a significant increase in velocity to a much lighter object upon collision.

Overall, the outcomes of these collisions can be summarized as follows: when two objects of equal mass collide, they exchange velocities; when a lighter object collides with a much heavier object, the lighter object rebounds with a speed close to its initial speed in the opposite direction, while the heavier object gains only a small amount of speed; and when a heavier object collides with a lighter one, the heavier object retains its speed, and the lighter object gains a substantial increase in speed.

Understanding these principles is crucial for analyzing elastic collisions in various physical contexts, such as billiards or other scenarios involving momentum transfer.

Here’s what students ask on this topic:

What is an elastic collision in physics?

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. This means that the total momentum and the total kinetic energy of the system remain constant before and after the collision. In mathematical terms, for two colliding objects, the conservation of momentum is expressed as:

$(m_1+m_2)(v_1+v_2)=(m_1+m_2)(v_1+v_2)$

Additionally, the conservation of kinetic energy is given by:

$K_{\mathrm{initial}}=K_{\mathrm{final}}$

Elastic collisions are often exemplified by collisions between billiard balls or gas molecules.

How do you solve elastic collision problems using conservation of momentum and kinetic energy?

To solve elastic collision problems, you need to use both the conservation of momentum and the conservation of kinetic energy. The steps are as follows:

1. Write the conservation of momentum equation:

$m_1{v}_1+m_2{v}_2=m_1{v}_1\text{'}+m_2{v}_2\text{'}$

2. Write the conservation of kinetic energy equation:

$0.5m_1{v}_1^2+0.5m_2{v}_2^2=0.5m_1{v}_1\text{'}^2+0.5m_2{v}_2\text{'}^2$

3. Solve the system of equations to find the final velocities $v_1\text{'}$ and $v_2\text{'}$.

These steps will help you determine the final velocities of the colliding objects.

What is the elastic collision equation and how is it used?

The elastic collision equation is a special equation used to solve problems involving elastic collisions. It is given by:

$v_1+v_1\text{'}=v_2+v_2\text{'}$

This equation is derived from the conservation of kinetic energy and is used alongside the conservation of momentum equation:

$m_1{v}_1+m_2{v}_2=m_1{v}_1\text{'}+m_2{v}_2\text{'}$

By using these two equations together, you can solve for the final velocities of the colliding objects. This method is particularly useful when dealing with two unknowns in elastic collision problems.

How do you solve elastic collision problems when one object is stationary?

When one object is stationary in an elastic collision, the problem simplifies significantly. The initial velocity of the stationary object is zero, which allows us to use simplified equations for the final velocities:

1. For the moving object:

$(m_1-m_2)/(m_1+m_2)\times v_{1}initial$

2. For the stationary object:

$(2m_1/(m_1+m_2)\times v_{1}initial$

These equations allow you to directly calculate the final velocities without solving a system of equations. This method is particularly useful for problems involving a moving object colliding with a stationary one, such as a sliding block hitting a stationary block.

What happens to the velocities of two objects of equal mass in an elastic collision?

In an elastic collision between two objects of equal mass, the objects essentially exchange their velocities. If object 1 with mass $m_1$ and initial velocity $v_1$ collides with object 2 with mass $m_2$ and initial velocity $v_2$, the final velocities will be:

$v_1\text{'}=v_2$

$v_2\text{'}=v_1$

This means that after the collision, object 1 will have the initial velocity of object 2, and object 2 will have the initial velocity of object 1. This behavior is often observed in billiard balls or other similar systems where the masses are equal.