PHYS101: Introduction to Mechanics

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.