Question

Block 1 of mass m1 slides from rest along a frictionless ramp from height h and then collides with stationary block 2, which has mass m2 = 3m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction is μk 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? Express your answer in terms of the variables given and g.

Answer 1

The first thing you have to do is compute the velocity of block 2 in each of the two cases: elastic and completely inelastic collision with block 1.

In the elastic case, both momentum and kinetic energy are conserved.

Vcm = Vo/4 is the center of mass velocity. This leads to
V2(final) = Vo/2
V1(final) = -Vo/2

In the inelastic case,
M1 Vo = (M1 + V2) Vfinal
= 4M1*Vfinal
Vfinal = Vo/4 for both blocks

The distance d of the slide is given by

(1/2) M Vfinal^2 = M g ìk X

Use that to solve for X. M cancels out

X = Vfinal^2/(2*g*ìk)

It slides four times farther in the elastic case, because Vfinal is twice as large.

Answer 2

A.) h/4μk
B.) h/16μk

Answer 3

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