PHYS 1211: Principles of PhysicsI (Mechanics), Cotten, Summer 2018 Chandler Whit

Question

PHYS 1211: Principles of PhysicsI (Mechanics), Cotten, Summer 2018 Chandler White(Student section: 52482) Main Menu Contents Grades Messages Courses Help Logout Timer? Notes valuatelde Feedback -Print rs! ? Course Contents » ...Homework Set 06Curved Surface 92200 Thcmaon BrocksGele 2 FIGURE P6.48 (c6p48) Consider a frictionless track as shown in Figure P6.48. A block of mass m1 = 8.0 kg is released from A, at a height h = 11.0 m. It makes a head-on elastic collision at B with a block of mass m2 = 17.0 kg that is initially at rest. Calculate the maximum height to which ml rises after the collision. Subrmit Answer Tries 0/8 Post Discussion Send Feedba

Explanation / Answer

First, what is the speed of m2 at the collision?

PE is converted to KE

PE = mgh = m x 9.8 x 11 = 107.8m

Kinetic Energy in J if m is in kg and V is in m/s

KE = ½mV² = 107.8m

V² = 215.6

V = 14.68 m/s

Elastic collisions

v is velocity after the collision, u before

u? = 14.68, u? = 0

m? = 8, m? = 17

v? = (u?(m?–m?) + 2m?u?) / (m? + m?)

v? = (u?(m?–m?) + 2m?u?) / (m? + m?)

v? = 14.68(–9) ) / (25) = –5.28 m/s

v? = (2*8*14.68) / (25) = +9.4 m/s

so we have the block rebounding with v = 5.28 m/s

again use KE = PE to determine height

½mV² = mgh

V² = 2gh

5.28² = 2*9.8h

h = 1.42 m

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