In the figure below, block 1 of mass m 1 slides from rest along a frictionless r
ID: 1777373 • Letter: I
Question
In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.95 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.545 and comes to a stop in distance d within that region.
(a) What is the value of distance d if the collision is elastic?
(b) What is the value of distance d if the collision is completely inelastic?
Explanation / Answer
a]speed of block 1 just before collision, v1 = sqrt(2gh) = sqrt(2*2.95*9.8) = 7.604 m/s
velocity of second block after collision v2' = 2m1v1/(m1+m2) = 2*1*7.604/(1+2) = 5.07 m/s
d = v^2/(2ug) = 5.07^2/[2*0.545*9.8] = 2.406m answer
b] in inelastic collision, v = m1v1*[m1+m2] = 7.604*1/3 = 2.535 m/s answer
d = v^2/(2ug) = 2.535^2/[2*0.545*9.8] = 0.602 m answer
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