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I was watching a video lecture on the topic "Center of Mass and Conservation of Linear Momentum", and this question came up there. You have to find:

  1. The maximum deformation of the spring

  2. The maximum deceleration of the wedge (the wedge has a mass $M$)

  3. Just before the mass $m$ touches the spring find the velocities of $m$ and $M$

What I thought is that if the block of mass "m" and the earth are considered as one system, I can apply conservation of mechanical energy, since the interaction forces between the two entities are conservative. Also, the work done by any external force on the system is zero, since the Normal force on the block by the wedge is always perpendicular to the block's displacement. Thus, the block's velocity can be simply found out by equating the loss in gravitational potential energy to the gain in kinetic energy.

But, in the video, it is explained that since the wedge accelerates to the left once the block starts falling down, the reference frame is non inertial and you can't apply conservation of mechanical energy.

I have trouble understanding what exactly it is that I am doing wrong, and about the implications of the frame of the wedge being non-inertial in this question. Why does it matter if the wedge is accelerating or not; like I said earlier the reaction force of the wedge on the block will always be perpendicular to the block's displacement. Hence work done is zero. Any help would be greatly appreciated.

(Also, I just need help in understanding the questions I have asked above, I just posted the question to provide context.)

EDIT- I asked this to my physics teacher and got the concept clarified. Thank you to those who have answered my question.

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  • $\begingroup$ If there is no net force acting on the acting on the wedge, then it must remain at rest according to Newton's first law, so what force is being applied to it? $\endgroup$ Dec 12, 2023 at 19:55
  • $\begingroup$ @Albertus Magnus It is written in the picture I have attached that friction is absent everywhere, so the ground that the wedge rests on is frictionless. And thus as the block slides down the wedge slides backwards. $\endgroup$ Dec 13, 2023 at 4:02
  • $\begingroup$ Can you provide a link to the video,the provided question is not completely clear as such. $\endgroup$
    – Blz
    Dec 26, 2023 at 6:03

2 Answers 2

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You say you want to choose $m$ as your system. Suppose you work in the non-inertial frame in which the wedge is at rest (left panel of the figure above). The normal force $\mathbf N$ on $m$ is perpendicular to the surface of the wedge, and its velocity $\mathbf v'$ is parallel to the surface, so $\mathbf N \perp \mathbf v'$, hence $\mathbf N$ does no work. However, there is a horizontal inertial (fictitious) force $\mathbf f$ since this is a non-inertial frame. $\mathbf f$ does do work, so energy is not conserved: you need to use the work-energy theorem to account for the work done by $\mathbf f$.

Suppose you want to work in the inertial "ground" frame instead (right panel). To get the correct velocities, we translate all the velocities in the non-inertial frame by $\mathbf V$, the leftward velocity of the wedge. You can see that $\mathbf v = \mathbf v' + \mathbf V \not\perp \mathbf N$ in this frame, so $\mathbf N$ does (negative) work on $m$. Therefore, energy is still not conserved, even though there is no inertial force here.

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Never use non-inertial frames to apply Newton's Laws.

Part 1.

For the system m, M, + spring the only external forces acting are the reaction from the ground and weight. These act in the vertical direction so horizontal momentum is conserved.

$$0=mv+MV$$

The max deformation of the spring is when $v=V$, and the only possibility allowed by conservation of momentum is $v=V=0$

Conservation of energy is then

$$ mgh= \frac{1}{2} k e^2$$

Part 2

As the m slides down M speeds up. M only slows down once $m$ hits the spring. The maximum deceleration of M will occur when the spring force is maximum, this is when the extension is $e$ given in part 1.

M acceleration is then

$$MA=ke$$

Part 3

$$\frac{1}{2}mv^2+\frac{1}{2}MV^2=mgh$$ $$mv=-MV$$ solve

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