Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
block 1 of mass 1 slides from rest along a frictionless ramp from height h=2.50m and then collides with stationary block 2, whi...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.
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
B.) h/16μk
Answered by
Samprit
Fhhdugm
There are no AI answers yet. The ability to request AI answers is coming soon!