Physics 11 Honours. x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin A Collision at an Intersection Example 1:

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1 Name: Physics 11 Honours Date: Unit 7 Momentum and Its Conservation 7.4 A perfectly inelastic collision in 2-D Consider a collision in 2-D (cars crashing at a slippery intersection...no friction). Because momentum is a vector, whenever we analyze a collision in two or three dimensions the momentum has to be split up into components. If no external force momentum is conserved. Momentum is a vector so px, py and pz x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin A Collision at an Intersection Example 1: A 1000 kg car is moving eastward at 20 m/s. It collides inelastically with a 1500 kg van traveling northward at 30 m/s. What is the velocity of the two vehicles immediately after the collision? It is an easy, straightforward problem to find the velocity of the center of mass of the two-car system immediately after the collision. First, the x- and y-components of the velocity of the center of mass

2 Elastic Collisions Elastic means that the objects do not stick. There are many more possible outcomes but, if no external force, then momentum will always be conserved Billiards: Consider the case where one ball is initially at rest. The final direction of the black ball will depend on where the balls hit x-dir Px : y-dir Py : Example 1: A 70.0 g golf ball is sitting at rest when a g super ball traveling west at 22 m/s strikes it, off center. If the super ball carries on at 16 m/s, 34 S of W after the collision, determine the post collision velocity of the golf ball. In this lesson you looked at collision that occur in two-dimensions. You saw that the total momentum of a system is conserved during a collision. Moreover, the total momentum in the x-direction and in the y-direction is also conserved during a collision. The key points you looked at are: Momentum is a vector and can be resolved into components

3 During a collision, the total momentum for the components is conserved: Name: Block: Unit 7 Momentum and Its Conservation 7.4 A perfectly inelastic collision in 2-D Assignment 1. A g hockey puck is sliding along the ice at 17 m/s due east when it hits a 1.4 kg water bottle sitting there at rest. The puck ends up going 5.2 m/s, 40.0 S of E after the collision. Calculate the velocity of the water bottle after the collision. 2. An 8.0 kg mass collides elastically with a 5.0 kg mass that is at rest. Initially, the 8.0 kg mass was travelling to the right at 4.5 m/s. After the collision, it is moving with a speed of 3.65 m/s and at an angle of 27 to its original direction. What is the final speed and direction of motion for the 5.0 kg mass? 3. A 0.40 kg model airplane is travelling 20 km/h toward the south. A 0.50 kg model airplane, travelling 25 km/h in a direction 20 east of south, collides with the first model airplane. The two planes stick together on impact. What is the direction and magnitude of the velocity of the combined wreckage immediately after the collision?

4 4. A 1500-kg car traveling at 90.0 km/h east collides with a 3000-kg car traveling at 60.0 km/h south. The two cars stick together after the collision. 5. A 1000 kg car traveling at 30 m/s, 30 south of east, collides with a 3000 kg truck heading northeast at 20 m/s. The collision is completely inelastic, so the two vehicles stick together after the collision. How fast, and in what direction, are the car and truck traveling after the collision?

5 To solve this problem, simply set up two conservation of momentum equations, one for the y- direction (positive y being north) and another for the x-direction (positive x being east). Setting up a vector diagram for the momentum is a good idea, too, like this: To set up the two momentum conservation equations, simply write down the equation for the momentum before the collision in the y-direction and set it equal to the momentum after the collision in the y-direction, and then do the same thing in the x-direction. The y equation can be rearranged to solve for the y component of the final velocity: Similarly, in the x-direction: It would be easy to figure out the final velocity using the Pythagorean theorem, but let's find the angle first instead, by dividing the y equation by the x equation: Now let's go back to get the final velocity from the Pythagorean theorem : This gives a final velocity of 18.4 m/s at an angle of 21.8 north of east.