Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Mastering Hor MasteringPhysics: HW#6 , Microsoft Edge HW#6 Question 5 teringphys

ID: 1658451 • Letter: M

Question

Mastering Hor MasteringPhysics: HW#6 , Microsoft Edge HW#6 Question 5 teringphysics.com/myct/item View?offset-next&tassignmentProblemD; 30020805 previous | 5 of 1 1 1 next MasteringPhysic Question 5 3 My Courses Submit My Answers Glve Up Course Home Assignments Scores eText On a frictionless horizontal air table, puck A (with mass 0.255 kg ) is moving toward puck B (with mass 0.366 kg ), which is initially at rest. After the collision, puck A has velocity 0.118 m/s to the left, and puck B has velocity 0.645 m/s to the right. Correct If you are required to use the answer obtained for a subsequent hint or part, use your unroundedifull precision answer Part B Study Area Calculate K, the change in the total kinetic energy of the system that occurs during the collision. Help Hints AK-0.0018 Submit My Answers Give Up Incorrect; One attempt remaining; Try Again ide Feedba Continue 7:45 PM Type here to search 6/20172

Explanation / Answer

m1 = 0.255 kg ; m2 = 0.366 kg ;

v1 = 0.118 m/s ; v2= 0.645 m/s

A)from the conservation of momentum

Pi = f

m1 va1 + 0 = m1 v1 + m2v2

va1 = (m1v1 + m2v2)/m1

va1 = (0.255 x 0.118 - 0.366 x 0.645)/0.255 = -0.808 m/s

Hence, va1 = 0.808 m/s

b)Intial KE of the system

KEi = 1/2 m1va1^2 = 0.5 x 0.255 x 0.808^2 = 0.0832 J

Final KE of system

KEf = 1/2 m1v1^2 + 1/2m2v2^2

KEf = 0.5 (0.255 x 0.118^2 + 0.366 x 0.645^2) = 0.0779 J

change in KE

delta-KE = KEf - KEi = 0.0779 - 0.0834 = -0.0053 J

Hence, delta KE = -0.0053 J