conservation of energy vs conservation of momentum different results

Question

A pendulum of length $1$ m and mass $100$ g attached to the end. Another 100 g mass move horizontally with speed 2 m/s. When collision happens this ball sticks with the pendulum and move together. Find the initial linear speed of the block.

conservation of energy :-

E(initially) = .5*m*v*v = .5*.1*2*2   = .2 joule
E(initially) = E(final) = .5*(.2)*v*v = .2 joule

v=sqrt(2)

conservation of momentum :-

m1 v1 + m2 v1 = m1 v2 + m2 v2
.1*0 + .1*2 = .1 (2v)`
v=1 m/s 

main question: why they are different 1 m/s and sqrt(2)

Answer 1

The before-energy is the kinetic energy of the second mass.

In such an inelastic collision, heat is produced. Hence the after-energy is the kinetic of the combined mass - plus some heat energy.

You may use the correct result from conservation of impulse to infer how much heat was involved.

Answer 2

As Hagen says, you need to remember that there are forms of energy other than mechanical. The total energy of a closed system will be conserved, the mechanical energy alone might not be. Another important difference between momentum and energy is that momentum is a vector but energy is a scalar.

Answer 3

$m_1=m_2=100\text{ g}$, $\ell=1\text{ m}$, $v_{1i}=0\,\mathrm{ms^{-1}}$, $v_{2i}=2\,\mathrm{ms^{-1}}$

Conservation of momentum $$ m_1v_{1i}+m_2v_{2i}=(m_1+m_2)v_{f}\qquad\Longrightarrow\qquad v_f=\frac{m_1v_{1i}+m_2v_{2i}}{m_1+m_2}=1\,\mathrm{ms^{-1}} $$

Conservation of the total energy $$ K_i+U_i=K_f+U_f\quad\Longrightarrow\quad \frac{1}{2}m_2v_{2i}^2=\frac{1}{2}(m_1+m_2)v_{f}^2+(m_1+m_2)gh\quad\Longrightarrow\quad h\approx 5.1\text{ cm} $$