Whats the difference between an object going down a dip and going up a hump, both of the same size? So this is a question from a past physics Olympiad paper. If a mass goes into a shallow dip or over a hump, both of the same size, how does the time to travel compare? 
My thought process was that since both paths have the same height, they lose and gain the same amount of gravitational potential energy during the dip/hump and so they have the same final velocity and hence also time.
The correct answer was A, shorter time via the dip. Can someone explain why this is?

 A: At any distance along the track, the speed of the body in the dip is greater than (or, on the flat, equal to) the speed of the body going over the bump, as its potential energy is less and its total energy is the same. So the  speed is faster everywhere and the time is less.
Same question as: if you throw a ball into the air and catch it, does it take longer - including the effect of air resistance - on the way up or down?
A: It would be easier to imagine the hill and dip as two straight lines. One going up and the other going down. The initial velocity is the same for both, and the velocity after the hill/dip is the same because of conservation of energy. Therefore, the velocity must be different when the ball rolls up the hill, or through the dip. 
When the ball rolls through the dip, it loses potential energy and gains kinetic energy. Therefore, the velocity at the bottom is greater than the initial velocity. Let that velocity be $v_1$.
When the ball rolls over the hill, it gains potential energy and loses kinetic energy. Therefore the velocity at the top is less than the initial velocity. Let that velocity be $v_2$.  Now, $$v_1>v_0>v_2$$
Because the distance for both is the same, we can use the formula $vt=d$ to solve for time. $$\frac dv=t$$
$$ \frac {d}{v_1}=t_1 \space\space \frac{d}{v_2}=t_2$$$$t_1<t_2$$
The actual velocity is given by $$PE+KE=PE_0+KE_0 \Rightarrow PE-PE_0+KE-KE_0 = 0 \Rightarrow mgh-mgh_0+\frac 1 2 mv^2-\frac 12 mv_0^2$$
