Otterbein University Department of PhysicsPhysics Laboratory 1500-9
Name: ______Partner’s Name: ______
EXPERIMENT 1500-9
TWO-DIMENSIONAL COLLISIONS
INTRODUCTION
In this lab we will consider two-dimensional collisions. When two objects collide, the resulting motion will almost always be two-dimensional. Exceptions are precisely head-on collisions, and collisions between objects that are confined to run on a one-dimensional track. Calculating the properties of two dimensional collisions is more involved than one-dimensional collisions for obvious reasons, so we will consider the simplest two-dimensional case where a ball (the projectile) is shot toward another ball (the target) which is initially at rest. The two collide, and each ball goes off in a different direction. A common example of this interaction is in a game of billiards.
In this experiment, we will have the two balls collide in mid-air, so they fall under the influence of gravity. Momentum is NOT conserved in the vertical (z) direction (unless you include the earth, which is more trouble than it’s worth). However, there is no net force on the balls in the horizontal plane, so momentum is conserved in the horizontal directions, which we take to be the x-y plane.
We are free tochoose the direction of the coordinate axes. A natural choice is to designate the direction of motion of the first ballto be the x-axis. That is, the momentum of this ball before collision is , which is the same as the total momentum. Remember that is a vector of length one (unit vector) is in the +x direction.
After the collision, the momentum is
,
where we decomposed the velocities into their x- and y-components:
Because the initial and final momenta must be the same, and because there is no momentum in the y-direction at the start, the total y-momentum at the end must be zero
.
In the x-directionwe must have:
We also know that if the collision is elastic, kinetic energy is conserved as well
.
where , etc.
PROCEDURE
Setup
1.Make sure that the projectile launcher is clamped securely to the table. Ensure that the launcher is set to fire exactly level, using the plumb line on the side of the launcher. You can also use a bubble level to check this.
2.Fire a test shot on the short range setting(one cleck). Make sure the ball is landing on the table.
3.Load and cock the launcher. Then swing the ‘tee’ (like a golf tee) into place in front of the muzzle and secure it. Don’t put it directly in front of the muzzle, but at a slight angle.
4.Put a second ball on the tee. Fire a test shot. Ensure that both balls landed on the table. If not, reposition the tee or the launcher and try both test shots again.
5.Cover the table with a large sheet of butcher paper. Tape the paper down. The paper should cover each landing spot, AND come right up underneath the launcher.
Initial Momentum
First we have to find the momentum of the launched ball in the absence of a collision.
- Use just one ball. Put a piece of carbon paper at the landing position and fire a single ball straight out of the launcher 5 times.
- Find a point at the center of the cluster of dots that mark the point-of-contact. Mark this point.
- Use a plumb bob to make a mark on the paper just under the spot where Ball #1 is going to hit Ball #2 on its way out, when Ball #2 is in position.
- Draw a line (with a ruler!) between the muzzle point and the center-of-cluster dot. This marks the average line followed by the ball. Find the distance the ball travelled along this line. From the scatter of the dots, estimate the uncertainty on this distance.
L: ______(average)
ΔL: ______(uncertainty)
- Measure the height of the gun above the table.
y0 = ______ - Use projectile motion to find the velocity of the ball as it leaves the muzzle. (Do your work on in the space provided below.)
v0 = ______
- Weigh the ball and find its mass.
m= ______
- Find the momentum and kinetic energy of the ball (show calculations below).
p0 = ______
K0 = ______
Elastic collision
- Cock the launcher one notch, load it, and add a second ball to the tee. Put carbon paper under the ball landing positions and fire the launcher, marking the positions of both balls when they fall.
- Repeat the procedure a total of 5 times, so that you have 5 dots in each landing cluster.
- Again, find the center of each cluster. Draw a line (with a ruler) from the cluster back to a point just under the point of collision.
- Measure the length of each line.
L1 = ______L2 = ______
ΔL1 = ______ΔL2 = ______
- As above, find the velocity, momentum, and kinetic energy of each ball. (Show your calculations on the next page. If you derived a general expression for v above, you can use it here without re-deriving it.)
v1 = ______v2 = ______
p1 = ______p2 = ______
KE1 = ______KE2 = ______
- Use a protractor to find the angle between each line and the original flight path of the single ball in the last section. (One of these should be negative.)
θ1 = ______θ2 = ______
- Decompose each of the above momenta into components. Be sure the signs are correct.
p1x = ______p2x = ______
p1y = ______p2y = ______
Use the following table to check whether momentum and kinetic energy are conserved in this collision.
Initial
/Final
px(kg m/s)
py(kg m/s)
KE(J)
Use the uncertainties of the distances found earlier to make reasonable estimates of the uncertainty in momentum and kinetic energy. Give your estimates below. (You can use the fact that Δp/p=ΔL/L and ΔKE/KE=2ΔL/L.)
Which quantities are conserved to within uncertainty? Is this what you expect?
Inelastic collisioN
Put a small piece of putty (or a loop of duct tape) on one of balls, oriented so that the balls will stick slightly when they hit.
Mass of ball with putty=______
Fire the launcher again using the bit of tape or putty. This can only be done once, since each trial will result in a different collision. Make sure you get both hits with the carbon paper.
As above, draw lines from the landing dots back to the collision point, and repeat the above calculations (show your calculations on next page).
L1 = ______L2 = ______
v1 = ______v2 = ______
p1 = ______p2 = ______
KE1 = ______KE2 = ______
θ1 = ______θ2 = ______
p1x = ______p2x = ______
p1y = ______p2y = ______
Initial
/Final
px (kg m/s)
py (kg m/s)
KE (J)
Calculations:
Which quantities are conserved to within your estimate of the uncertainty? Is this what you expect?
How elastic was the collision? A fully elastic collision conserves KE. A fully inelastic collision is one where the two balls stick together, so you get minimum KE.
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