A block of mass M1 = 3:5 kg moves with velocity v1 = 6:3 m=s on a frictionless s

Question

A block of mass M1 = 3:5 kg moves with velocity v1 = 6:3 m=s on a frictionless surface. It collides with block of mass M2 = 1:7 kg which is initially stationary. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1:85 m . (Please, only use formulas of Momentum not use the Energy formulas).

a. What is the coecient of kinetic friction on the rough surface?
b. How far would M2 travel if the coecient of restitution was r = 0:6

Explanation / Answer

after collision both block stick together so collision is inelastic

from the conservation momentum

M1v1 + M2v2 = (M1+M2)v
v=M1v1/(M1+M2) = (3.5*6.3)/(3.5+1.7)=4.24m/s

from the work energy theorem

KE=Net Work

0.5(M1+M2)v^2=(M1+M2)*g**1.85

0.5 ( 3.5 + 1.7) ( 4.24)^2 = ( 3.5 + 1.7) ( 9.8) *1.85

=0.495

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.