Conservation of momentum & energy

[sevesham]

Homework Statement

Hi,
I do not know how to resolve question 1d and 2d at all. I hope the rest is well resolved, not sure about 1c and 2c. Can someone check my work and help me resolve problem with this questions, please?
Task
A hammer of mass 1700kg is dropped 2.5m under its own in fluency of gravity and hits a pile of mass 2700kg pushing it into the ground 200mm with each stroke. The hammer does not bounce back.

1) Calculate using the conservation of Momentum:
a) the common velocity immediately after impact,
b) the useful work done at each stroke,
c) the ground resistance,
d) the energy lost at each impact.

2) Calculate using the conservation of Energy:
a) the common velocity immediately after impact,
b) the useful work done at each stroke,
c) the ground resistance,
d) the energy lost at each impact.

Homework Equations

v=total momentum (p) / total mass (m)

The Attempt at a Solution

Conservation of Momentum
1 a) velocity after impact = total momentum (Tp) / total mass (Tm)
m1=1700kg
m2=2700kg
u=0
s1=2.5m
s2=0
a=g=9.81m/s^2
v1=? v1^2=u^2+2*a*s1 = 0^2+2*9.81*2,5=49.05 v1=√49.05=7.0036m/s
v2=0 c'ouse s2=0
Hammer p1=1700*7.0036=11906.07kgm/s Pile p2=2700*0=0
v2=p1+p2/m1+m2=1907.07/4400=2.7059m/s

1b) W=F*s=Tm*a*s
s3=200mm=0.2m
u2=2.7059m/s
v=0
a1=? a1=v^2-u2^2/2s3=0^2-2.7059^2/2*0.2=-18.3047m/s^2
W=4400*(-18.3047)*0.2=-16108.17J -16.11kJ/stroke

1c) Fg=(m1+m2)*g+(m1+m2)*a1=Tm*(g+a1)
Tm=m1+m2=4400kg
u2=2.7059m/s
v=0
s3=0.2m
g=9.81m/s^2
a1=18.3047m/s^2
Fg=4400*(9.81+18.3047)=123704N 123.7kN

1d) ?

Conservation of Energy
2a) I'm not sure about this calculation.
m1=1700kg
m2=2700kg
u=0
s1=2.5m
s2=0
a=g=9.81m/s^2
v1=? v1^2=2*PE/2=2*m1*g*s1=2*1700*9.81*2.5=49.05 v1=√49.05=7.0036m/s
v2=0 c'ouse s2=0
Velocity after impact
v3=(m1*v1)+(m2*v2)/m1+m2=1907.07/4400=2.7059m/s

2b) W=F*s=m*a*s PE=m*a*s W=PE
Tm=4400kg
s3=0.2m
u2=2.7059m/s
v=0
a1=? a1=v^2-u2^2/2s3=0^2-2.7059^2/2*0.2=-18.3047m/s^2

W=Tm*a1*s3=4400*(-18.3047)*0.2=-16108.17J

2c) Fg= change of PE + change of KE/ change of distance
Tm=4400kg
g=9.81m/s^2
s3=0.2m
v3=2.7059m/s
PE=Tm*g*s3
KE=1/2*Tm*v3^2

Fg=Tm*g*s3+1/2*Tm*v3^2/s3=4400*9.81*0.2+1/2*4400*2.709/0.2=123704N

2d) ?

 

[kuruman]

Parts 1 a-c are correct except that the useful work done in 1b must be positive.
Part 1d: The energy lost at each impact is the difference between the energy just before the impact and the energy immediately after the impact.
Part 2a is correct and should be no different from 1a.
Parts 2b does not appear to use energy conservation as the question asks. You should do this without finding the acceleration first.
Part 2c is correct.
Part 2d, same comment as 1d.

 

[haruspex]

I do not understand question 2. Is it asking you to calculate these matters as though energy (work) is conserved and momentum is not? If so, all of the answers in that part will be factually wrong, and 2d becomes trivial.

 

[kuruman]

I do not understand question 2. Is it asking you to calculate these matters as though energy (work) is conserved and momentum is not? If so, all of the answers in that part will be factually wrong, and 2d becomes trivial.

Precisely my assessment, however I thought that these questions may only be part of the question, another part (not posted) perhaps being a critical comparison of the answers obtained by the two methods.

 

[sevesham]

Hi,
This is whole exercise. Thanks KURUMAN for your tips. I found resolution for 2b, 2d but still strugling with 1d. I do not know how to find initial energy in conservation of momentum.

 

[kuruman]

To find the energy just before the collision, calculate the sum of kinetic energies of all parts of the system that are moving. If you take the zero of potential energy to be the point of collision, you don't have to worry about potential energy. So what is the energy before the collision? Don't get too hung up on using momentum conservation to find the initial energy; you can't because as the hammer drops, momentum is not conserved because the external force of gravity acts on it.

