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Consider a block sliding without friction on a wedge as in the following picture:

enter image description here

Given the height of the block of mass $m$ as a function $h(t)$ of time $t$ with initial height $h(0) = h_0$, is the following method to determine the acceleration of wedge $M$ correct?

Using conservation of energy, the initial potential energy of the block is equal to the final kinetic energy plus potential energy after a time $t$: $$mgh_0=\frac{1}{2}m(v_1\cos(30^\circ)-v_2)^2+\frac{1}{2}M(v_2)^2+\frac{1}{2}m(v_1\sin(30^\circ))^2 +mgh(t),\tag{1}$$ where $v_1$ is the velocity of the block in frame of the wedge and $v_2$ is the velocity of the wedge w.r.t ground. Using conservation of momentum in the frame where the ground is at rest, we also have $$m(v_1\cos(30^\circ)-v_2)=Mv_2\tag{2}.$$

The second equation looks perfectly fine to me as it just momentum conservation. However, the first equation might be incorrect. I applied this technique to solve this problem, which asks for the acceleration of the wedge if the block's height varies as $h(t) = 1.5\ \mathrm{m} + 1.5\frac{\mathrm{m}}{\mathrm{s}^2} t^2$. I solved for $v_2(t)$ from the two equations and then differentiated it to get the final result. The answer I get using this technique (putting $M=m=1$ as the result is independent of mass) is $3 \mathrm{ms^{-2}}$. However, the correct result should have been $\frac{2}{\sqrt{3}} \mathrm{ms^{-2}}$. I'm certain that there is no trivial calculation error. I've checked it several times.

So, why is equation (1) not applicable in this case?

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    $\begingroup$ Here is a duplicate? physics.stackexchange.com/q/277412/104696 $\endgroup$
    – Farcher
    Commented Apr 17, 2017 at 11:01
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    $\begingroup$ @Farcher That one doesn't really help. I still don't see the error in my method. $\endgroup$
    – user139621
    Commented Apr 17, 2017 at 11:59
  • $\begingroup$ As I understand the site "homework" policy, "Check my work" questions are off topic. The chatrooms are a better place to ask such questions - as you have done. $\endgroup$
    – sammy gerbil
    Commented Apr 17, 2017 at 21:02
  • $\begingroup$ @sammygerbil I see a big difference between a typical check-my-work question that goes "is this answer right?" and this question, which has identified where in the reasoning process the error has occurred but not why the asker's reasoning is mistaken. We can discuss further in Physics Chat or on Physics Meta if you like. $\endgroup$
    – rob ♦
    Commented Apr 18, 2017 at 17:34
  • $\begingroup$ @rob Yes, let us discuss in the Physics Meta chatroom. $\endgroup$
    – sammy gerbil
    Commented Apr 18, 2017 at 23:19

1 Answer 1

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As I cannot comment (min 50 reputation) I will write it as the answer, All the equations framed for conservation of energy, momentum were correct. The point where you were wrong was that you assumed the result is independent of mass. In fact the particular acceleration of block $m$ height was caused only for that set of masses. What I am trying to say is the question states that the acceleration of m is $\rm3\ m/s^2$. As you took $m=M=1$, for this set of masses the acceleration will not be $\rm3\ m/s^2$.

Equations one and two are correct, and would have given correct answer if you knew the masses of blocks

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  • $\begingroup$ Where does the question state that the acceleration of block is $3 m/s^2$ ? It doesn't. $\endgroup$
    – user139621
    Commented Apr 17, 2017 at 20:56
  • $\begingroup$ Differentiate h(t) twice for acceleration. Or the given eqn is of form x-x0 = ut +(0.5)at^2(Displacement in uniform accelerated motion) $\endgroup$
    – Jay N
    Commented Apr 17, 2017 at 20:59
  • $\begingroup$ Ok, I see! In that case are my two equations sufficient to find $dv_2/dt$ without putting any values for $m$ and $M$. I think there are too many variables. Will they all get eliminated? $\endgroup$
    – user139621
    Commented Apr 17, 2017 at 21:02
  • $\begingroup$ No, without putting values of m & M, the equation 1 & 2 are not sufficient. $\endgroup$
    – Jay N
    Commented Apr 17, 2017 at 21:08
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    $\begingroup$ Hi, welcome to Physics.SE. We try to answer conceptual questions without becoming a homework help service; I've edited out your photograph of a "complete solution." $\endgroup$
    – rob ♦
    Commented Apr 17, 2017 at 21:40

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