Does momentum conservation imply energy conservation? I was trying to figure out the situation in which energy is conserved and momentum is not and it was quite easy to find out one case which is of a stone tied to a string moving in a uniform circular motion.
Then I thought to consider the reverse situation in which momentum is conserved but energy is not. To me it seems that as soon as we chose a system in which momentum is conserved then it automatically is implied that in such system energy is conserved too. But looking at the case how these two conservations come into existence due to  two different symmetries (one  related to the invariance in physical laws due to translation in space and the other in time). 
So it would be quite helpful 


*

*if someone can point out to a case in which momentum is conserved but energy is not.

*otherwise if the above is not possible then explain why that is the case?
[Note that I am considering every form of energy of the given system.]

This is to summarise my question so that no further confusion occurs to future visitors. The question in short is:


*

*can we lose time symmetry and retain translation symmetry? (Give an appropriate example for the case)

 A: Yeah, it's interesting to contemplate the relation between energy and momentum.
Energy can morph from form to form, with kinetic energy being only one of them. Momentum, on the other hand, deals with a subset of all forms of energy: motion.
So let's set up a system with the energy oscillating between potential energy and kinetic energy: let's say a vibrating string (for example a guitar string).
The momentum of the string is oscillating, while the total energy is conserved.
However, looking at it that way is cheating. The string is part of a system: the system as a whole is the guitar with the string. To aid this thought demonstration, take a guitar with one string, and make that string really heavy. Have that system be suspended, such that it cannot exchange momentum with surroundings. Then the string and the body are continuously exchanging momentum with each other, and the momentum of that system is conserved.

In your question, you submitted the case of an object connected to a central axis that is infinitely rigid, with the object moving in a circular motion.
That example doesn't actually offer what you want.
You need to consider the system as a whole. This means that you also need to consider the momentum of the support that maintains the circular motion of the circumnavigating object.
Consider for example the motion of the planet Jupiter around the Sun. Jupiter and the Sun are both orbiting the common center of mass of the Sun/Jupiter system. That common center of mass lies somewhat outside the Sun, not even inside the surface of the Sun.

About the title of your question:
"Does momentum conservation imply energy conservation?"
In my opinion: from an axiomatic point of view: no.  
I think that in terms of formal logic the two are independent axioms. In other words, I think that in terms of formal logic neither can be derived from the other.

