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

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?