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I had a quick question regarding this homework problem, I can't seem to understand why they chose to use the conservation of momentum in vertical direction and not the horizontal direction as you would get a different end result if you chose the horizontal direction to work with. Would anyone kindly explain this to me ? Is this because the question states that there can be no tangential force on the plate surface so that means horizontal forces and thus working in the horizontal direction is neglected ?? Thank you very much in advance!

Problem + Solution

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  • $\begingroup$ Could you elaborate what you mean by "as you would get a different end result if you chose the horizontal direction to work with"? $\endgroup$
    – NMech
    Jun 29 at 16:58
  • $\begingroup$ if i chose to work with horizontal direction then, h2/h = -1/(2*sin(theta)) +1/2 $\endgroup$
    – jjrrrmkaas
    Jun 29 at 17:08
  • $\begingroup$ I am not sure how you obtained from y-momentum $$\frac{h_2}{h}= -\frac{1}{2\sin\theta}+\frac{1}{2}$$ However, from what I understand the x-y Coordinate system is selected because the x direction is parallel to the frictionless surface. Therefore the $R_x$ force is equal to zero. However, the $R_y$ is not equal to zero, its a non zero force, which is not possible to calculate without the x-momentum coordinate. $\endgroup$
    – NMech
    Jun 29 at 18:21
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The water hits the surface and the moving directions of the water particles get changed. Their speeds are assumed to stay.

Let's follow what happens when a particle with mass M hits the surface and splits to pieces. Piece 1 with mass M1 continues to the +X direction in the solution drawing and piece 2 which has the rest of the mass M2 = (M-M1) turns to the negative X direction.

Piece 1 has momentum V * M1 and it's directed to positive X.

Piece 2 has momentum -V(M - M1) if we calculate it to the +X direction.

The total momentum of the incoming particle is not conserved. The Y direction momentum is used to make the normal force to the surface. The sum of the X-direction momentums of the particles stay because the plate causes no force to the particles in X-direction (=frictionless)

The X-direction momentum of the incoming particle is V * M *sinθ

The X-direction conservation equation:

V *M * sinθ = V * M1 -V(M - M1)

V can be dropped off. After reordering we get M1 = M(1 + sinθ)/2 or equivalently

(M1/M) = (1 + sinθ)/2

If we assume the water is non-compressible the jet thicknesses must have the same ratio and that's the same result as presented in the original solution.

Your "working in the horizontal direction -idea is a misconception". The conservation of the momentum is only true for the "along the surface component". The rest of the momentum is lost because the surface stops the motion in the surface normal direction.

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