How to find the energy of an object that was at rest?

In summary: Then the momentum of the gas should be the same as the momentum...of the rocket?No, the momentum of the gas is not the same as the momentum of the rocket.
  • #1
Davidllerenav
424
14

Homework Statement


In two rockets, one of which moves and the other is at rest, the motors are connected for a short time. During their operation they throw the same mass of gas (small in comparison with the mass of the rocket) at the same speed with respect to the rockets. The kinetic energy of the rocket in motion was ##E_0##, increase by 4 percent. Determine the energy of the second rocket

Homework Equations


##E_k=\frac{1}{2}mv^2##

The Attempt at a Solution


I first wrote the final kinetic energy of the second rocket: ##E_k=E_0+\frac{4}{100}E_0##, then by using that equation: ##\frac{1}{2}mv_f^2=\frac{1}{2}mv_0^2+\frac{1}{25}(\frac{1}{2}mv_0^2)=mv_f^2=mv_0^2+\frac{1}{25}mv_0^2\Rightarrow m_vf^2=\frac{25mv_0^2+mv_0^2}{25}=\frac{26mv_0^2}{25}\Rightarrow 25v_f^2=26v_0^2##. After that I don't know what to do.
 
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  • #2
Davidllerenav said:

Homework Statement


In two rockets, one of which moves and the other is at rest, the motors are connected for a short time. During their operation they throw the same mass of gas (small in comparison with the mass of the rocket) at the same speed with respect to the rockets. The kinetic energy of the rocket in motion was ##E_0##, increase by 4 percent. Determine the energy of the second rocket

Homework Equations


##E_k=\frac{1}{2}mv^2##

The Attempt at a Solution


I first wrote the final kinetic energy of the second rocket: ##E_k=E_0+\frac{4}{100}E_0##, then by using that equation: ##\frac{1}{2}mv_f^2=\frac{1}{2}mv_0^2+\frac{1}{25}(\frac{1}{2}mv_0^2)=mv_f^2=mv_0^2+\frac{1}{25}mv_0^2\Rightarrow m_vf^2=\frac{25mv_0^2+mv_0^2}{25}=\frac{26mv_0^2}{25}\Rightarrow 25v_f^2=26v_0^2##. After that I don't know what to do.
What relates the change in speed of the rocket to the mass and relative speed of the exhaust?
 
  • #3
haruspex said:
What relates the change in speed of the rocket to the mass and relative speed of the exhaust?
I don't know.
 
  • #4
Davidllerenav said:
I don't know.
A conservation law perhaps?
 
  • #5
haruspex said:
A conservation law perhaps?
Oh, the conservation of the energy?
 
  • #6
Davidllerenav said:
Oh, the conservation of the energy?
No, there is unknowable energy input from the combustion of the fuel. What does that leave?
 
  • #7
haruspex said:
No, there is unknowable energy input from the combustion of the fuel. What does that leave?
the
Potential energy?
 
  • #8
Davidllerenav said:
the
Potential energy?
No, not energy. What other conservation laws do you know?
 
  • #9
haruspex said:
No, not energy. What other conservation laws do you know?
Momentum conservation.
 
  • #10
Davidllerenav said:
Momentum conservation.
Right.
You are told
Davidllerenav said:
they throw the same mass of gas ...at the same speed with respect to the rockets.
What does that tell you about the changes in momentum?
 
  • #11
haruspex said:
What does that tell you about the changes in momentum?
That momentum increases?
 
  • #12
Davidllerenav said:
That momentum increases?
Well, yes, ejectng the exhaust gases increases the momentum of the rocket - that's how rockets work.
But by how much is the momentum of the rocket increased if a mass M of gas is exhausted at speed v relative to the rocket?
 
  • #13
haruspex said:
Well, yes, ejectng the exhaust gases increases the momentum of the rocket - that's how rockets work.
But by how much is the momentum of the rocket increased if a mass M of gas is exhausted at speed v relative to the rocket?
It is increasing by 4%, so it would be ##P_f=P_i+\frac{1}{25}P_i##, right?
 
  • #14
The kinetic energy of the moving rocket has increased by 4%. That does not mean that its momentum has increased by 4%.

You had already correctly calculated that:
Davidllerenav said:
##25v_f^2=26v_0^2##
That should allow you to compute by how much the velocity of the moving rocket had increased.
 
  • #15
jbriggs444 said:
The kinetic energy of the moving rocket has increased by 4%. That does not mean that its momentum has increased by 4%.

You had already correctly calculated that:

That should allow you to compute by how much the velocity of the moving rocket had increased.
From that I would have ##v_f=\sqrt{\frac{26}{25}v_0\Rightarrow v_f=1.01v_0##, am I right?
 
