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Assumptions: No friction , $m_2$=5kg and $z_1$=2kg. $z_2$ is fixed at its place.

Here, there is a mass of $5kg$ on a movable wedge of mass $2kg$. So, my question is that can we say that acceleration of the mass $5kg$ along with the x-axis = acceleration of $2kg$ wedge along the x-axis? Also, the net acceleration of $5kg$ is in the southwest direction but its acceleration along the x-axis is what I ask about.

  • My points as to why they are equal :

If the accelerations become unequal, then the mass and wedge would lose contact.

  • My point as to why they are not equal :

We can say force applied on $2kg$ mass = force applied on $5kg$ mass. So, since their masses are different, acceleration will be no different.

I wish to know what should the real solution and pls correct me if there is any mistake in my solution.

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  • $\begingroup$ Is this the initial position? What is the full text of the question? $\endgroup$
    – nasu
    Commented May 5, 2021 at 11:34
  • $\begingroup$ Yes. The full text of Q is to find the relation between acceleration of 5kg mass along x axis and 2kg along x axis? $\endgroup$
    – Rider
    Commented May 5, 2021 at 11:42
  • $\begingroup$ No, this just the question. The full text should describe the system. $\endgroup$
    – nasu
    Commented May 5, 2021 at 11:46
  • $\begingroup$ What is it that you’re not getting from the Q above , you tell me that. I’ll tell you. This Q has no text actually. So , about the movements . I can tell you that the right wedge will not move at all. The 5kg mass and 2kg mass only moves. @nasu $\endgroup$
    – Rider
    Commented May 5, 2021 at 11:49
  • $\begingroup$ Why would anything move? $\endgroup$
    – nasu
    Commented May 5, 2021 at 11:51

2 Answers 2

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If we take, b, as the length of the flat bottom of $m_2$, and consider the motion from the time the wedges are in contact to when, b, hits the lower surface; then $m_2$ shifts to the left a distance, (b/2) = (1/2)${a_2}{t^2}$, and $m_1$ shifts a distance, b = (1/2)${a_1}{t^2}$ . Divide these and you get $a_1 = 2 a_2$ (where the, a's, are the horizontal components of acceleration). Given that, write force equations for each mass and solve.

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Acceleration will be equal only normal to the surface of contact. It will not be equal along X direction as you have mentioned.

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    $\begingroup$ Can you elaborate on your answer more and also give a proof maybe using Newton laws. When will they be equal ? An example of it as well which can tell that what I thought was wrong. Also , about the points I made. If two bodies are touching the surfaces , if one is faster and the other less , then they would lose contact. $\endgroup$
    – Rider
    Commented May 5, 2021 at 14:59

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