Collisions in Two Dimensions (velocity).docx

8/10/2019 Collisions in Two Dimensions (velocity).docx

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Collisions in Two Dimensions Name ______________________________AP Physics - Becke

Question:What is the evidence that momentum is conserved in a two-dimensional collision? In what manner does theevidence support the law of momentum conservation?

Purpose:To conduct an analysis of a two dimensional collision in an effort to gather convincing evidence that the totalsystem momentum is conserved AND to describe how the evidence supports the law of momentumconservation.

Background:In this lab, we will launch a moving incident sphere A that will collide with astationary target sphere B, as shown.

If the collision is not head on, the twospheres will go in different directions

after the collision.

If momentum is being conserved, then when we combine the two momentum vectors after the collision ( pa' and

pb'), they should be equal to the initial momentum vector ( p

a).

The velocity of the incident sphere when it leaves the ramp is horizontal; therefore, the momentum that we areconcerned with is also horizontal. This means that it is only the horizontal components of the velocities that willbe included in any calculations of the momentum.

In this lab, the actual collision will occur at a certain height above the floor. Since each sphere has the sameinitial vertical velocity (0 m/s) and falls the same height, the spheres will be in the air for the same amount of

time. During this time they will be traveling horizontally as they accelerate towards the ground.

Materials: two dimensional collision apparatus, carbon paper, meter sticks, spheres, masking tape, paper,protractor

Procedure: Set up your apparatus so that the sphere that rolls from the launcher launches horizontally beyond the

edge of the table and lands on the floor. Measure the vertical distance from the floor to the point where the sphere will leave the launcher. Hold the sphere at the top of the launcher against the guide screw. Allow the sphere to roll down a

couple of times from this height to determine the general area that it is landing.

Note that the hanging mass is located directly under the point where the collision will later take place. Tape a large piece of paper on the floor that will include both where the sphere lands and the locationof the collision.

Mark the spot on the floor directly under the hanging mass. This is the point from which you willmeasure the horizontal distance travelled.

Place a piece of carbon paper over the taped paper (shiny side up). Launch the sphere ten times so that it lands on the carbon paper. Lift the carbon paper and circle these dots.


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Before collecting any more data, it is necessary to ensure that the screw holding the target sphere is atthe proper height.

o Place the target sphere (of equal mass to the incident sphere) on the screw directly in front ofthe launcher so that there will be a head on collision.

o Adjust the screw so that the target sphere is at the same height as the incident sphere at thetime of collision.

o Carry out several head on collisions (releasing the incident sphere from the same height asbefore). If the screw is set to the proper height, the target sphere should land in the same spotas the original group of dots. If it does not, adjust the height of the screw until the targetsphere does land in the same area as the original incident sphere by itself .

Equal Mass Collision:Set-up:

Adjust the location of the target sphere so that when the sphere rolling down the apparatus strikes theone at rest there will be a glancing collision (the more glancing, the better!). Be careful not to disturbthe location of the ramp!

Data collection: Run a couple of trials of the glancing collision (without carbon paper), using the same launch height (of

the incident sphere) as the other trials. Tape an additional piece of paper to the floor where the twospheres are landing if they do not land on your big piece of paper.

Next place two pieces of carbon paper at theselocations so that the spheres will leave a markwhen they hit the paper.

Carry out ten trials of this glancing collision.Remove the carbon paper and label the dots.

Draw lines from the mark representing thepoint underneath the launch position to thecenter of each of the three groups of dotscreated. Measure and record the length of

these lines and the angles (using 0 as thedirection of the original displacement of theincident sphere).

Unequal Mass Collision: Replace the metal target sphere with the glass marble and repeat the Data collection procedure

above, using a different color pencil or pen to circle the landing positions after the collision and to drawthe lines between launch position and the landing positions.

Data :

Mass of steel sphere: 7.60 g = 0.00760 kg Height of Launcher: _______ m

Incident ballwith no collision

Incident ballafter collision

Target ballafter collision

Length (m) () Length (m) () Length (m) ()

0 Equal mass collision

Unequal mass collision

x

Incident ball landingspot after collision

Incident ball landingspot without collision

Target ball land ingspot after collision

Location ofcollision

(hanging mass)


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Analysis:

1. Determine the time it takes for the spheres to reach the ground:

2. Calculate the following (note that the x and y subscripts do not indicate vertical motion, but insteadrepresent the two different axes of motion within the horizontal plane)

velocity = x/t v x = v cos (0) px = mv x vy = v sin (0) py = mvy *Incident Sphere

before collision0 0

Total momentumbefore

0

(*The reason you marked the landing point of the incident sphere without a collision was to be able tofill in the first row in the above table)

Equal Mass Collision3. Calculate the final (post-collision) velocity and momentum of the incident sphere and the target sphere.

velocity = x/t v x = v cos px = mv x vy = v sin py = mvy Incident Sphere

after collisionTarget Sphereafter collision

Total momentumafter

Errorbefore vs after

Percent Errorbefore vs after


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Unequal Mass Collision4. Assuming momentum is conserved; determine the mass of the target sphere.

Initial Momentum (from question 2)

velocity = x/t v x = v cos px = mv x vy = v sin py = mvy

Incident Sphereafter collision

velocity = x/t v x = v cos vy = v sin Target Sphere after

collision

Expected momentum ofTarget Sphere after collision

Total (Pythagorized) momentum of

Target Sphere after collision

Calculated Mass ofTarget Sphere

Actual Mass ofTarget Sphere

2.42 10 kg

Percent Error

Collisions in Two Dimensions (velocity).docx

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