# 0.4 6.5 elastic collisions in one dimension  (Page 2/4)

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Solution

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

${p}_{1}=p\prime {}_{1}+p\prime {}_{2}$

or

${m}_{1}{v}_{1}={m}_{1}{v\prime }_{1}+{m}_{2}{v\prime }_{2}.$

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

$\frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{1}{v\prime }_{1}{}^{2}+\frac{1}{2}{m}_{2}{v\prime }_{2}{}^{2}.$

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

${v\prime }_{2}=\frac{{m}_{1}}{{m}_{2}}\left({v}_{1}-{v\prime }_{1}\right).$

Substituting this expression into the second equation (internal kinetic energy equation) eliminates the variable ${v\prime }_{2}$ , leaving only ${v\prime }_{1}$ 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

${v\prime }_{1}=4\text{.}\text{00 m/s}$

and

${v\prime }_{1}=-3\text{.}\text{00 m/s}.$

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 $\left({v\prime }_{1}=-3\text{.}\text{00 m/s}\right)$ is negative, meaning that the first object bounces backward. When this negative value of ${v\prime }_{1}$ is used to find the velocity of the second object after the collision, we get

${v\prime }_{2}=\frac{{m}_{1}}{{m}_{2}}\left({v}_{1}-{v\prime }_{1}\right)=\frac{0\text{.}\text{500 kg}}{3\text{.}\text{50 kg}}\left[4\text{.}\text{00}-\left(-3\text{.}\text{00}\right)\right]\phantom{\rule{0.25em}{0ex}}\text{m/s}$

or

${v\prime }_{2}=1\text{.}\text{00 m/s}.$

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.

## Phet explorations: collision lab

Investigate collisions on an air hockey table. Set up your own experiments: vary the number of discs, masses and initial conditions. Is momentum conserved? Is kinetic energy conserved? Vary the elasticity and see what happens.

## Section summary

• An elastic collision is one that conserves internal kinetic energy.
• Conservation of kinetic energy and momentum together allow the final velocities to be calculated in terms of initial velocities and masses in one dimensional two-body collisions.

## Conceptual questions

What is an elastic collision?

## Problems&Exercises

Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.

Professional Application

Two manned satellites approach one another at a relative speed of 0.250 m/s, intending to dock. The first has a mass of $4\text{.}\text{00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ , and the second a mass of $7\text{.}\text{50}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{kg}$ . If the two satellites collide elastically rather than dock, what is their final relative velocity?

0.250 m/s

A 70.0-kg ice hockey goalie, originally at rest, catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. Suppose the goalie and the ice puck have an elastic collision and the puck is reflected back in the direction from which it came. What would their final velocities be in this case?

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(a) Find the recoil velocity of a 70.0-kg ice hockey goalie, originally at rest, who catches a 0.150-kg hockey puck slapped at him at a velocity of 35.0 m/s. (b) How much kinetic energy is lost during the collision? Assume friction between the ice and the puck-goalie system is negligible. (Figure 8.9)
Calculate the velocities of two objects following an elastic collision, given that m1 = 0.500 kg, m2 = 3.50 kg, v1 = 4.00 m/s, and v2 = 0.
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