collisions2D

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University of Pennsylvania Department of Physics COL2D.1

Conservation of Momentum in Two Dimensional Collisions

Goals of this lab

Test the conservation of momentum in two dimensional collisionsTest the conservation of energy in two dimensional collisions between objects of equal massTest your predictive skills on the landing point of a 1 2 inch steel target ball

Overview

One of the principles that governs any collision is that the total momentum does not change in acollision, even though the individual momenta certainly do. The momentum of a body is, by

definition, the mass times the velocity of the body. Momentum is a vector since velocity is a vector.

If expressed in components, the principle does not yield one scalar equation but actually three -- one

for each dimension: the x momentum is conserved, the y momentum is conserved, and the zmomentum is conserved. We find the momentum of each ball before and after the collision and

then check to see if the vector sum of the individual momenta is in fact unchanged by the collisionbetween the balls.

We will test this prediction by colliding two balls. The projectile ball rolls down a ramp andcollides with a stationary target ball. They both then fall a known distance into a box where their

landing positions are marked. The height z in Figure 1 is the vertical distance from the top of the

ball suppport or tee at the end of the track to the surface of the paper on which the landing points are

marked. We may take the +x direction to be the direction the projectile ball is moving when itstrikes the target. The ramp is constructed so that the projectile ball is moving horizontally at the

end of the ramp, when it strikes the target ball. Thus, the motion of the two balls just before and

after the collision is strictly horizontal.

Figure 1: Side view of the balls before collision


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Figure 2 shows the positions of the projectile and target balls before and after collision as theyappear when looking down on the setup. Point A is the center of the projectile ball and also the

pivot point of the tee holder. Point B is the center of the target ball and also the center of the tee

holder. After the collision the target and projectile balls move off in different directions (think of

billiards!).

Figure 2: View from above of the balls before and after collision

The initial momentum of the projectile ball is found by measuring the distance the projectile ball

travels from point A to the point where it lands when there is no collision with a target ball. Thefinal momenta of the projectile ball and the target ball are found by measuring the x and y distances

travelled by the balls after collision to their landing points. For example, after the collision theprojectile ball travels the distance AD. In Figure 2 px and yp are the x- and y-components of the

distance AD travelled by the projectile.

PreLab Question 1: Pool would be a much less interesting game if the balls were merelypoints, rather than extended objects. Angled shots with glancing impacts would not be

possible. A point particle cue ball could only make a point particle eight ball continue in its

original direction.

Figure PreLab-1


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Real pool balls have a diameter of 2 1/4 inches (about 6 cm), and can collide in many ways.In Figure PreLab-1 we see a cue ball coming in from the left and striking a stationary 8 ball.

The amount by which the collision is not head-on can be quantified with a measurement called

the impact parameter. This is determined by drawing a line through the center of mass of theprojectile ball and in the direction of motion. A second line, parallel to the first, is drawn

through the center of mass of the stationary target ball. The impact parameter is the distancebetween the two parallel lines, which is the separation of the centers of mass on a line

perpendicular to the line of motion.

The impact parameter for this collision of two 6 cm pool balls is 2.5 cm.

1. At what angle with respect to the line of motion of the cue ball will the 8 ball travel after

impact? (Hint: Keep in mind that the force from the cue ball must be perpendicular to the

surfaces of the balls, that it must act along a line drawn between the centers of both balls.)

2. At what value of the impact parameter does the collision fail to happen?

Finding the final momenta of the projectile and target balls is a little more complicated. Consider

the motion of the balls after collision when they fall into the tray. As you might recall fromstudying projectile motion, you can break down the motion of a falling ball into a horizontal and a

vertical motion. The balls move in the horizontal direction with constant velocities, 'pV and'tV , and

in the vertical direction with the familiar acceleration due to gravity.

Since you know the balls fall from a height z above the tray, you know how long it takes them to

fall, i.e. their time of flight. Thus, when you measure the horizontal distances travelled by the

balls to their landing points, you can use the time of flight to calculate their velocities (which we

cannot measure) just after collision. This can be done algebraically, rather than numerically withactual times.

Prelab Question 2: A 30 g projectile ball rolls down a ramp and collides with a target ball atthe end of the ramp. The projectile ball lands at a point 20 cm in the x- direction and 14 cm in

the y-direction from the point of collision. Suppose the ramp is 10 cm above the landingsurface. Calculate the projectile balls momentum immediately after impact.

Thus, for the collision, conservation of linear momentum gives

''0ttpppp

VmVmVm +=+ (1)

where mp and mt are the projectile ball and target ball masses, respectively. The projectile velocity

before the collision is pV and the target velocity before collision is zero. The final horizontal

velocities of the projectile and target balls are 'pV and'tV , respectively. Please note that the

velocities are all vectors in the XY plane with X and Y components. We cannot measure thevelocities directly but we can measure the distances travelled in flight and calculate the velocities.


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Figure 3 shows the horizontal displacement of the projectile after the collision. Point A is itsposition before the collision and the dark circles in the box indicate where the projectile landed in

different trials. The best value of the displacement is found from the distances xp and yp . The

uncertainties in these values can be approximated by half the length of the dashed box.

Check to make sure the track and receiving tray youre using for this lab are both level before you

do the lab. The bubble level at the end of the track must show the bubble centered in the circle.Use the carpenters level to check the levelling of the receiving tray. If either apparatus is not level,

detach the C-clamp and level the apparatus. If the track is not level, adjust the three leveling feetunderneath the track until the bubble is centered. If the receiving tray is not levelled, adjust the

three leveling feet supporting it until it is. When both the track and the tray are level, reattach the

C-clamp.

