Asked by Isabel

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.

Answers

Answered by drwls
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.
Answered by Zach
A.) h/4μk
B.) h/16μk
Answered by Samprit
Fhhdugm
There are no AI answers yet. The ability to request AI answers is coming soon!

Related Questions