Speed of an object placed on a wave If I place say a small paper on top of a water wave it moves slower than the wave speed. Can anyone explain why that is the case?
 A: Linear water waves in general transport energy, not mass. Depending on the form of the wave, the paper will take on different orbits if we think of it as being a passive tracer. For instance, for linear waves the paper's motion is circular in deep water, elliptical in intermediate water and horizontal in shallow water. We now make this more rigorous. 
Recall, for irrotational inviscid 2 dimensional water waves the governing equation is
$$\nabla^2 \phi = 0$$
where $\nabla^2 \equiv \partial_{xx}+\partial_{zz}$ and $\phi$ is the velocity potential. We require particles on the free surface to remain there, and for the pressure to be continuous across the air-water interface, which gives us (assuming the atmospheric pressure to be 0) our surface boundary conditions 
$$\left.\eta_t+\phi_x\eta_x = \phi_z\right|_{z=\eta}; \quad \quad \left. \phi_t+\frac{1}{2}(\nabla \phi)^2 +gz=0\right|_{z=\eta},$$
where $\eta(x,t)$ is the free surface displacement. Finally, we require the bottom to be impermeable, hence
$$\phi_z = 0 \quad @ z= -h$$
for $h$ the depth of the water. These are a closed set of equations. Note, the governing equation, ie Laplace's equation, is linear and hence simple to solve. However, the free surface conditions are nonlinear, and more significantly, evaluated at a moving interface. This last condition makes these equations very difficult to solve. 
Therefore, to make progress, we generally take asymptotic expansions of these equations. Here, we'll just linearize the equations about some small parameter $\epsilon = ak$ for characteristic wave amplitude $a$ and wavenumber $k$. We will also consider periodic monochromatic waves, so that one can show that the following $\eta,\phi$ satisfy the governing equations to first order in $\epsilon$: 
$$\eta = a \cos \theta; \quad \phi = a\omega \sin \theta \frac{\sinh k(z+h)}{\sinh kh},$$
where $\omega$ is the angular frequency given by the dispersion relationship
$$\omega = \sqrt{gk \tanh kh},$$
and $\theta\equiv kx-\omega t$. 
To answer your question, we would like to know the behavior of a particle with position $(\xi+x_o,\zeta+z_o)$ originally located at some position $(x_o,z_o)$.This is the Lagrangian description of the fluid. Recall, we can relate the Eulerian and Lagrangian descriptions of a fluid through the velocity field, that is 
$$\phi_x = \frac{d \xi}{d t}; \quad \phi_z = \frac{d\zeta}{dt}.$$
Substituting in the relations for $\phi$ found above, and then integrating in time, we find 
$$\xi =-a\frac{\cosh k(z_o+h)}{\sinh kh} \sin \theta_o; \quad \quad \zeta = a\frac{\sinh k(z_o+h)}{\sinh kh} \cos \theta_o,$$
where $\theta_o =kx_o - \omega t$. We note that the particle displacements from the origin scales with the wave amplitude $a$. 
Now, we can eliminate $\theta_o$ by squaring the expressions for the particle position to find 
$$\frac{\xi^2}{\left(a\frac{\cosh k(z_o+h)}{\sinh kh} \right)^2 } +\frac{\zeta^2}{\left(a\frac{\sinh k(z_o+h)}{\sinh kh} \right)^2 }= 1$$
which describes different motion dending on the value of $kh$. For deep water waves $kh\gg 1$ and we have an an ellipse, while for shallow water waves $kh\ll1$ and we have purely horizontal motion. I've attached a figure from Kundu and Cohen which illustrates this. 
So we see the particles don't translate (actually the orbits are not closed, which you can see by examining the transport at second order, which is called Stokes drift). Therefore, mass is not transported, while the energy is.
Now, the story can be very different for nonlinear and breaking waves, but this is a different story for a different day. 
I hope this helps. Feel free to ask any other questions,
Nick
