In the figure, block 1 of mass m 1 slides from rest along a frictionless ramp fr
ID: 1345111 • Letter: I
Question
In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.4 m and then collides with stationary block 2, which has mass m2 = 5m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction ?k is 0.2 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?
Can't quite seem to get part one down, but nailed part 2. The answer I have for part one is 4.25 m/s.
Your answer is partially correct. Try again.Explanation / Answer
speed of block 1 as it comes 3.4m down.
using mv^2 /2 = mgh
v = sqrt(2 x 9.81 x 3.4) = 8.17 m/s
using equation for elastic collision,
velocity of approach = velocity of seperation
8.17 = v2 + v1
v1 = 8.17 - v2
now using momentum conservtaion,
m1 x 8.17 + m2 x 0 = -m1v1 + m2v2
8.17m1 = - m1(8.17 - v2) + 5m1 v2
v2 = 2.72 m/s .............speed of block 2 just after collision.
d = 2.72^2 / 0.2x9.81 x2 = 1.89 m