In the diagram, an object is falling down a frictionless ramp. let us neglect air resistance

at point no. 2 in the diagram, let the object accelerates to a velocity v1 (just along the direction of the slope) due to rolling down the ramp. Now, v1 has a x-component as well as y-component.

Now, my question is:

In the diagram, In between the points no. 2 and 3, the object passes over a horzontal distance. During this part of motion, which statement of the following is true:

i) the magnitude of the velocity remains constant, i.e. equal to v1, and only the direction changes.

or, ii) the magnitude of the velocity changes, in such a way, that only the x-component of v1 remains still; whereas, the y-component of v1 completely vanishes, as soon as the object leaves the ramp and gets onto the horizontal trajectory(between the points marked 2 and 3).


Neither of the 2 answers is true. You ask about the velocity between the points 2 and 3.

Answer 1. is false because the direction of the velocity is horizontal and to the left at all points between 2 and 3.

Answer 2 is false because the magnitude of velocity is constant between 2 and 3.

I think what you are really concerned about is how the velocity changes at point 2. This cannot be decided without further information about the apparatus.

If there is a smooth change in the direction of the ramp at point 2 then the magnitude of velocity will be constant, only the direction changes (Answer 1). The velocity between 2 and 3 is $v_1$, the same as it was at the bottom of the incline.

If there is a sudden change in direction, then there is a collision between the object and the horizontal surface. If the object does not rebound and jump up after colliding with the horizontal surface at point 2, then we must assume that its vertical component of momentum is absorbed by the horizontal surface. Only the horizontal component of velocity remains, so the velocity between 2 and 3 is $v_1 \cos\theta$, where $\theta$ is the angle of inclination of the slope between 1 and 2. So in this case both the magnitude and direction of velocity change at 2 (Answer 2).

Judging by the gap I would be inclined to think there is a sudden change in direction, in which case Answer 2 is correct.

  • $\begingroup$ Thank you so so much, Sammy! It was such a fantastic explanation. But I had one question about a sentence you wrote, "If there is a smooth change in the direction of the ramp at point 2 then the magnitude of velocity will be constant, only the direction changes (Answer 1)." Here, although if there is a smooth slope, whether, the change in "magnitude" of the velocity is absolutely zero, or is the change so negligible, that we assume it to be zero? $\endgroup$ Dec 11 '16 at 15:04
  • $\begingroup$ If the ramp changes direction smoothly then the normal reaction force of the ramp on the object is always perpendicular to the object's velocity, because the direction of velocity is always along the tangent to the ramp. When force is perpendicular to velocity then there is a change of direction but no change in magnitude. This is the same as when a particle is moving in a circle. The change in magnitude is exactly zero; it is not a small but negligible amount. There is of course an increase in the magnitude due to gravity, but that is not caused by the change in the slope of the ramp. $\endgroup$ Dec 11 '16 at 17:46

It is answear 1: After the acceleration due to the vertical component of motion the magnitude of the velocity stays constant. You can try it out by bicycling down a hill. The energy has to be conserved, so if we have no friction or deformation of the material to absorb or transform it, it has nowhere to go. If the angle is too sharp or the velocity too high the ball might also bounce back (for example if it hits the ground with a 90° angle) but as long as the kinetic energy is not absorbed or converted into potential energy on a way back up it stays constant.


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