I am interested in seeing how you resolved 2b and 2c as well as the answers you got.

@haruspex:
I now think that "conservation of energy" for part 2 means "total" energy, not just mechanical energy. One can write $E_i=E_f+E_{lost}$ and use the answer in 1d for $E_{lost}$. Nevertheless, the answer to 2d is still trivial.

 

[sevesham]

Hi,

Answer 2c presented in the first post.
I improved 2b according to your instruction, use conservation of energy.

2b) W=PE PE=KE KE=Tm*v3^2/2
Tm=4400kg
v3=2.7059m/s

W=Tm*v3^2/2=4400*2.7059^2/2=16108.17J
Same answer I received in momentum 1b. Hope is ok.

2d)
m1=1700kg (hammer)
m2=2700kg (pile)
s1=2.5m
s2=0
g=9.81m/s^2
PEi=(m1*g*s1)+(m2*g*s2)=(1700*9.81*2.5)+(2700*9.81*0)=41692.5J
velocity calculated in 2a
v1=7.0036m/s
v2=0
v3=2,7059m/s

KE=PE
Initial KE=(m1*v1^2/2)+(m2*v2^2/2)=(1700*7.0036^2/2)+(2700*0/2)=41692.85J
Final KE=Tm*v3^2/2=4400*2.7059^2/2=16108.17J KEf=W

Elost=KEi-KEf=41692.85-16108.17=25584.68J

I hope this is correct if not let me know, please.

1d) No idea how find Ei in conservation of momentum. For Ef I can use calculated work done from 1b and subtract from Ei.

 

[haruspex]

I now think that "conservation of energy" for part 2 means "total" energy, not just mechanical energy.

But then "calculate using conservation of energy" is impossible. You would need information on the heat generated.

 

[kuruman]

For 2b you assumed that v3 is known from part 1 where it was found using momentum conservation. I think the problem is asking you to find v3 again using energy conservation. Read my note to @haruspex in post #6 about how to do this.

No idea how find Ei in conservation of momentum. For Ef I can use calculated work done from 1b and subtract from Ei.

You can't. As I explained in post #6 momentum is not conserved from the time when the hammer is released to the time it hits the mass. Momentum conservation means that the momentum does not change but stays the same. When the hammer is released from rest its momentum is zero, when it collides with the mass its momentum is non zero. Although its mechanical energy (kinetic plus potential) is conserved and stays the same, its momentum is not conserved and does not stay the same. Therefore you cannot use momentum conservation to find anything while the hammer is falling.

So when can you use momentum conservation? The answer is only during the short time of the collision. You can pretend that the collision happens very fast, so fast that the combined system of hammer and mass do not move appreciably over the time interval of the collision. This is another way of saying that one can neglect the effect of the external forces during the collision on the two-mass system while the collision takes place.

By contrast, you can always use total energy conservation as long as you can account for all the transformations of energy from one form to another as forces do work on one part of the system or another. Here you have a two-mass system consisting of a hammer and a mass. While the hammer is dropping, the only transformation of energy is potential to kinetic which means that mechanical energy is also conserved. When the hammer hits the mass and drives it into the ground, you have the kinetic energy of the hammer being transformed into kinetic energy of the hammer plus the target mass, heat of friction with the ground, sound energy, energy of vibration of the masses and vibration of the ground, kinetic and potential energy of the dust that might be raised, to name a few. You can find what the kinetic energy of the hammer and the mass are immediately after the collision because there is an equation that you can use. Unless you have equations that permit to find the other transformed energies, the only other thing you can do is lump them all together under one name and call it $E_{lost}.$

 

[kuruman]

But then "calculate using conservation of energy" is impossible. You would need information on the heat generated.

Yes you would. I suspect the problem expects one to use the answer for E_lost in part 1d. Using m₁gh = ½(m₁+m₂)v₃² + E_lost gives a consistent answer for v₃ and maybe that's the purpose of this exercise.

 

[haruspex]

No idea how find Ei in conservation of momentum

You are not asked to. You find the velocity just before impact and hence E_i in the usual way; using conservation of momentum you find the velocities after impact, and hence E_f.

I'm uncertain what is expected in parts b. There are two different E_is involved in this process, the KE just before impact (which is certainly appropriate in parts a), and the total initial energy that goes into driving down the pile; the second reasonably includes the GPE lost in descending 200mm.

 

[kuruman]

I'm uncertain what is expected in parts b.

Here is my reasoning on that. There are two kinds of work done on the pile after the collision, work done by conservative forces and work done by non-conservative forces. It will require a stretch of the imagination to call the work done by non-conservative forces "useful" wheras there is only one conservative force acting on the pile as it descends. Therefore ...