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Connected Particles on an inclined plane

image.jpg This system is initially held at rest. Once released, which way would the connected particles move and why? I answered the question thinking that it would fall down the plane (as A is heavier) but I got a negative acceleration (right answer wrong sign). The mark scheme implies that motion is up the plane. Why?
Original post by bubblegum21
image.jpg This system is initially held at rest. Once released, which way would the connected particles move and why? I answered the question thinking that it would fall down the plane (as A is heavier) but I got a negative acceleration (right answer wrong sign). The mark scheme implies that motion is up the plane. Why?


Component of weight of A down plane is 4g so for A to move down the plane we would require 4g>T but for B to move upwards we would need T>6g But this is impossible so A must move up the plane.
But it doesn't matter which direction you assume as the sign of your answer will tell you whether you arec right or wrong.
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Original post by bubblegum21
image.jpg This system is initially held at rest. Once released, which way would the connected particles move and why? I answered the question thinking that it would fall down the plane (as A is heavier) but I got a negative acceleration (right answer wrong sign). The mark scheme implies that motion is up the plane. Why?


Well, if you got the negative sign then the motion would be opposite your positive direction, which means it will move upwards. Maths is never wrong :smile:

The reason for this upwards motion is that the component of the weight along the plane (which is responsible for pulling the whole system 'downwards to the left' ) is smaller than the weight of the ball (which is responsible for pulling the whole system 'downwards to the right' ). If we increase the slope of this plane, eventually the component of weight along the plane will be larger and the system will fall down.

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