Collision #1: Two disks slide on an air hockey table so that friction can be ignored. Disk A, with mass = 0.100 kg, is sliding to the right at a speed of 7.50 m/s directly toward Disk B, which is sliding to the left at a speed of 3.00 m/s and has a mass of 0.360 kg. Before collision Disk A BA Disk B The two disks have an elastic, or "bouncy" collision. After the collision, Disk B slides to the right at a speed of 1.57 m/s. All motion takes place in 1 dimension, so we can define "to the right" as the +x direction and "to the left" as the -x direction. Our goal is to calculate the velocity (magnitude and direction) of Disk A after the collision. The total momentum for the system will be conserved. In class, you learned two ways to apply conservation of momentum when the system consists of two interacting objects. One way to express this is: APA APB. We'll use this approach for the first part of this problem. Enter values for vector quantities using + and - to indicate direction. (a) What is the change in momentum for Disk B? APB = kg-m/s (b) What is the change in momentum for Disk A? APA = kg.m/s (c) What is the momentum of Disk A after the collision? Pt,A= kg-m/s (d) What is the velocity of Disk A after the collision? Uf,A= m/s Another way to express conservation of momentum is: Pi,total = Pf,total. We'll use this approach for calculations involving a second collision between Disks A and B. Collision #2: This time, Disk A (mA = 0.100 kg) has an initial speed of 7.90 to the right, and Disk B (mB = 0.360 kg) has an initial speed of 2.60 to the left. After the collision, Disk B has a final speed of 1.97 m/s to the right. (e) What is the total momentum of the system of Disk A and Disk B before the collision? Ptotal = kg-m/s (f) What is the momentum of Disk B after the collision? Pf,B = kg-m/s (g) What is the momentum of Disk A after the collision? Pt,A= kg.m/s (h) What is the velocity of Disk A after the collision? Uf,A= m/s Both of these approaches work they are just two different ways to apply conservation of momentum for the system. Working with total momentum allows you to consider systems of more than two interaction objects. Collision #3: Our disks are going to have one more collision: Disk A (mA = 0.100 kg) has an initial speed of 2.20 m/s moving to the right and Disk B (mB = 0.360 kg) has an initial speed of 6.50 m/s moving to the left. This time the disks magically stick together after the collision, as shown below. - Both of these approaches work they are just two different ways to apply conservation of momentum for the system. Working with total momentum allows you to consider systems of more than two interaction objects. Collision #3: Our disks are going to have one more collision: Disk A (mA = 0.100 kg) has an initial speed of 2.20 m/s moving to the right and Disk B (mB moving to the left. This time the disks magically stick together after the collision, as shown below. after collision BLA-B =? Disk A Disk B (i) What is the total momentum of the system? Ptotal kg.m/s (j) What is the velocity of the stuck-together disks after the collision? VA and B m/s = 0.360 kg) has an initial speed of 6.50 m/s