Both pucks start at the same speed and time on a frictionless track. One goes down and up a large dip and then returns to the same starting height. Which one arrives to a finish line first?
Break it down to vertical and horizontal components and you will see that they arrive at the same time
If you break down the components, the puck going down the dip will have an increased horizontal velocity while travelling at the bottom of the dip (because of potential energy being converted to kinetic energy). When the puck returns to the same starting height, it will have the same starting velocity. So the puck going down the dip will reach the end faster, because it travelled at an increased velocity for a given time.
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This is wrong, the change in height only translates to energy gain in the vertical direction, so both pucks arrive at the same time assuming frictionless
How do you mean by “energy gain in the vertical direction” ? Energy gained is not dependent on the direction- energy gained is energy gained.
So, if the puck slides to a lower surface than what it was originally moving on, and starts travelling parallel to the original surface, are you saying there is no gain in speed?
If yes, what happened to the difference in potential energy at the two different heights?
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I looked into this more and your actually correct and I’m wrong.
The puck that dips will never go below the starting horizontal velocity, but will have an increase from the dip.
The only case where they arrive at the same time would be if the puck was just dropped (no smooth curve/dip)
To clarify during the interview I said the dip one would arrive first and then the interviewer told me to think more, then told me the correct answer was they would arrive at the same time lol 
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Thanks for correcting me, I should reach out to the company/manager and tell them to brush up on their technical skills lmao
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Breaking down the question, let’s separate the pucks as two distinct scenarios:
Puck 1
- Travels horizontally along a frictionless surface with no change in potential energy (PE) or kinetic energy (KE)
- Initial velocity will be the same as final velocity (no friction)
v_i = v_f
Let’s assume puck 1 initial velocity is equal to 2 m/s.
Puck 2
- Travels horizontally along frictionless surface, (1)
v_i = 2 m/s
- Travels down the ramp, PE converted into KE (2)
PE_0 + KE_0 = PE_1 + KE_1
mgh + 1/2mv_i^2 = 0 + 1/2mv_1^2
v_1 = \sqrt{2(gh + 1/2v_i^2)}
Let’s plug in the following numbers as h = 1 m, g = 9.81 m/s^2, and we’ll find that v1 is equal to 4.86 m/s .
- Travels back up the ramp, experiences negative PE and returns to same starting velocity (3)
v_f = v_i = 2 m/s
Because puck 2 experiences increased horizontal velocity while traveling down the ramp, it will reach the end of the track first.
This video here provides a great visual explanation, and how the increased horizontal velocity from the downward dip (potential energy → kinetic energy) translates into the puck with the dip reaching the end of the track first.
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