0

A block of mass m is kept at the lowest point on a wedge of mass $M$ of height $h$ and it has an angle of inclination $\theta$. The wedge accelerates with an acceleration $a$ such that the mass $m$ is accelerated up the incline.

What will the trajectory of the block be with respect to ground when it has reached the highest point of the wedge and is just about to lose contact with it?

All the surfaces are smooth. Additional variables to solve this question may be defined if needed, as I myself don't know if additional info is required.

enter image description here

Matti P.
  • 6,012
sandy
  • 3
  • 3
    please include $dx/dt$ and $d^2x/dt^2$ for each object so that one doesn't have to solve it like a physics problem – Anindya Prithvi Aug 04 '20 at 09:41
  • Unless you just pose the mathematical part in it, it will be transferred to PSE – Arjun Aug 04 '20 at 09:57
  • i was told to ask this question from PSE – sandy Aug 04 '20 at 09:58
  • 1
    You should still do as much of the problem as you can. If possible, solve the problem yourself so you're just asking for confirmation or correction. Edit the question to show your work. Use MathJax: https://math.stackexchange.com/help/notation – David K Aug 04 '20 at 12:00

1 Answers1

0

Hint.

Calling

$$ \cases{ F_a = m(a,0)\\ W=m(0,-g)\\ N=n(-\sin\theta,\cos\theta) } $$

we have

$$ F_a + W + N = m b(\cos\theta,\sin\theta) $$

where $b$ is the acceleration module along the ramp.

Now solving for $n,b$ we obtain

$$ b = a\cos\theta-g\sin\theta $$

so we need $a\cos\theta>g\sin\theta$

now along the ramp, the movement follows

$$ s = \frac 12 b t^2 $$

Cesareo
  • 33,252