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# Elastic Collisions

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Elastic Collisions

### Key Equations

$\sum p_{\text{initial}} = \sum p_\text{final} \; \; \text{The total momentum does not change in closed systems}$

$\sum KE_{\text{initial}} = \sum KE_\text{final} \; \; \text{The total kinetic energy does not change in elastic collisions}$

Guidance
1. An elastic collision is a collision where both momentum and kinetic energy are conserved. The only true case of elastic collisions are in particle physics. However, pool balls, bumper cars, etc. loose very little of their kinetic energy during the collision and approximate an elastic collision. These types of collisions are used as elastic collision problems.
2. In elastic collisions the objects can not stick together (this would suck kinetic energy out of the system)
3. In elastic collisions one can use conservation of momentum and conservation of energy to solve the problem

#### Example 2

Question : Chris and Ashley are playing pool. Ashley hits the cue ball into the $8$ ball with a velocity of $1.2\mathrm{m/s}$ . The cue ball $(c)$ and the $8$ ball ( $e$ ) react as shown in the diagram. The $8$ ball and the cue ball both have a mass of $.17\mathrm{kg}$ . What is the velocity of the cue ball? What is the direction (the angle) of the cue ball?

Answer : We know the equation for conservation of momentum, along with the masses of the objects in question as well two of the three velocities. Therefore all we need to do is manipulate the conservation of momentum equation so that it is solved for the velocity of the cue ball after the collision and then plug in the known values to get the velocity of the cue ball.

$m_{c}v_{ic}+m_{e}v_{ie}&=m_{c}v_{fc}+m_{e}v_{fe}\\v_{fc}&=\frac{m_{c}v_{ic}+m_{e}v_{ie}-m_{e}v_{fe}}{m_{c}}\\v_{fc}&=\frac{.17\mathrm{kg}\times 2.0\mathrm{m/s}+.17\mathrm{kg}\times 0\mathrm{m/s}-.17\mathrm{kg}\times 1.2\mathrm{m/s}}{.17\mathrm{kg}}\\v_{fc}&=.80\mathrm{m/s}\\$

Now we want to find the direction of the cue ball. To do this we will use the diagram below.

We know that the momentum in the y direction of the two balls is equal. Therefore we can say that the velocity in the y direction is also equal because the masses of the two balls are equal. $m_cv_cy=m_ev_ey \rightarrow v_cy=v_ey$

Given this and the diagram, we can find the direction of the cue ball. After 1 second, the $8$ ball will have traveled $1.2\mathrm{m}$ . Therefore we can find the distance it has traveled in the y direction. $sin25^o=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{x}{1.2\mathrm{m}} \rightarrow x=sin25\times 1.2\mathrm{m}=.51\mathrm{m}$ Therefore, in one second the cue ball will have traveled $.51\mathrm{m}$ in the y direction as well. We also know how far in total the cue ball travels in one second $(.80\mathrm{m})$ . Thus we can find the direction of the cue ball. $sin^{-1}\frac{\text{opposite}}{\text{hypotenuse}}=sin^{-1}\frac{.51\mathrm{m}}{.80\mathrm{m}}=40^o$

### Simulation

Note: move the elasticity meter to 100% for perfectly elastic collisions.

### Time for Practice

1. You are playing pool and you hit the cue ball with a speed of $2\;\mathrm{m/s}$ at the $8$ -ball (which is stationary). Assume an elastic collision and that both balls are the same mass. Find the speed and direction of both balls after the collision, assuming neither flies off at any angle.
2. A $0.045\;\mathrm{kg}$ golf ball with a speed of $42.0\;\mathrm{m/s}$ collides elastically head-on with a $0.17\;\mathrm{kg}$ pool ball at rest. Find the speed and direction of both balls after the collision.
3. Ball $A$ is traveling along a flat table with a speed of $5.0\;\mathrm{m/s}$ , as shown below. Ball $B$ , which has the same mass, is initially at rest, but is knocked off the table in an elastic collision with Ball $A$ . Find the horizontal distance that Ball $B$ travels before hitting the floor.
4. Students are doing an experiment on the lab table. A steel ball is rolled down a small ramp and allowed to hit the floor. Its impact point is carefully marked. Next a second ball of the same mass is put upon a set screw and a collision takes place such that both balls go off at an angle and hit the floor. All measurements are taken with a meter stick on the floor with a co-ordinate system such that just below the impact point is the origin. The following data is collected:
1. no collision: $41.2 \;\mathrm{cm}$
2. target ball: $37.3 \;\mathrm{cm}$ in the direction of motion and $14.1 \;\mathrm{cm}$ perpendicular to the direction of motion
1. From this data predict the impact position of the other ball.
2. One of the lab groups declares that the data on the floor alone demonstrate to a $2$ % accuracy that the collision was elastic. Show their reasoning.
3. Another lab group says they can’t make that determination without knowing the velocity the balls have on impact. They ask for a timer. The instructor says you don’t need one; use your meter stick. Explain.
4. Design an experiment to prove momentum conservation with balls of different masses, giving apparatus, procedure and design. Give some sample numbers.
5. A 3 kg ball is moving 2 m/s in the positive $x-$ direction when it is struck dead center by a 2 kg ball moving in the positive $y-$ direction at 1 m/s. After collision the 3 kg ball moves at 3 m/s 30 degrees from the positive $x-$ axis. Find the velocity and direction of the 2 kg ball.

#### Answers to Selected Problems

1. $8 \;\mathrm{m/s}$ same direction as the cue ball and $0 \;\mathrm{m/s}$
2. $\mathrm{v}_{golf} =-24.5 \;\mathrm{m/s}; \mathrm{vpool} = 17.6 \;\mathrm{m/s}$
3. $2.8\;\mathrm{m}$
4. .
5. $1.5 \ m/s \ 54^\circ$