College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- 7. A 1.000 kg particle moves at a speed of 0.500 m/s, as shown in Figure 2a. It collides with a 2,000 kg particle at rest at origin. What is the system's total momentum of the system in the x and y direction before the collision? +y m2) V4 30 m m2 +X +X 45° (a) (b) Fig. 2: Collision of mass m, and m, 8. Figure 2b shows that, after a collision, m, travels at a speed v3 at an angle e, = 315.0° with respect to the horizontal axis, while m, moves at a speed v, at an angle O, = 30.0° with respect to the horizontal axis. Write an equation for the system's total momentum in the x and y direction after the collision. 9. Using the equations produced in question 8, solve the value of v3 and v4.arrow_forwardWhat is the momentum of a garbage truck that is 14500 kg and is moving at 10.0 m/s ? momentum: kg⋅m/s At what speed would a 9.35 kg trash can have the same momentum as the truck? speed: m/sarrow_forwardA pitcher throws a baseball of mass m = 0.150m=0.150 kg with a velocity of 43.7 m/s. The momentum of the baseball is a numerical answer below. Accepted formats are numbers or "e" based scientific notation e.g. 0.23, -2, 1e6, 5.23e-8 ..... k x xg /sarrow_forward
- Table 1a: Elastic Collisions and Change in Momentum Trial vf (m/s) vi (m/s) Dv (m/s) Dp (kg•m/s) cart 0.89 -0.97 1.86 0.5022 cart with one mass 0.65 -0.80 1.45 0.754 cart with two masses 0.51 -0.63 1.14 0.878 Table 1b: Elastic Collisions and Impulse Trial I (N•s) cart 0.632 cart with one mass 0.959 cart with two masses 1.066 Table 1c: Elastic Collisions and Impulse Trial % difference between I and Dp (I -Dp)/ ((Dp+ I)/2) * 100 cart 22.89 % cart with one mass 23.93 % cart with two masses 19.34 % Q1. What can you conclude about Table 1c (Impulse I and change in Momentum Dp) for Elastic Collisions?arrow_forward1. What is the change in momentum of a 26.3 g bird flying with the speed of 9.25 m/s due east then due north?arrow_forwardQuestions 1.-4. refer to a head-on collision between two balls with the following initial values: Before Collision m2 V Ix 3.5 kg 4.5 kg 5.0 m/s -2.0 m/s 1. What is the total momentum (in kg m/s) of the system? 2. If the two balls stick together, what is their common speed (in m/s) after they collide? 3. How much kinetic energy (in joules, positive number) was lost in this collision? 4. If the two balls do not stick together, and the first ball moves at vịx = -1 m/s after they collide, what is the speed (in m/s) of the second ball?arrow_forward
- I. A lump of clay (m = 3.01 kg) is thrown towards a wall at speed v = 3.15 m/s. The lump sticks to the wall. (a) What kind of collision is it? Is momentum conserved during this collision? Why or why not? (b) Calculate the impulse imparted on the lump by the wall. (c) Calculate percent of initial kinetic energy lost during this collision. II. Same lump is thrown towards the same wall, but this time it bounces off the wall at speed of 3.15 m/s. (a) What kind of collision is it? Is momentum conserved during this collision? Why or why not? (b) Calculate the impulse imparted on the lump by the wall. (c) Calculate percent of initial kinetic energy lost during this collision. III. Same lump is thrown towards the same wall, but this time it bounces off the wall at speed of 2.24 m/s. (a) What kind of collision is it? Is momentum conserved during this collision? Why or why not? (b) Calculate the impulse imparted on the lump by the wall. (c) Calculate percent of initial kinetic…arrow_forwardmeteor moving through the Earth's atmosphere has a momentum of 1.15 x 104 kgm/s. As it falls, friction with the atmosphere slows it to 1/4 its original speed, as its mass shrinks to 2/9 of its original mass. Determine its new momentum. A. 320 kgm/s B. 639 kgm/s C. 10350 kgm/s D. 207000 kgm/sarrow_forwardPhysics: Unit: Momentum and collisions Consider a collision in one dimensional that involves two objects of masses 4.5 kg and 6.2 kg. The larger mass is initially at rest, and the smaller mass has an initial velocity of 16 m/s (E). The final velocity of the larger object is 10.0 m/s (E). Calculate the final velocity of the smaller object after the collision.(Hint: = m1Vi1 + m2Vi2 = m1Vf1 + m2Vf2)arrow_forward
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