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enter image description here

In the above image let us assume that block slides along the plane of the wedge and the wedge moves towards left (on a smooth surface).

I know that the velocity of the block can be found by simply using conservation of energy and conservation of momentum (in the horizontal direction)

My question is as follows:

When the block slides to the bottom most point it's velocity in the horizontal direction is the horizontal component of it's velocity minus the velocity of the wedge. This results in a decrease of Kinectic Energy (With respect to the case of a fixed wedge). I know that this energy went to the kinetic energy of the wedge but I am unable to understand how. I know that while conserving mechanical energy of closed systems of two or more blocks there is usually an action-reaction pair of forces which decreases one objects energy while increasing the others. Which forces are acting in the horizontal direction in this case?

(Edit: New Question) When writing the equations for conservation of energy at the bottom most point (in order to find the velocity of the wedge) why do we write the horizontal component of the velocity of the small block as v cos(a)-V ? v:velocity of block with respect to wedge a:Angle of inclination of the wedge V:velocity of the wedge.

I know that the block is on the wedge but can’t understand what is it about the interaction between the wedge and the block which requires us to account for the velocity of the incline while calculating the horizontal velocity of the block. An analogy of a person sitting in a car can be used to better understand my question. A person sitting in the car at any point has the velocity as well as the acceleration of the car due to the normal reaction between the person and the car. What is the ‘Normal Reaction’ in this case. (Note:The wedge is frictionless)

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  • $\begingroup$ Kinetic energy is not decreasing.In fact it has increased. $\endgroup$ – RunMachine_Kohli Sep 8 '19 at 15:20
  • $\begingroup$ @Unique Can You please elaborate..(Edit:The question has been changed) $\endgroup$ – Aditya Ahuja Sep 8 '19 at 15:47
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The block & wedge system has a reduced amount of gravitational potential energy whilst a corresponding increase in kinetic energy.
The gravitational force $mg$ does work on the block & wedge system as the block moves downwards.

enter image description here

The horizontal component (green) of the force on the wedge due to the block $N’$ (red) does work on the wedge and increases the kinetic energy of the wedge. The vertical component of the force on the block due to the wedge $N$ does negative work on the block but the horizontal component of force $N$ does positive work on the block.

The net effect is that the work done by the gravitational force on the block increases the kinetic energy of the block and the wedge.


The vector acceleration (and velocity and displacement if the system starts from rest) diagram looks like this with $\theta$ being the angle of the wedge.

enter image description here

In the diagram $\vec a_{\rm bw}$ is the acceleration of the block relative to the wedge, $\vec a_{\rm bg}$ is the acceleration of the block relative to the ground and $\vec a_{\rm wg}$ is the acceleration of the wedge relative to the ground, such that $\vec a_{\rm bw}+ \vec a_{\rm wg} = \vec a_{\rm bg}$.

The direction of $\vec a_{\rm bg}$ is the direction of motion of the block relative to the ground.

If the wedge is much more massive than the block then $\vec a_{\rm bg}\approx \vec a_{\rm bw}$ as $\vec a_{\rm wg}\approx 0$ ie the wedge does not move relative to the ground and the block slides down the slope with an acceleration of $g\sin \theta$

If the wedge is much less massive than the block then $\vec a_{\rm bg}\approx \vec g$ ie the block falls down almost vertically and the block gains hardly any horizontal component of velocity when compared with the gain in the vertical component of velocity.

So as the direction of the acceleration of the block relative to the ground, $v_{\rm bg}$, becomes closer to the vertical there is less acceleration of the block in the horizontal direction as compared with the acceleration in the vertical direction.

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  • $\begingroup$ Okay so I get what you’re saying, but what I want to know is why the block has less horizontal velocity when the wedge is moving as compared to the horizontal velocity in case of a fixed wedge. $\endgroup$ – Aditya Ahuja Sep 9 '19 at 3:46
  • $\begingroup$ @AdityaAhuja I have added to my answer in terms of accelerations. $\endgroup$ – Farcher Sep 9 '19 at 10:50
  • $\begingroup$ Can you also explain this in terms of reduced kinetic energy of the block.The block pushes the wedge (not a massive wedge) and gives it some of its kinetic energy.The wedge in turn must do some negative work on the block and reduce its kinetic energy .Which force will do this negative work? $\endgroup$ – Aditya Ahuja Sep 9 '19 at 10:57
  • $\begingroup$ In your answer you have mentioned that the vertical component of the normal does negative work. This should have reduced the vertical velocity of the block but instead the horizontal component decreases (vcos(a)-V).How is this possible ? $\endgroup$ – Aditya Ahuja Sep 9 '19 at 11:07
  • $\begingroup$ @AdityaAhuja If the acceleration is smaller, the gain in velocity is smaller and so the gain in kinetic energy is smaller. $\endgroup$ – Farcher Sep 9 '19 at 11:09

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