## Elastic Collisions in One Dimension

As you read, pay attention to Example 8.4. In this example, one of the objects is initially at rest (its velocity equals zero), so it does not have an initial momentum. This lets us simplify the conservation of energy momentum equations. Then, by using the equations for conservation of energy and momentum, we can solve for final velocity after collision.

Now, to solve problems involving one-dimensional elastic collisions between two objects we can use the equations for conservation of momentum and conservation of internal kinetic energy. First, the equation for conservation of momentum for two objects in a one-dimensional collision is

$p_{1}+p_{2}=p_{1}^{\prime}+p_{2}^{\prime} \quad\left(F_{\text {net }}=0\right)$

or

$m_{1} v_{1}+m_{2} v_{2}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime} \quad\left(F_{\text {net }}=0\right)$,

where the primes (') indicate values after the collision. By definition, an elastic collision conserves internal kinetic energy, and so the sum of kinetic energies before the collision equals the sum after the collision. Thus,

$\frac{1}{2} m_{1} v_{1}^{2}+\frac{1}{2} m_{2} v_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{\prime 2}+\frac{1}{2} m_{2} v_{2}^{\prime} 2 \quad(\text { two-object elastic collision) }$

expresses the equation for conservation of internal kinetic energy in a one-dimensional collision.

#### Example 8.4 Calculating Velocities Following an Elastic Collision

Calculate the velocities of two objects following an elastic collision, given that

$m_{1}=0.500 \mathrm{~kg}, m_{2}=3.50 \mathrm{~kg}, v_{1}=4.00 \mathrm{~m} / \mathrm{s}, \text { and } v_{2}=0$.

#### Strategy and Concept

First, visualize what the initial conditions mean – a small object strikes a larger object that is initially at rest. This situation is slightly simpler than the situation shown in Figure 8.6 where both objects are initially moving. We are asked to find two unknowns (the final velocities $v_{1}^{\prime}$ and $v_{2}^{\prime}$). To find two unknowns, we must use two independent equations. Because this collision is elastic, we can use the above two equations. Both can be simplified by the fact that object 2 is initially at rest, and thus $v_{2}=0$. Once we simplify these equations, we combine them algebraically to solve for the unknowns.

#### Solution

For this problem, note that $v_{2}=0$ and use conservation of momentum. Thus,

$p_{1}=p_{1}^{\prime}+p_{2}^{\prime}$

or

$m_{1} v_{1}=m_{1} v_{1}^{\prime}+m_{2} v_{2}^{\prime}$.

Using conservation of internal kinetic energy and that $v_{2}=0$,

$\frac{1}{2} m_{1} v_{1}^{2}=\frac{1}{2} m_{1} v_{1}^{\prime}{ }^{2}+\frac{1}{2} m_{2} v_{2}^{\prime}{ }^{2}$.

Solving the first equation (momentum equation) for $v_{2}^{\prime}$, we obtain

$v_{2}^{\prime}=\frac{m_{1}}{m_{2}}\left(v_{1}-v_{1}^{\prime}\right)$.

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable $v_{2}^{\prime}$, leaving only $v_{1}^{\prime}$ as an unknown (the algebra is left as an exercise for the reader). There are two solutions to any quadratic equation; in this example, they are

$v_{1}^{\prime}=4.00 \mathrm{~m} / \mathrm{s}$

and

$v_{1}^{\prime}=-3.00 \mathrm{~m} / \mathrm{s}$.

As noted when quadratic equations were encountered in earlier chapters, both solutions may or may not be meaningful. In this case, the first solution is the same as the initial condition. The first solution thus represents the situation before the collision and is discarded. The second solution $\left(v_{1}^{\prime}=-3.00 \mathrm{~m} / \mathrm{s}\right)$ is negative, meaning that the first object bounces backward. When this negative value of $v_{1}^{\prime}$ is used to find the velocity of the second object after the collision, we get

$v_{2}^{\prime}=\frac{m_{1}}{m_{2}}\left(v_{1}-v_{1}^{\prime}\right)=\frac{0.500 \mathrm{~kg}}{3.50 \mathrm{~kg}}[4.00-(-3.00)] \mathrm{m} / \mathrm{s}$

or

$v_{2}^{\prime}=1.00 \mathrm{~m} / \mathrm{s}$.

#### Discussion

The result of this example is intuitively reasonable. A small object strikes a larger one at rest and bounces backward. The larger one is knocked forward, but with a low speed. (This is like a compact car bouncing backward off a full-size SUV that is initially at rest.) As a check, try calculating the internal kinetic energy before and after the collision. You will see that the internal kinetic energy is unchanged at 4.00 J. Also check the total momentum before and after the collision; you will find it, too, is unchanged.

The equations for conservation of momentum and internal kinetic energy as written above can be used to describe any one-dimensional elastic collision of two objects. These equations can be extended to more objects if needed.

#### Making Connections: Take-Home Investigation—Ice Cubes and Elastic Collision

Find a few ice cubes which are about the same size and a smooth kitchen tabletop or a table with a glass top. Place the ice cubes on the surface several centimeters away from each other. Flick one ice cube toward a stationary ice cube and observe the path and velocities of the ice cubes after the collision. Try to avoid edge-on collisions and collisions with rotating ice cubes. Have you created approximately elastic collisions? Explain the speeds and directions of the ice cubes using momentum.

Source: Rice University, https://openstax.org/books/college-physics/pages/8-4-elastic-collisions-in-one-dimension