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So you have two tracks of different inclines meeting at a point. Two stones are released from this point each one along the direction of one incline, from rest. Which stone reaches the ground faster? The one on the steeper incline? I don't understand why. By the kinematic equations of motion displacement in the y direction is equal to -gt^2 for both the motions, and since they are released from the same height, they should reach at the same time. I feel like i'm missing something very important...

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  • $\begingroup$ I have used the fact that the s=ut+1/2(at^2) is a vector equation with three independent equations satisfying the x, y and z directions respectively. I also wish to know how we can account for the horizontal velocity of the ball when it has no acceleration in the horizontal direction and starts with initial velocity 0? $\endgroup$ – kimi Jan 14 at 16:31
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You are forgetting the fact that here normal reaction acting from the track changes the direction of motion the object and only a smaller component of acceleration due to gravity is acting along the direction of motion ($\leq g$).

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  • $\begingroup$ Yes, thank you, this is what I was forgetting! $\endgroup$ – kimi Jan 15 at 3:34
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With or without friction, the rock on the less steep track has a lower acceleration because its y-acceleration is mg(sin(theta) and this theta is smaller than that of the steeper slope. If these tracks have friction, then the track with the smaller angle will allow for more non-conservative work to be done on the rock and in the end, its final velocity will be lower than the rock traveling down the steeper slope, making it take even longer to make it all the way down the track.

Hope this helps, -W

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  • $\begingroup$ The friction part of your answer was pretty helpful too, thank you. $\endgroup$ – kimi Jan 15 at 3:40
  • $\begingroup$ glad I could help $\endgroup$ – William Cooper Jan 15 at 18:52

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