1 2 2 blocks of ice are sliding on frictionless ice in opposite directions collide. The 1st block's mass is 3kg. It is moving to the right at 3 m/s. The 2nd block's mass is 2kg. It is moving to the left at 2 m/s. After the impact, the 2nd block moves to the right with a speed of 2 m/s What is the coefficient of restitution? m1 = 3 vla= 3 m2 = 2 v2a= -1 After collision: v2b=2 Momentum_a= m1*vla + m2*v2a v1b = (Momentum_a - m2*v2b) / m1 % Restitution equation e = (v2b - v1b) / (v1a - v2a)

1 2 2 blocks of ice are sliding on frictionless ice in opposite directions collide. The 1st block's mass is 3kg. It is moving to the right at 3 m/s. The 2nd block's mass is 2kg. It is moving to the left at 2 m/s. After the impact, the 2nd block moves to the right with a speed of 2 m/s What is the coefficient of restitution? m1 = 3 vla= 3 m2 = 2 v2a= -1 After collision: v2b=2 Momentum_a= m1vla + m2v2a v1b = (Momentum_a - m2*v2b) / m1 % Restitution equation e = (v2b - v1b) / (v1a - v2a)

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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Question

1
2
2 blocks of ice are sliding on frictionless ice in opposite directions collide.
The 1st block's mass is 3kg. It is moving to the right at 3 m/s.
The 2nd block's mass is 2kg. It is moving to the left at 2 m/s.
After the impact, the 2nd block moves to the right with a speed of 2 m/s
What is the coefficient of restitution?
m1 = 3
vla= 3
m2 = 2
v2a= -1
After collision:
v2b=2
Momentum_a= m1*vla + m2*v2a
v1b = (Momentum_a - m2*v2b) / m1
% Restitution equation
e = (v2b - v1b) / (v1a - v2a)

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