# Conservation of momentum & energy

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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.

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=49.0 5 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) ???

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.

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=49.0 5 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) ???

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#2

I'm not 100% sure what they're getting at BUT if you compare the useful work done with the initial GPE of the hammer perhaps (or the KE of the hammer just as it strikes the pile)? It's the only thing I can think of which hasn't been asked already. The difference between this and the useful work would be lost energy

Hope this helps.

Hope this helps.

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#3

(Original post by

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.

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=49.0 5 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) ???

**sevesham**)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.

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=49.0 5 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) ???

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(Original post by

For (d), you need to find the kinetic energy before and after the inelastic collision and find the difference between the kinetic energy before and after the inelastic collision.

**Eimmanuel**)For (d), you need to find the kinetic energy before and after the inelastic collision and find the difference between the kinetic energy before and after the inelastic collision.

**2d)**

m1=1700kg (hammer)

m2=2700kg (pile)

s1=2.5m

s2=0

g=9.81m/s^2

PE=(m1*g*s1)+(m2*g*s2)=(1700*9.8 1*2.5)+(2700*9.81*0)=

**41692.5J**

velocity calculated before in 2a

v1=7.0036m/s (for hammer)

v2=0 (for pile)

v3=2,7059m/s (after impact)

PE=KE

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 have no idea how to find the initial energy. Energy final = work done.

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#5

(Original post by

I resolved 2d for conservation of energy, but not sure is ok.

m1=1700kg (hammer)

m2=2700kg (pile)

s1=2.5m

s2=0

g=9.81m/s^2

PE=(m1*g*s1)+(m2*g*s2)=(1700*9.8 1*2.5)+(2700*9.81*0)=

velocity calculated before in 2a

v1=7.0036m/s (for hammer)

v2=0 (for pile)

v3=2,7059m/s (after impact)

PE=KE

Initial KE=(m1*v1^2/2)+(m2*v2^2/2)=(1700*7.0036^2/2)+(2700*0/2)=

Final KE=Tm*v3^2/2=4400*2.7059^2/2=

Elost=KEi-KEf=41692.85-16108.17=

I have no idea how to find the initial energy. Energy final = work done.

**sevesham**)I resolved 2d for conservation of energy, but not sure is ok.

**2d)**m1=1700kg (hammer)

m2=2700kg (pile)

s1=2.5m

s2=0

g=9.81m/s^2

PE=(m1*g*s1)+(m2*g*s2)=(1700*9.8 1*2.5)+(2700*9.81*0)=

**41692.5J**velocity calculated before in 2a

v1=7.0036m/s (for hammer)

v2=0 (for pile)

v3=2,7059m/s (after impact)

PE=KE

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=WElost=KEi-KEf=41692.85-16108.17=

**25584.68J**I have no idea how to find the initial energy. Energy final = work done.

Don’t need to find the initial energy. Your answer (

**25584.68J**) seems to be ok.

Actually you don’t need to go through all this stuff.

From (a) you know the common velocity after impact say

*v*

_{2}.

Initial KE of hammer is just

*m*

_{1}

*gh*

_{1}where

*h*

_{1}is 2.5 m.

KE of the hammer and pile after impact is ½ (

*m*

_{1}+

*m*

_{2})

*v*

_{2}

^{2}.

Find the difference and you will get the energy lost after impact.

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(Original post by

Don’t need to find the initial energy. Your answer (

Actually you don’t need to go through all this stuff.

From (a) you know the common velocity after impact say

Initial KE of hammer is just

KE of the hammer and pile after impact is ½ (

Find the difference and you will get the energy lost after impact.

**Eimmanuel**)Don’t need to find the initial energy. Your answer (

**25584.68J**) seems to be ok.Actually you don’t need to go through all this stuff.

From (a) you know the common velocity after impact say

*v*_{2}.Initial KE of hammer is just

*m*_{1}*gh*_{1}where*h*_{1}is 2.5 m.KE of the hammer and pile after impact is ½ (

*m*_{1}+*m*_{2})*v*_{2}^{2}.Find the difference and you will get the energy lost after impact.

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