Please help! Show all steps fully. Part b only. I will rate :) In the figure bel

Question

Please help! Show all steps fully. Part b only.

I will rate :)

In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.00 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 mu k is 0.400 and comes to a stop in distance d within that region. What is the value of distance d if the collision is elastic? m What is the value of distance d if the collision is completely inelastic? m Did you conserve mechanical energy to find the speed of block 1 at the bottom? Did you use one of the equations for a one-dimensional elastic collision to get the speed of block 2 just after the collision? Did you then calculate how far the block must slide for the kinetic frictional force to stop it? The procedure for the inelastic collision is similar except now only momentum is conserved in the collision.

Explanation / Answer

if inelastic v of block just before collision v = sqrt(2 g h) collsion m1 v = (M1 + 2 M1 ) V V = 1/3 sqrt(2 gh ) now energy Ei + wf = Ef 1/2 (3 M1) * (1/3 sqrt(2 gh ))^2 - 0.4* (3 M1 ) *9.81*d = 0 1/2 (3) * (1/3 sqrt(2 *9.81*3))^2 - 0.4* (3 ) *9.81*d = 0 d=0.83 m

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