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enter image description here (Image updated for clearance)

Hello,

Assuming that there is no friction anywhere and both box 1 and ramp 2 start at rest, I was wondering why this is an isolated system in terms of momentum calculation. My professor approached calculating the velocities of the box and the ramp assuming that the momentum is conserved before and after the release of box 1. However, I don't understand how the system can be treated as if there is no net force acting on the system. If we draw the two axes horizontal and vertical to the surface of ramp 2, then the vertical component of gravity is canceled out by the normal force but the horizontal component of gravity remains unhindered and this is precisely why box 1 will be moving at all. I'm having a hard time understanding how the conservation of momentum approach can be applied here.

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I am sorry but I have had to write this as an answwer as it is too long as a comment.

There is no external horizontal force acting on the system comprising the two blocks so for the system of two blocks the total horizontal momentum must stay constant and it is zero if the blocks start from rest.

There is no horizontal component of the weights of the blocks as those forces act vertically downwards and so the weights of the two blocks cannot influence the horizontal momentum of the system.

Block 2 has the normal force (yellow) and the vertical force (red) mg acting on it and block 1 has a normal forces equal in magnitude and opposite in direction to the yellow force on block 2, a vertical weight force acting on it and a vertical (there is no friction) upward force due to the surface on which it rests.

This arrangement is discussed in this answer.

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