Practice 6F: | 1 | 2 |
3 | Go up

**Kinetic Energy in perfectly inelastic collisions - by
Lauren Bly, 2001**

**1. A 0.25 kg arrow with a velocity of 12 m/s to the west
strikes and pierces the center of a 6.8 kg target.**

** a. What is the final velocity of the combined mass?**

** b. What is the decrease in kinetic energy during the
collision?**

a. Here's what you know: The arrow has a mass of 0.25 kg and initial velocity of 12 m/s. The target has a mass of 6.8 kg and an initial velocity of 0 m/s. Together, the arrow and the target have a mass of 7.05 kg and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the arrow before plus the p of the target before equals the p of the arrow and target together. So .25 x 12 + 6.8 x 0 = 7.05 x v final. 3 + 0 = 7.05 v, and

v = .43 m/s to the west. The final velocity is positive, and so is the velocity of the arrow, and so the velocity of the combined pair is also west.b. Here’s what you know: The kinetic energy measured before and after will not be the same, and to find the difference you need to calculate the before and the after and subtract them. The before and after kinetic energies are calculated with KE =

^{1}/_{2}mv^{2 }. The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before. The value of KE before is (0.5)(0.25)(12)^{2}+ (0.5)(6.8)(0)^{ 2}. Therefore, the KE before is 18 J. The KE afterward is (0.5)(7.05)(0.43)^{ 2}, which equals 0.65 J. 18 J minus .65 J is about17 J, counting significant figures.

(Table of contents)

**2. During practice, a student kicks a .40 kg soccer ball
with a velocity of 8.5 m/s to the south into a 0.15 kg bucket lying on its
side. the bucket travels with the soccer ball after the collision.**

** a. What is the final velocity of the combined
mass?**

** b. What is the decrease in kinetic energy during the
collision? **

a. Here's what you know: The soccer ball has a mass of 0.40 kg and an initial velocity of 8.5 m/s. The bucket has a mass of 0.15 kg and an initial velocity of 0 m/s. Together, the ball and the bucket have a mass of 0.55 kg and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the ball before plus the p of the bucket before equals the p of the ball and bucket together. So 0.4 x 8.5 + 0.15 x 0 = .55 x v final. 3.4 + 0 = .55 v, and

v = 6.18 m/sto the south. The velocity is positive, and so is the velocity of the ball initially, so they are in the same direction, which is south.b. Here’s what you know: The kinetic energy measured before and after will not be the same, and to find the difference you need to calculate the before and the after and subtract them. The before and after kinetic energies are calculated with KE =

^{1}/_{2}mv^{2 }. The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before. The value of KE before is (0.5)(0.4)(8.5)^{2}+ (0.5)(0.15)(0)^{ 2}. Therefore, the KE before is 14.5 J. The KE afterward is (0.5)(0.55)(6.18)^{ 2}, which equals 10.5 J. 14.5 J minus 10.5 J is4 J.

(Table of contents)

**3. A 56 kg ice skater traveling at 4.0 m/s to the north
suddenly grabs the hand of a 65 kg skater traveling at 12.0 m/s in the opposite
direction as they pass. Without rotating, the two skaters continue skating
together with joined hands.**

** a. What is the final velocity of the two skaters?**

** b. What is the decrease in kinetic energy during the
collision? **

a. Here's what you know: The ice skater has a mass of 56 kg and an initial velocity of 4.0 m/s. The second ice skater has a mass of 65 kg and an initial velocity of -12 m/s. Together, the two skaters have a mass of 121 kg and we are looking for the final velocity. Using the formula p = mv, and the concept that p before equals p after, the p of the first skater before plus the p of the second skater before equals the p of the two skaters together. So 56 x 4 + 65 x -12 = 121 x v final. 224 + -780 = 121 v, and

v = -4.6 m/s to the south. The velocities of the two skaters have different signs (positive and negative) because they are in opposite directions. The answer to the problem is negative, so the skaters are traveling south.b. Here’s what you know: The kinetic energy measured before and after will not be the same, and to find the difference you need to calculate the before and the after and subtract them. The before and after kinetic energies are calculated with KE =

^{1}/_{2}mv^{2 }. The values of the masses and the velocities can be found in section a. You add the values of the two objects in question, and get the total before. The value of KE before is (.5)(56)(4)^{2}+ (.5)(65)(12)^{ 2}. Therefore, the KE before is 5128 J. In this case, both velocities are positive because the result calculation is energy, which is positive. The KE afterward is (.5)(121)(4.6)^{ 2}, which equals 1280 J. 5128 J minus 2180 J is 3847 J, which is3.8 x 10.^{ 3}J

(Table of contents)