Is momentum along the line of collision conserved when a ball falls on an inclined plane

Question

A ball of mass 1kg falling vertically with a velocity2m/s strikes a wedge of mass 2kg. Wedge lies a smooth horizontal surface and the coefficient of resitution between the ball and the wedge is 1/2. Find the velocity of the wedge and the ball immediately after collision.

I obtained the correct equation involving the coefficient of restitution. However, for the second equation involving momentum, I obtained an incorrect equation.

Method 1

Momentum along normal to wedge is conserved.

Initial momentum = Final momentum

$(-2\cos 30)(1) = (V_y)(1) + (-V_w \sin 30)(2)$

$V_w=V_y+\sqrt{3}$

Method 2

Impulse of ball, which is along normal, = $J = V_y - (-2\cos 30) = V_y + \sqrt{3}$

Impulse of wedge, which is along horizontal $=(V_w-0)(2)=2V_w$

Horizontal component of J which acts to the right = Impulse of wedge which acts to the left

$J \sin 30 =2V_w $

$4V_w= V_y+\sqrt{3}$

Why do the equations from the 2 methods differ? More specifically, why is Method 1 is incorrect and Method 2 is correct. Isn't impulse essentially a change in momentum and since duration of collision is $0$, shouldn't the final and initial momentum should be same?

Answer 1

The ground provides a force that has a non-zero component perpendicular to the direction of the incline. Therefore, momentum is not conserved in the direction perpendicular to the incline.

Answer 2

Consider the ball & wedge system.
For that system no horizontal external forces act and so the linear momentum of the system in the horizontal direction must be conserved.
This will immediately give you a relationship between the horizontal components of the velocities of the ball and the wedge immediately after collision.