Have you ever played pool or seen someone playing pool? As you play, you hit a ball at a billiard table and that ball hits the other balls on the table, causing them to move as well. A similar situation can be seen while playing volleyball. When you get ready to play, you practice with the ball. When you hit it to the ground, the ball bounces back. These situations are examples of elastic collision.
While playing pool, you hit a ball at a billiard table, and that ball hits the other balls. This is an example of elastic collision. Wikimedia Commons
Elastic Collision Definition
Elastic collisions are collisions in which the total kinetic energy stays the same before and after the collision. However, the kinetic energies of individual objects can change.
Elastic Collision: A collision in which the total kinetic energy of the objects stays the same throughout the collision.
Types of Elastic Collision
Stationary Target
As in the example of playing pool, a moving ball can hit a stationary ball and make it move as well. A moving ball has a certain velocity and thus also has kinetic energy. When it hits a stationary ball, it transfers some of its energy to the other ball. This means that the kinetic energy of the moving ball decreases, and the kinetic energy of the stationary ball increases.
Moving Target
When both objects are moving and hit each other, an elastic collision can still happen. The total kinetic energy and momentum will still be conserved, but the formulas for the final velocities will differ from the stationary target case.
Elastic Collisions Formula
Stationary Target Case
After a moving ball hits a stationary ball, it transfers some of its kinetic energy to the other. They both may have a final velocity after the collision. StudySmarter Originals
When an object with mass \(m_1\) and an initial velocity \(V_{1i}\) hits a stationary ball with mass \(m_2\), the object with mass \(m_1\) has a final velocity \(V_{1f}\) and the object with mass \(m_2\) has a final velocity \(V_{2f}\). The net linear momentum is conserved, that's why the total momentum stays the same before and after the collision.
The equation for the linear momentum conservation can be written as:
Equation 1: \(m_1V_{1i}=m_1V_{1f}+m_2V_{2f}\)
The equation for the kinetic energy conservation can be written as:
Equation 2: \(\frac 1 2 m_1(V_{1i})^2=\frac 1 2 m_1(V_{1f})^2+\frac 1 2m_2(V_{2f})^2\)
To find the final velocities, we can do some algebra. We can convert the momentum equation to:
Equation 3: \(m_1V_{1i}-m_1V_{1f}=m_2V_{2f}\)
Since \(m_1\) is common on the left side, we can rearrange the equation:
Equation 4: \(m_1(V_{1i}-V_{1f})=m_2V_{2f}\)
We can rearrange the energy equation as well. First of all, we can multiply the equation by 2.
Equation 5: \(m_1(V_{1i})^2=m_1(V_{1f})^2+m_2(V_{2f})^2\)
We can move \(m_1(V_{1f})^2\) to the left side.
Equation 6: \(m_1(V_{1i})^2-m_1(V_{1f})^2=m_2(V_{2f})^2\)
Since \(m_1\) is common on the left side, we can now rearrange the equation:
Equation 7: \(m_1((V_{1i})^2-(V_{1f})^2)=m_2(V_{2f})^2\)
Because we are subtracting two items that are in the square form, we can rewrite it as:
Equation 8: \(m_1(V_{1i}+V_{1f})(V_{1i}-V_{1f})=m_2(V_{2f})^2\)
After dividing Equation 8 by Equation 4 and then rearranging, we can find the elastic collision formulas for the stationary target case.
$$V_{1f}=\frac {m_1-m_2}{m_1+m_2}V_{1i}$$
$$V_{2f}=\frac {2m_1}{m_1+m_2}V_{1i}$$
Moving Target Case
Both moving balls with velocities can have different final velocities after an elastic collision. The total kinetic energy is still conserved. StudySmarter Originals
When an object with mass \(m_1\) and an initial velocity \(V_{1i}\) hits a ball with mass \(m_2\) and an initial velocity \(V_{2i}\), the object with mass \(m_1\) has a final velocity \(V_{1f}\) and the object with mass \(m_2\) has a final velocity \(V_{2f}\).