When you take your data for the collision between balls of equalmass, you might notice that the landing points of the balls fall on

a circle, regardless the angle of the tee holder. This circularshape follows from the geometry of inscribing a right triangle in

a circle. Figure 4 shows two different hypothetical collisions

between two balls of equal mass. In the figure point P is the

collision point for both collisions, X1 and Y1 are the landing

points of the two balls for the first collision and X2 and Y2 are

the landing points for the second collision. In general, if we are

considering only one collision, X is the landing point of the

projectile ball, and Y is the landing point of the target ball.

The kinetic energy of the projectile ball after collision

is ( )2'2

1ptp vmKE = and the kinetic energy of the target ball after the collision is ( )

2'

2

1ttt vmKE = .

To express these kinetic energies in terms of the lengths XP, PY, and XY in Figure 4, consider the

definition of velocity as distance divided by time . The kinetic energy of the projectile ball is2

2

1

=

t

PXmKE pp and the kinetic energy of the target ball is

2

2

1

=

t

PYmKE tt .

Figure 3

Figure 4


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The total kinetic energy after the collision is

22

2

1

2

1

+

=

t

PYm

t

PXmKE tp (2)

We can simplify equation (2) for collisions of two balls of equal mass. We know that the balls willfall the same distance to the receiving tray, so that is the same for both balls. This makes the total

kinetic energy after the collision

+

=

22

2

1

t

PY

t

PXmKE (3)

The Pythagorean Theorem tells us that PX2+ PY

2= XY

2, so the total kinetic energy after the

collision is

2

2

1

= t

XY

mKE (4)

If energy is conserved, the total energy before the collision must equal the total energy after thecollision. The total energy before the collision is just the kinetic energy of the projectile. Energy

conservation can thus be expressed by

22

2

1

2

1

=

t

XYmmvp (4)

If we cancel like terms, we gett

XYvp = . Thus, the diameter of the circle formed by the landing

points from the collisions equals the horizontal distance the projectile ball travels when there is nocollision.

PreLab Question 3: Consider a triangle inscribed in a circle as shown in Figure PreLab 3.

Point P represents the collision point between two balls of equal mass. Point X is the landing

point of the projectile ball and point Y is the landing point of the target ball.

This arrangement can be modeled using a compact disk and its

case. Hold one corner of the case on point P on the edge of the

CD, and let the case turn about that point.

Figure PreLab 3

P

XY


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Part 2: Collisions Between Balls of Equal Mass

1. Remove the tee holder and find point A, the position of the center of the projectile ball at the

instant of collision (see Figure 2). Use the plumb bob to mark the point which is directly below

the end of the ramp and exactly where the tee holder pivots.

2. To set up a collision, replace the tee holder, making sure the flat side of the tee holder is facingup and the side with the ledge is facing down. Place the target 5/8-inch ball on the tee. Study

collisions with the target ball in 3 different positions (angles). Point B is found by removing the

tee, sliding the brass rod through the hole with the pointed tip down, and marking the spot usingthe carbon paper. Points C and D are obtained by placing carbon paper where the balls will

land, just like in Part 1. For each position of the tee holder, do enough trials to estimate the

uncertainty in your measurements of C and D.

3. Construct point A on your paper, from your collision data. We assert that point A lies on the

extrapolation of the straight line BC. While each trial has a different pair of points B and C, the

extrapolation of the line BC for every trial should pass through the same point A. Therefore,

point A is at the intersection of the extrapolations of the BC lines.

4. Calculate the fall time of the balls and the velocities for one collision position of the target ball.

5. Do one head-on collision and analyze your results.

Part 3: Collisions with Balls of Unequal Mass

1. Use the 1/2-inch diameter ball as the target. Turn the tee holder over, so that the flat side isdown and the side with the step is up, and replace the tee. This step compensates for the size

difference of the balls to ensure that their centers line up correctly at collision.

2. Flip over the large sheet of white paper to mark your landing positions. You dont want toconfuse your data with that of Part 2. You may need to change the starting point for the

projectile ball which you picked in Part 1.

3. Record data for 3 different positions of the tee holder. Again, as in Part 2, do enough trials for

each position of the tee to find the best location for the landing points and the uncertainty in thatlocation.

Questions

1) Compare the x and y momenta before and after collision, including uncertainties. Is momentumconserved?

2) For collisions with balls of equal mass do your results from the two methods of determining

point A agree? That is, do most of your BC lines intersect at the same place and is that

intersection point close to the point A you determined with the plumb bob?

3) For equal size balls with the same mass, the displacement lengths can be used in place of themomenta. Why does this work? Explain mathematically why we can make this assumption by

calculating the momenta and looking at the conservation equation.


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4) After a collision between balls of the same mass, if energy is conserved then the angle between

the projectile and target distance vectors should be 90. What was your measured angle? Whyis there a discrepancy?

5) The distance DC should be the same for the 3 positions of the tee holder, and it should be the

diameter of the circle made by your landing points. Also, the distance DC should be the sameas the initial projectile distance. Measure the distance DC for the 3 different positions of the tee

holder and average them. Are they within error of the initial projectile distance?

6) What happens in head on collision?

7) How do the fall times of the balls affect our results in this experiment?

8) Extra Credit: After the collision the target ball moves in the direction of BC in Figure 2. This

direction is asserted to be an extrapolation of the line AB, assuming there is no friction betweenthe two balls. Show in a diagram the direction of the force that the projectile ball exerts on the

target ball when they collide. Is this assertion justified?

9) ExtraCredit: For collisions between balls of unequal masses, show that conservation of

energy and momentum predicts a geometrical shape other than a circle. What is this shape?

HS 5/15/2006 collision2D.doc

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