Energy Conservation or Momentum?

[azn4life1990]

Homework Statement

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-h frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

a) suppose the packages stick together. what is their common speed after the collision?


b) suppose the collision between the packages is perfectly elastic. to what height does the package of mass m rebound?

Homework Equations

KE = PE
KE = 1/2 mv²
PE = mgh

The Attempt at a Solution

V_i = 0m/s

Ek = Ep

1/2mv² = mgh
The masses cancel each other out, leaving me with

v = $$\sqrt{}2gh$$

v = 58.8m/s, this is the velocity at which the box 1 hits box 2

From this point i decided to use momentum to find the final velocity
(2m+m)v_f = mv_i

This simplifies to 3V_f = V/3 = 19.6m/s
Why would using momentum be flawed?

When I use KE is conserved throughout this system I get another answer... that is why one of the methods I'm using is wrong

1/2mv² = 1/2mv²

1/2(2m+m)v² = 1/2mv²

Doing some algebra I get that V² = V²/3

For this answer the Vf = 33.95m/s

 

[Doc Al]

The Attempt at a Solution

V_i = 0m/s

Ek = Ep

1/2mv² = mgh
The masses cancel each other out, leaving me with

v = $$\sqrt{}2gh$$

v = 58.8m/s, this is the velocity at which the box 1 hits box 2

You forgot the square root.

 

[D H]

When I use KE is conserved throughout this system I get another answer... that is why one of the methods I'm using is wrong

Using conservation of kinetic energy for the first problem will give you a different answer than conservation of momentum. You can't use conservation of kinetic energy here because kinetic energy is not conserved. The collision is inelastic.

 

[azn4life1990]

thanks for the input D H
so, for all collisions momentum is conserved.
for elastic collisions momentum + KE is conserved
for inelastic collisions momentum is conserved but KE isn't
so the square root of 58.8 = 7.668m/s
and the final velocity of boxes stuck together is 2.56m/s

 

[D H]

Whoa! That doesn't conserve momentum.

 

[azn4life1990]

sorry? i didnt understand what you meant by that

 

[Doc Al]

so, for all collisions momentum is conserved.
for elastic collisions momentum + KE is conserved
for inelastic collisions momentum is conserved but KE isn't
so the square root of 58.8 = 7.668m/s
and the final velocity of boxes stuck together is 2.56m/s

Looks good to me. Now solve part b).

 

[azn4life1990]

alright for the next part we have perfect elastic collision

Ek = conserved
P = conserved

This is my first equation for Momentum:
m₁v₁+m₂v₂ = m₁v_f1+m₂v_f2

This is my 2nd equation for kinetic energy
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v_f1²+1/2m₂v_f2²

This should take some algebra :sLooking into my textbook I found these simplified equations

V_f1 = V_i*(m₁-m₂/m₁+m₂)
For this part I found that the Vf of the first object = -2.56m/s

V_f2 = V_i*(2m₁/m₁+m₂)
For this part I found that the Vf of the object = 5.11m/s

 

[Delphi51]

It checks out - momentum conserved, energy conserved.
Also did it without the simplified equations and got the same answers.