We can write the momentum conservation equation as:
Equation 10: \(m_1V_{1i}+m_2V_{2i}=m_1V_{1f}+m_2V_{2f}\)
We can write the kinetic energy conservation as:
Equation 11: \(\frac 12m_1(V_{1i})^2+\frac 12m_2(V_{2i})^2=\frac 12m_1(V_{1f})^2+\frac 12m_2(V_{2f})^2\)
We can rearrange Equation 10:
Equation 12: \(m_1V_{1i}-m_1V_{1f}=m_2V_{2f}-m_2V_{2i}\)
Equation 13: \(m_1(V_{1i}-V_{1f})=-m_2(V_{2i}-V_{2f})\)
Also, we can rearrange Equation 11:
Equation 14: \(m_1(V_{1i}-V_{1f})(V_{1i}+V_{1f})=-m_2(V_{2i}-V_{2f})(V_{2i}+V_{2f})\)
We can divide Equation 14 by Equation 13 and rearrange it to get the formulas for the final velocities of the objects:
$$V_{1f}=\frac{m_1-m_2}{m_1+m_2}V_{1i}+\frac{2m_2}{m_1+m_2}V_{2i}$$
$$V_{2f}=\frac{2m_1}{m_1+m_2}V_{1i}+\frac{m_2-m_1}{m_1+m_2}V_{2i}$$
Elastic Collision Equation
For the stationary target case, our velocity equations are:
$$V_{1f}=\frac {m_1-m_2}{m_1+m_2}V_{1i}$$
$$V_{2f}=\frac {2m_1}{m_1+m_2}V_{1i}$$
For the moving target case, our final velocity equations are different:
$$V_{1f}=\frac{m_1-m_2}{m_1+m_2}V_{1i}+\frac{2m_2}{m_1+m_2}V_{2i}$$
$$V_{2f}=\frac{2m_1}{m_1+m_2}V_{1i}+\frac{m_2-m_1}{m_1+m_2}V_{2i}$$
Elastic Collision Examples
A ball with a mass of 2 kg is moving to the right with a velocity of 4 m/s and hits a stationary ball with a mass of 1 kg at the billiard table. What are the final velocities of the balls?
Solution:
To find the final velocities, we can use Equation 9. As given in the example, \(V_{1i}=4 \frac ms\) and \(m_1=2 kg\). We can insert these values into the equation.
$$V_{1f}=\frac {2 kg - 1 kg}{2 kg + 1 kg}\times 4 \frac ms = \frac 1 3 \times 4 \frac ms = \frac 4 3 \frac ms$$
$$V_{2f}=\frac {2 \times 2 kg}{2 kg + 1 kg}\times 4 \frac ms = \frac 4 3 \times 4 \frac ms = \frac {16} 3 \frac ms$$
A ball with a mass of 6 kg and an initial velocity of 4 m/s hits another ball with a mass of 4 kg and an initial velocity of 2 m/s. They both are moving to the right. What are their final velocities after the collision?
Solution:
We can use our formula from above to find the final velocities. Here, \(m_1=6 kg\) and \(m_2=4 kg\), \(V_{1i} = 4 \frac ms\) and \(V_{2i}=2 \frac ms\).
$$V_{1f}=\frac{6 kg-4 kg}{6 kg+4 kg}\times 4 \frac ms + \frac{2\times 4 kg}{6 kg+4 kg}\times 2 \frac ms$$
$$V_{1f}= \frac 2{10}\times 4+ \frac 8{10}\times 2 = \frac 8{10}+\frac{16}{10}=\frac{24}{10}\Rightarrow 2.4 \frac ms$$
$$V_{2f}=\frac{2\times6 kg}{6 kg + 4 kg}\times 4\frac ms + \frac{4 kg - 6 kg}{6 kg + 4 kg}\times 2\frac ms$$
$$V_{2f} = \frac {12}{10}\times 4 + \frac{-2}{10}\times 2 = \frac{48}{10}-\frac 4{10} = \frac{44}{10}\Rightarrow 4.4 \frac ms$$
Elastic Collisions - Key takeaways
- Elastic collisions are collisions in which the total kinetic energy stays the same before and after the collision.
- The kinetic energies of individual objects can change after the collision.
- The net linear momentum is conserved.
- Playing pool and other situations involving bouncing balls can be examples of elastic collision.
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