homework and exercises - When to use Conservation of Energy vs Conservation of Momentum

I am confused as to when each should be used to solve problems.I am not looking for the answer to this question, but I will include it here so you have all the information about what I am asking about.

A 15.0 kg block is attached to a very light horizontal spring of force constant 400.0 N/m and is resting on a frictionless horizontal table. (See (Figure 1) .) Suddenly it is struck by a 3.00 kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.

I see that the idea suggested here (at Yahoo Answers) is to use conservation of momentum to solve this problem. But I wanted to solve using conservation of energy in the following way:

$K.E =\frac{mv^2}{2}$

$Spring-P.E=\frac{kx^2}{2}$

$KE_i = KE_f + PE_f$

$KE_i - KE_f = PE_f$

Solving the equations, you get

$x = 0.670m$

The correct answer given is 0.387m.

Answer

When to use which law

Your assumption that conservation of energy (considering only kinetic energy) works while dealing with the collision in the above question is not correct.

Conservation of energy (kinetic energy) doesn't appear to work in all kinds of collision. Some of the initial kinetic energy of the bodies are lost as heat and/or part of it is stored in the form of potential energy of the bodies (deformed body). These kind of collisions are called inelastic collisions.

Hence, direct application of conservation of energy with just kinetic energy terms is not possible. In these cases, the problem cannot be solved with just conservation of momentum. You need some experimental input (usually the coefficient of restitution is given).

However, there are cases where conservation of energy (initial kinetic energy = final kinetic energy) is applicable. Such collisions are called elastic collisions.

Conservation of momentum is always valid and safe whereas conservation of energy requires all forms of energy including heat, sound, light, etc to be considered (which ever stated)

Solution to the given problem

In the above problem, the final velocity of the block is already given. By using the conservation of momentum, you can calculate the final velocity of the block.

Initial Momentum of the stone= $mv$ = $24$ $kg.ms^{-2}$

Applying the law of conservation of momentum,

$24 = m_{stone}v_{final-stone} + m_{block}v_{block-final}$

Substituting the given values in the equation, you get the final velocity of the block to be $2 ms^{-2}$.

Before the collision, the block was at its mean position. After the collision the block will begin to oscillate with the same mean position.

The total energy of the system is equal to the kinetic energy supplied by the moving stone. When the block reaches its extreme position, all the energy will exist in the form of potential energy of the spring. Therefore, by applying the law of conservation of energy,

$\frac{mv^2}{2} = \frac{kx^2}{2}$

You get x to be $\sqrt{\frac{3}{20}}$