# Container filled with fluid move with a acceleration a

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My try :

My option B is not getting matched.
Answer is given as B, C.

Your calculation is correct so far but it is not complete.

In the vertical direction there is a gravitational field of strength $g$ acting downwards. This is equivalent to an acceleration upwards at rate $g$. The acceleration of the ball relative to the container is $a_y=3g$ upwards.

In the horizontal direction the ball accelerates to the right at $a_x$ relative to the container. This is different from the acceleration $a$ of the container itself, which is equivalent to a gravitational field acting to the left. So we can think of the container as stationary and acted upon by gravitational fields downwards and to the left.

Horizontally the ball must cover twice the distance as vertically but in the same time. Since $s \propto a$ and $a_y=3g$ then we must have $a_x=6g$.

The final step which you have missed is to relate $a_x$ (the acceleration of the ball relative to the container) to the acceleration $a$ of the container itself. As in the vertical direction we have $a_x=3a=6g$. Therefore $a=2g$.

The time for the ball to reach the edge of the container is $\sqrt{\frac{L}{3g}}$ as you calculated. So the options B, C are correct.

answered Apr 19 by (28,466 points)
edited Apr 21
@SammyGerbil I don't get how you relate the acceleration of the ball relative to the container $a_x$ to the acceleration of the container $a$. You used $a_x=3a=6g$ but I don't see why $a_x=3a$
In the vertical direction $a_y=3g$. The number $3$ arises from the difference in density between the ball and the liquid.

In the horizontal direction, in the lab frame the container accelerates at rate $a$ to the right. In the frame of the container this is equivalent to an effective gravitational field of strength $a$ acting left. The difference in density is the same horizontally as it is vertically. So we must write $a_x=3a$ in the same way we wrote $a_y=3g$.