Physics 30: Module 2: Lesson 31
Module 2: Lesson 3 Component Collision Analysis WorksheetComponent analysis can be used to solve inelastic and elastic collisions in two dimensions. Below is an example problem and solution for each type of collision. All directions are in the polar positive specification.
Inelastic Collision in Two Dimensions (Complete the even numbered questions)
Example ProblemA car with a mass of 1250 kg is travelling 18.0 m/s at 315°. It has an inelastic collision with a truck of mass 1850 kg that was southbound at 15.0 m/s. Both the car and the truck stick together after the impact. What was the velocity of the wreckage immediately after the collision?
First, find the x and y components of the car’s momentum.
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Second, find the y component of the truck’s momentum. (The x component will be zero because it is not moving in the x direction.)
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Now, total the x components and y components of the momentum before the collision.
Note: The most common mistake in these problems is forgetting to specify the positive (+) or negative (–) directions in the addition! / Px total = px1 + px2
Px total = (+1.59 x 104 kgm/s) + (0)
Px total = +1.59 x 104 kgm/s
Py total = py1 + py2
Py total = (–1.59 x 104 kgm/s) + (–2.78 x 104 kgm/s)
Py total = – 4.37 x 104 kgm/s
Add the total x and y components as vectors.
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polar positive direction = 360° – 70.0° = 320°
Finally, divide the total momentum by the total mass to determine the total velocity. /
Elastic Collision in Two Dimensions
Example Problem
A car with a mass of 1350 kg is travelling 20.0 m/s at 125°. It has an elastic collision with a truck of mass 2650 kg that was travelling at 18.0 m/s at 205°. If the truck has a velocity of 17.0 m/s at 155° immediately after the collision, what is the velocity of the car?
To solve this problem, follow the same general steps as in an inelastic collision—there are just more momentums to add. The example problem solution is based on the following steps:
- Find the x and y components of momentum for each object both before and after the collision.
- Apply the conservation principle to each direction independently:
- x momentum before = x momentum after
- y momentum before = y momentum after
- Manipulate the conservation principle in terms of the unknown variables and solve.
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Step 2: Determine the x and y momentum of the truck before the collision.
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Step 3: Determine the x and y momentum of the truck after the collision.
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Step 4: Apply the conservation principle, and manipulate for the unknown variables. /
Step 5: Determine the x and y components of the car’s momentum after the collision by applying the conservation principle from step 4.
Step 6: Add the x and y components of the car’s momentum after the collision as vectors to determine the overall momentum:
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polar positive direction = 180° + 43.7° = 224°
Step 7: Finally, divide the car's final momentum by the car's mass to determine its final speed, and note the direction for the final velocity. /
Obviously, this is a very long process. But if one of the objects involved in the collision is at rest or travelling in only an x or a y direction, the process gets much shorter. This example problem illustrates the most complex scenario. The problems on the worksheet can be solved by an application of the process. Most problems will not, however, always require so many calculations.
Component Collision Analysis Worksheet
As part of your Module 2: Lesson 3 Assignment. Complete the odd numbered questions. Show all your work. Demonstrate that you have used the GRASP form. Of course, these questions provide you with the practice that you need to successfully complete this course. You should respond to all of the questions and place the answers in your course folder.
The answer(s) for each question follows in brackets.
1. / Add the following momenta to determine the resultant momentum for the following collisions:
- 50 kg•m/s at 45° and 59 kg•m/s at 90°
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- 175 kg•m/s at 100° and 144 kg•m/s at 180°
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- 1.6 kg•m/s at 280° and 5.5 kg•m/s at 190°
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- 1659 kg•m/s at 260° and 3125 kg•m/s at 190°
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- 3.9 kgm/s at 10° and 3.7 kgm/s at 70°
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2. / A billiard cue ball with a mass of 0.60 kg and an eight ball with a mass of 0.55 kg are rolled toward each other. The cue ball had a velocity of 3.0 m/s at 360°, and the eight ball had a velocity of 2.0 m/s at 90o. After the collision, the cue ball moves off at a velocity of 2.8 m/s at 105o. What is the velocity of the eight ball?Answer:
3. / A proton heading due north is struck by a neutron heading at 320°. The neutron bounces off with a heading of 50°. The initial speed of the proton was 2800 m/s, and the neutron was travelling at 2900 m/s. After the collision, the speed of the neutron is unchanged. If a neutron has the same mass as a proton, what is the final velocity of the proton?Answer:
4. / A car is entering the freeway. The merge lane has a heading of 260°. The car has an inelastic collision with a truck that had a velocity of 80 km/h at 270°. The car had a mass of 1400 kg and a speed of 75 km/h. The mass of the truck was 1800 kg.- What is the initial common velocity of the car and truck after the collision? (
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- If the wreckage takes 6.0 s to come to rest, how far from the point of impact will they come to rest if we assume a uniform negative acceleration?
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- What force did each vehicle apply to the other if the actual time of impact was 0.010 s?
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5 marks / 5. / Air hockey puck A, travelling at a velocity of 0.33 m/s at 360o, elastically strikes air hockey puck B, which was travelling at 0.66 m/s at 295°. After the collision, puck A has a velocity of 0.17 m/s at 288°. If the mass of A was 628.0 g and the mass of B was 529.0 g, what was the final velocity of B?Answer:
5 marks / 6. / Two identical objects, each with a mass of 367 g, collide elastically. Object A was travelling at 30 cm/s at 145°. Object B was at rest. After the collision, object A has a velocity of 30 cm/s at 176°. What is the velocity of B?Answer:
Once you have completed all of the questions, submit your work to your teacher.