we use conservation of momentum when momentum is transferred by one thing to another to conserve it. first object loses its momentum while another gets its momentum so finally momentum is conserved.in your example as bullet strikes to wooden block,it loses its momentum but now wooden block gets its momentum so finally as much momentum was lost by bullet, same amount momentum is gained by wooden block. even its also of two types-linear and angular momentum of conservation.

some another examples will help you to understand better-

linear momentum conservation-

-When atoms collide with each other the law of conservation of momentum is followed.

-When the bullet hits the wooden block the law of conservation of momentum is followed and the block rises to a particular height.

-When two cars hit each other in an accident then the law of conservation of momentum is followed.

-When a baseball is hit by the bat the law of conservation of momentum is followed.

-When two runners collide in a race then the law is followed.

-Two billiard balls colliding with each other.

angular momentum conservation-

-During skating if the skater folds his or her hands then he is able to rotate at a greater speed as compared to when he opens his hands wide. This is on account of conservation of angular momentum.

now about energy conservation,in every field and everywhere total energy is going to be conserved as its fundamental law of nature. in your case kinetic energy of bullet is going to be converted in kinetic energy of wooden block and some heat energy also due to friction between bullet and wooden block, so finally total energy is conserved.

It goes is necessary to depend on the problem has analyzed, in case of the force of the flow in an elbow: it is necessary to to apply both the conservation of the energy and the conservation of the quantity of movement. As for the difference of result in both he is that the variations with regard to the time so much of the terms of energy as the quantity of movement are despised in general.

Both conservation of energy and conservation of momentum always apply, but are not always equally convenient to use. In the case of the bullet and block, heat is generated by the collision. The amount of heat may be sufficient to melt the bullet, and/or develop burn marks on the bullet hole. In order to apply conservation of energy, the heat energy must be accounted for. That is very inconvenient. It is hard work to obtain all of the parameters needed to calculate the heat energy directly. Conservation of momentum is an easier route to solve the problem in such an "inelastic collision", i.e., any collision in which energy is transferred to heat. When the final state of motion is calculated by conservation of momentum, in this problem, one can add up the energies. Subtraction of the total final energy from the total initial energy will then provide you with the amount of energy lost, in a much easier way..

In theory, momentum is never lost, and never converted to anything else but more momentum. Energy, also, is never lost, but can be converted to many difficult-to-quantify forms. I would say this is the reason why conservation of momentum is almost always less problematic than conservation of energy for describing mechanical situations. I can't think of a contrary example, but would love to hear one!

My prescription for students: I don't think I would ever use conservation of energy for collisions (and any other situation involving non-conservative forces that can't be neglected).

My prescription for first-year students:

Look at the FORCES in the problem:

No EXTERNAL forces - momentum conservation.

No NON-CONSERVATIVE forces - energy conservation.

(By non-conservative forces I mean forces that cannot be derived from a potential).

An additional prescription (a more basic one, but completely equivalent to the first):

Symmetry in continuous spatial translations - momentum conservation.

Symmetry in continuous temporal translations - energy conservation.

(This is a consequence of the Noether theorem).

This problem can you see in case transformation energy study. Thanks.

The work-energy theorem typically used in mechanics courses does not include any heat term. "Conservation of energy" in this context should then be read "conservation of mechanical energy within the considered mechanical system". In your example with the log+bullet system, *mechanical* energy is not conserved, so conservation of momentum or angular momentum should be considered.

To explain this to the students, I usually throw something (not a student) into the wall creating a load bang, and explain that any impact creating noise, mechanical waves that eventually dissipate into heat removes mechanical energy from the system, inferring that mechanical energy is not conserved.

Hello,

Is this question still active? The proposed solution preference and comments are incorrect. Conservation of energy can be applied without a problem between the two end states.

In general, and answering the question,  conservation of momentum is used whenever vectors must be used to describe a dynamic problem. Conservation of mass and conservation of energy apply to scalar quantities  (in classical mechanics) and may not provide a full description when velocity  must be used as a vector quantity. Note that the principle of virtual work can be applied in static situations also when vectors must be used, and provide the correct solution simply from a conservation of energy viewpoint.

Cheers,  Ed Gerck 

I would like to put the question/problem  slight differently.

Say there is Object mass m1 moving with velocity v1

Momentum1=m1v1

kinetic energy1=0.5*m1*sqr(v1)

and there is another Object mass m2 moving with velocity v2

Momentum2=m2v2

kinetic energy2=0.5*m2*sqr(v2)

Assume object m1 is space station 1 & object m2 is space station 2 being colinear

If m1 is to be attached to m2, then v1=v2=v

Combined momentum= (m1+m2)*v

Combined Energy=0.5(m1+m2)*sqr(v)

Here both momentum & energy are conserved.

Thanks..