In the figure, block 1 of mass m_1 slides from rest along a frictionless ramp fr
ID: 1413210 • Letter: I
Question
In the figure, block 1 of mass m_1 slides from rest along a frictionless ramp from height h = 3.1m and then collides with stationary block 2, which has mass m_2 = 3m_1. After the collision, block 2 slides into a region where the coefficient of kinetic friction mu_k is 0.55 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic? Number Unit Number Unit By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your Instructor.Explanation / Answer
Apply conservation of energy for for block 1
m1 g h top + 1/2 m v top^2 = m1 g h bot + 1/2 m v bot^2
v bot^2 = 2gh top
v bot = sqrt 2(9.8) ( 3.1) = 7.79 m/s
(a) elastic
m1v + m2 v = m1v1+m2v2
m(7.79) + 0= m v1+ 3mv2
7.79=v1+ 3v2
1/2 m1v^2 + 1/2 m2 v^2 = 1/2 mv1^2 + 1/2 m2 v2^2
m( 7.79)^2 + 0 = mv1^2 + 3mv2^2
60.6891 = v1^2 + 3v2^2
60.6891 = (7.79-3v2)^2 + 3v2^2
60.6891 = 60.6841-46.74 v2 + 9 v2^2 + 3 v2^2
12 v2^2 =46.74 v2
v2 = 3.895
v1 = 7.79-3(3.895) = -3.895
1/2 m2 v^2 - F1 d = 1/2 m v1^1
1/2 * 3m ( 3.895)^2 - 0.55 ( 3m) g d = 0
22.75 - 16.17d = 0
d = 1.41 m
(b)
in elastic
m1v + m2v ( m1+ m2) v
mv + 0 = ( m+ 3m) v
m( 7.79) = 4m v
v = 1.9475 m/s
1/2 ( m+ 3m) v^2 - F1 d = 1/2 mv1^2
1/2 * 4m( 1.9475)^2 - 0.55(4m) d = 0
7.58 - 21.56 d = 0
d = 0.351 m