However, if both energy conservation and momentum conservation are laws that hold good universally, and presumably they do, then in any thought demonstration we come up with we will always see that energy and momentum are both conserved.
I think: in terms of formal logic the fact that both hold good universally does not logically imply that they are interdependent (but thought-provoking for sure).
A: Due to the principle of energy conservation, energy is always conserved for an isolated system in an inertial reference frame. Therefore, considering the conservation of all forms of energies (the sum of them), the question does not make much sense. 
It makes sense if referring to the mechanical energy only, whose conservation is a theorem and not a principle. 
Here it is possible to show elementary counterexamples where the angular momentum is conserved whereas the mechanical energy is not. 
A simple example is a point of the matter of mass $m$ attached to one endpoint an ideal stick with negligible mass and the other endpoint of the stick is fixed to the origin $O$  of the Cartesian axes and the point can freely rotate in the plane xy (without friction) of an inertial reference frame. 
The trick is supposing that the length of the stick varies in time with a given function $d=d(t)$.
The motion of the point is described in terms of the angle $\theta(t)$ determined by the stick with respect to the axis $x$.
The  equation of motion are 
$$\frac{d}{dt}\left(m d(t)^2 \dot{\theta}(t)\right)=0$$
The mechanical energy is 
$$E= \frac{1}{2} md(t)^2 \dot{\theta}^2$$
and it is easy to see from the equation above that it is not conserved on every non-trivial solution of the equation of motion if the assigned function $d=d(t)$ is not constant.
Conversely, the angular momentum is directed along the 
axes $z$ and has a numerical value which exactly coincides with
$$L=\left(m d(t)^2 \dot{\theta}(t)\right).$$
Therefore it is conserved (no matter how you choose the function $d$) on every solution of the equation of motion.
A: If you’re just considering mechanical energy, consider the case of a firework or other exploding object. The total momentum is conserved, but the kinetic energy is increased from the chemical energy of the explosion.   
More formally, momentum conservation is associated with isotopic space while energy conservation is associated with no changes over time: the explosion happens at a moment and energy changed; you pull the circular motion in different directions and momentum changes. So you can develop different situations to do different things. 
A: While there are some good answers, no one has yet mentioned Noether's theorem. Fundamentally, the conservation of momentum and conservation of energy have different characters. Notably conservation of linear momentum is linked with the laws of physics being independent of where something happens and conservation of energy is linked with the laws of physics being independent of when something happens.
Simply: Linear momentum conservation is linked to position invariance.
Energy conservation is linked to time invariance.
More rigorously, Noether's theorem demonstrates that conservation of linear momentum arises from a Lagrangian that has no explicit dependence on position and conservation of energy from a Lagrangian that has no explicit dependence on time. Furthermore, other symmetries imply other conservation laws. A lack of explicit dependence on the angle in the Lagrangian gives conservation of angular momentum, suggesting that angular momentum can be considered fundamentally different than linear momentum (which is backed up by spin in quantum mechanics not being interpretable as linear motion). As an aside, purely (Newtonian) 2-body gravitational systems have another conserved vector owing to symmetry in the Lagrangian of the inverse square force law. 
Hence, it is easy to write down models where energy is conserved, but momentum is not, and systems where momentum is conserved and energy is not. Furthermore, these models can be good approximations of many real situations as described in other answers. You can have models of collisions where momentum is conserved but energy is "lost" to internal degrees of freedom or radiation. You can have thermodynamic models where you have a container of effectively infinite mass (which leads to momentum not being conserved when gas molecules collide with the walls) but with conserved energy. In the end, however, linear momentum, angular momentum, and energy seem to be conserved in fundamental interactions and maybe in the universe as a whole.
As an aside, Special Relativity treats position and time in a similar way and likewise wraps up momentum and energy in the 4-momentum, which shows energy as the time component of a relativistic 4-momentum. However, although relativity shows some symmetry between position and time and consequently momentum and energy, they still have different characters owing to different signs in the signature ($\mathrm{d}s^2 = -c^2\,\mathrm{d}t^2 + \mathrm{d}x^2 + \mathrm{d}y^2 + \mathrm{d}z^2$).
A: In Newtonian theory, you're dealing with only gravity and EM forces which are conservative and central so there is no example of a closed system where either energy or momentum are not conserved. So energy and linear and angular momentum conservation are always implied by Noether's theorem. Your example is wrong even if you only consider linear momentum for the whole system. 
In QM  Hamiltonian systems are always space- and time-symmetric so again the same thing. 
In general( relativity) this symmetry doesn't have to be there. one can derive some similar conserved quantities but they don't necessarily equate to momentum or energy. So they are not conserved 
