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there is some confusion to me in the case of "motion of block on a frictionless wedge"

Below is a simple diagram!

enter image description here

Let us consider a situation as above in which there is a block of mass $m$ moving with velocity $v$ in positive $x$ direction on a frictionless wedge as given above.

Now since block doesn't have any velocity in $y$ direction initially so it moves horizontally as long as it is on table But when at the instant it encounters the wedge it it gains some some velocity in upward direction and starts to move upward (i.e in positive $y$ direction) with velocity decreasing with time. it approaches some height and after that comes back down the wedge.

  1. What gives the block that upward velocity (i.e which force(s) changes the velocity in $y$ direction.) because of which it goes to some height and after that comes back?

  2. What is the velocity of Block (net velocity which is along wedge). At the Instant it just starts to slide on the wedge?

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When the block collides with the wedge, there will be a normal impulsive force from the wedge, in a direction perpendicular to the surface of the wedge.

If you notice, this normal impulsive force has a component in the vertical direction. This is what provides the block with the vertical velocity.

As for your second question, the magnitude of the velocity will not change. This is because work has not been done by any of the forces(Normal force and gravity) during the time the block starts sliding on the wedge.

The normal force doesn't do work because its direction is perpendicular to the displacement.
Gravity won't do work because there is no sufficient displacement in the vertical direction, in the small time it takes to just start sliding on the wedge.

So the speed of the block at the base of the wedge will be the same as it was on the horizontal plane. Of course, it will decrease as the block goes up on the wedge.

Note: this is assuming the block is particle like. If it has finite dimensions, there will be rotational motion when it reaches the wedge.

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After the block enters onto the inclined surface of the wedge, the initial velocity of the block can be resolved into a sliding velocity, S and, perpendicular to that, a velocity P. The initial velocity of the wedge is zero. As the inclination of the wedge surface increases, the initial sliding velocity, S must diminish. If we let that angle approach 90 degrees, it is clear that the block will not even slide up the wedge, because it is an impact. This would be like the collision of two hockey pucks. The diagram shows the initial sliding velocity, S. This answers your second question. The vertical component of S is also shown. This answers your first question. The velocity P is purely impact. enter image description here

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  • $\begingroup$ Welcome to Physics.SE! I suggest the following: 1) Take the tour! 2) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. 3) If you have a good question, ask it! Just search for duplicates and follow the help center rules. $\endgroup$ – heather Aug 3 '16 at 19:16
  • $\begingroup$ The second question asked for the velocity at the instant it just slides onto the wedge. An earlier answer stated "As for your second question, the magnitude of the velocity will not change. " That answer is incorrect, because the sliding velocity must diminish as the inclination of the wedge surface increases. When the inclination reaches 90 degrees, the block does not even slide onto the wedge. $\endgroup$ – Dennis Stevens Aug 15 '16 at 18:49

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