  • #16
Davidllerenav said:
From that I would have ##v_f=\sqrt{\frac{26}{25}}v_0\Rightarrow v_f=1.01v_0##, am I right?
Yes, except that 1.01 is rather inaccurate. Anyway, don't bother about the arithmetic for now. What does momentum conservation allow you to deduce?
 
  • #17
haruspex said:
Yes, except that 1.01 is rather inaccurate. Anyway, don't bother about the arithmetic for now. What does momentum conservation allow you to deduce?
Well, since the momentum of the system is constant, so the sum of the both rockets momentum must be constant, right?
 
  • #18
Davidllerenav said:
Well, since the momentum of the system is constant, so the sum of the both rockets momentum must be constant, right?
The two rockets are in different systems. The untethered rocket, together with its exhaust gases, is an isolated system.
 
  • #19
haruspex said:
The two rockets are in different systems. The untethered rocket, together with its exhaust gases, is an isolated system.
Then the momentum of the gas should be the same as the momentum increase.
 
  • #20
Davidllerenav said:
Then the momentum of the gas should be the same as the momentum increase.
Right, and what do we know about the momentum of the gas from the other rocket?
 
  • #21
haruspex said:
Right, and what do we know about the momentum of the gas from the other rocket?
Well, since the rocket was at rest, the momentum of the system must be zero, so after the exhaust, the momentum of the gas must be the same as the momentum of.tje rocket but negative.
 
  • #22
Davidllerenav said:
Well, since the rocket was at rest, the momentum of the system must be zero, so after the exhaust, the momentum of the gas must be the same as the momentum of.tje rocket but negative.
Yes, but can you connect this with the momentum change of the first rocket?
 
  • #23
haruspex said:
Yes, but can you connect this with the momentum change of the first rocket?
Yes. Since both rockets connect their engines at the same time and thy exhaust the same mass of gas, then the change of momentum of the first rocket must be the same as the change of the rocket at rest, right?
 
  • #24
Davidllerenav said:
Yes. Since both rockets connect their engines at the same time and thy exhaust the same mass of gas, then the change of momentum of the first rocket must be the same as the change of the rocket at rest, right?
Yes.
 
  • #25
haruspex said:
Yes.
So, to calculate the momentum of the rocket at rest I just need to use the change in the velocity of the second rocket, which was ##1.01v_i##? ##P=mv_f=m(1.01v_i)##, and so the kinetic energy from the rocket at rest would be ##E_k=\frac{1}{2}m(1.01v_i)^2##. But why did you said that it is rather inaccurate?
 
  • #26
Davidllerenav said:
the change in the velocity of the second rocket, which was ##1.01v_i##?
That is not the change in velocity.
Davidllerenav said:
why did you said that it is rather inaccurate?
What is √1.04?
 
  • #27
haruspex said:
That is not the change in velocity.
It is the new velocity. So what velocity should I use?
haruspex said:
What is √1.04?
It is ##1.0198...##.
 
  • #28
Davidllerenav said:
It is the new velocity.
Right, so what is the change in velocity?
Davidllerenav said:
It is 1.0198...
So what would be the right way to express that to two decimal places?
 
Last edited:
  • #29
haruspex said:
Right, so what is the change in velocity?
it would be ##\Delta v= v_f-v_0## so the change would be ##\Delta v= 1.012v_0-v_0=0.012v_0##, right?

haruspex said:
Right, so what is the change in velocity?

So what would be the right way to express that to two decimal places?
1.012?
 
Last edited:
  • #30
Davidllerenav said:
it would be ##\Delta v= v_f-v_0## so the change would be ##\Delta v= 1.012v_0-v_0=0.12v_0##, right?1.012?
No, 1.0198 would not be abbreviated to 1.012; and even if it were, subtracting 1 does not yield 0.12.
 
  • #31
haruspex said:
No, 1.0198 would not be abbreviated to 1.012.
Then what would it be?
haruspex said:
and even if it were, subtracting 1 does not yield 0.12.
Yes sorry, I meant 0.012.
 
  • #32
Davidllerenav said:
Then what would it be?
You will kick yourself when you realize the error you are making.
Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.
 
  • #33
haruspex said:
Repeatedly increment the final nonzero digit of 1.0198 until there are only two decimal places.
Oh, I think I see it know, woudl it be 1.02?
 
  • #34
Davidllerenav said:
Oh, I think I see it know, woudl it be 1.02?
Yes.
 
  • #35
haruspex said:
Yes.
So ##\Delta v=0.02## and the kinetic energy of the second rocket, the one that was at rest is ##E_k=m\Delta v^2= m(0.02)^2v_0^2##? Is t right?
 

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