For example, two laser light sources/receivers are some large distance (10 mpc or more) apart shooting light at each other as they are  receding from each other. A distant observer will see the energy of light drop, in time at each receiver because of respective redshifts but momentum will be conserved. Or simply if they were two point masses as the space between them is increasing potential energy is lost while momentum stays the same. 
A: Example of momentum conservation, while energy is not conserved (sticky collision):
Just let two pieces of sticky stuff with opposite momenta (so in total zero momentum), collide frontally. The end result is a piece of this stuff that has also zero momentum but the kinetic energy has conversed into internal heat (energy is alway conserved, just like total momentum, which in this case enters into the internal of the stuff). It's just an inelastic encounter. 
So, also in the example you give, both energy and momentum (angular momentum) are always conserved. 
So to answer your question: No. Momentum conservation doesn't imply energy conservation (when potential energy is not considered). More generally they don't imply each other both ways (again, not considering potential energy). 
They are separate conservation laws that can't be connected.
Momentum by itself nevertheless is connected with (kinetic) energy by itself:
$$p=mv,$$
and E (leaving out the potential energy)
$$E=\frac{1}{2} m v^2,$$
so 
$$\frac{dE}{dv}=mv$$
Taking potential energy into account. 
In this case yes. Both imply the other (momentum conservation and total energy conservation), as kinetic energy is transformed into internal energy (potential energy), like heat, chemical potential, etc.
A: For a single isolated particle (with constant mass) conservation of momentum does indeed imply conservation of kinetic energy (and also of mechanical energy, as if it is isolated there is no force applied to it and therefore there is no potential energy). This is easy to prove:
Let $\vec{p}=m\vec{v}$ be the momentum and $K=\frac{1}{2}mv^2$ the kinetic energy, then conservation of momentum reads:
$$\frac{d\vec{p}}{dt}=\vec{0}$$
expanding it
$$\frac{d\left(m\vec{v}\right)}{dt}=\vec{0}$$
$$\frac{dm}{dt}\vec{v}+m\frac{d\vec{v}}{dt}=\vec{0}$$
$$m\frac{d\vec{v}}{dt}=\vec{0}$$
and multiplying it by $\vec{v}$ we have
$$m\frac{d\vec{v}}{dt}\cdot\vec{v}=\vec{0}\cdot\vec{v}=0$$
$$\frac{d\left(\frac{1}{2}mv^2\right)}{dt}=0$$
$$\frac{dK}{dt}=0$$
$$\square$$
However, for a general isolated system of particles, this is not true in general. Say you have a completely inelastic collision(one where both particles end up stuck together) between to particles with masses $m_1$ and $m_2$ and velocities $\vec{v}_1$ and $\vec{v}_2$. 
Before the collision the total momentum is:
$$\vec{p} = \vec{p}_1 + \vec{p}_2 = m_1\vec{v}_1+m_2\vec{v}_2$$
and the total kinetic energy is
$$K=K_1 + K_2 = \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2$$
After the collision both particles move with the same velocity, and the conservation of momentum dictates that this should be such that:
$$\left(m_1+m_2\right)\vec{v} = m_1\vec{v}_1+m_2\vec{v}_2$$
$$\vec{v} = \frac{m_1\vec{v}_1+m_2\vec{v}_2}{m_1+m_2}$$
However, if we compute the kinetic energy after the collision we have
$$ K = \frac{1}{2}\left(m_1+m_2\right)v^2$$
which in general is not the same as before the collision, take for example $m_1=m_2=m$ and $\vec{v}_1=\vec{v}=-\vec{v}_2$ (that is, to equal particles moving with the same speed towards each other, which, after the collision remain still), and we have a constant momentum of $\vec{0}$ before and after the collision, however, the total kinetic energy goes from $K=mv^2$ to $0$.
Of course, inelastic collisions exist because energy dissipates in other forms of energy as acoustic, deformation, thermal... but the principle of conservation of momentum and energy talking about dynamics (that is kinetic and potential energy).
A: The total energy (internal + kinetic) of  hot body  isotropically emitting radiation while moving in the vacuum far from other bodies, will decrease, while momentum is conserved.
A: Mechanical energy is conserved in a system where only the conservative forces do work to the system..When you considered the vertical circular motion of an object tied with a string, the only forces acting on the object was the tension and gravity out of which tension does not do any work on the system as the motion is always perpendicular to the direction of tension and hence only gravity does work(which is a conservative force) and hence the total mechanical energy of the system is conserved...When you talk about conservation of momentum, no external force must be acting on the system.Here(vertical circular motion of an object) the gravitational force is an external force and it changes the momentum of the system. But in the case of collisions, we can say that momentum is conserved unless an external impulsive force acts on the system(Since collisions happen in an infinitesimally small time, non-impulsive force won't change the momentum of a system to an appreciable extent). So we can say that in a system, where the internal forces are non-conservative momentum will be conserved, but mechanical energy will not be conserved  IF THE INTERNAL NON CONSERVATIVE FORCE DO WORK ON THE SYSTEM. And in a non-rigid system, internal conservative forces do work on the system and hence energy will not be conserved but since there is an absence of external force momentum will be conserved! Now if you want an example of such a system, it's very simple-just think of a block sliding over a rough plank and the system is placed on smooth ground. The system is non-rigid and kinetic friction will do work on the system,and momentum will be conserved and energy will not be conserved in this case.:D
