**Title: Elastic Collision Analysis Between a Soccer Ball and a Golf Ball** **Introduction:** Consider a scenario where a 0.80-kg soccer ball traveling at 20 m/s makes a head-on elastic collision with a 0.050 kg golf ball initially at rest. For the purpose of this analysis, we will ignore friction, gravity, and any other external forces. **a) Relative Velocity After Collision:** To find out by how much the golf ball is moving faster than the soccer ball after the collision, we use the relative velocity formula: \[ U_{12} = v_1 - v_2 = 20 \text{ m/s} - 0 \text{ m/s} = 20 \text{ m/s} \] **b) Speed of the Soccer Ball After Collision:** The velocity of the soccer ball immediately after the collision is calculated using the formula: \[ v_3 = \frac{mv}{m+M} \] Substituting the values: \[ v_3 = \frac{0.80 \text{ kg} \times 20 \text{ m/s}}{0.80 \text{ kg} + 0.050 \text{ kg}} = \frac{16 \text{ kg m/s}}{0.85 \text{ kg}} = 18.8235 \text{ m/s} \] **c) Speed of the Golf Ball After Collision:** The velocity of the golf ball immediately after the collision is calculated as: \[ v_3 = \frac{mv}{m+M} \] Substituting the values: \[ v_3 = \frac{0.050 \text{ kg} \times 20 \text{ m/s}}{0.050 \text{ kg} + 0.80 \text{ kg}} = \frac{1 \text{ kg m/s}}{0.85 \text{ kg}} = 1.17647 \text{ m/s} \] **Conclusion:** Through these calculations, we determine the outcome of an elastic collision between a soccer ball and a golf ball. The velocities indicate how both balls respond post-collision in an ideal scenario where external forces are negligible.

Let A= (3.0 m, 200south of east), B = (2.0 m, north), and C= (5.0 m, 700south of  west).  Draw and label A, B, and C with their tails at the origin. Use a coordinate system with the x-axis to the east.  b. Write A, B, and C in component form, using unit vectors.  c. Find the magnitude and the direction of D = A + B+ C.

Can you please answer question 2 and all of the sub problems and show all of the steps to the solution

How exactly would you solve this problem??