Question

A cubical container is filled with a liquid of density $\sigma$. A small ball mass $M$ of density $\frac{\sigma}{4}$ is released at point O on a wall from position as shown, when the container starts accelerating with horizontal acceleration a. Assuming non-viscous liquid

• A. The ball will strike at point R if a = 2g
• B. The ball will strike at point R if a = 3g
• C. The ball will strike at point P.
• D. The ball, if strikes at point R it will take time $t = \sqrt{\frac{d}{3g}}$

Answer 1

Answer: (A), (D)

The ball will strike at point R if a = 2g

Answer 2

The problem describes the motion of a ball in an accelerating liquid-filled container. We analyze the motion in the non-inertial frame of the container, which is accelerating horizontally with acceleration $a$.

In the non-inertial frame, a pseudo force $F_{pseudo} = Ma$ acts on the ball in the direction opposite to the container's acceleration.

The gravitational force on the ball is $F_g = Mg$ downwards.

The buoyant force in an accelerating fluid is given by $\vec{F}_B = -\rho_{liquid} V (\vec{g} - \vec{a}_{container})$, where $V$ is the volume of the ball, $\rho_{liquid}$ is the density of the liquid, $\vec{g}$ is the acceleration due to gravity, and $\vec{a}_{container}$ is the acceleration of the container.

The volume of the ball is $V = \frac{M}{\rho_{ball}} = \frac{M}{\sigma/4} = \frac{4M}{\sigma}$.

Let the x-axis be horizontal to the right and the y-axis be vertical upwards. Then $\vec{g} = -g \hat{j}$ and $\vec{a}_{container} = a \hat{i}$.

The buoyant force is $\vec{F}_B = -\sigma \left(\frac{4M}{\sigma}\right) (-g \hat{j} - a \hat{i}) = 4M (g \hat{j} + a \hat{i})$.

The pseudo force is $\vec{F}_{pseudo} = -M \vec{a}_{container} = -Ma \hat{i}$.

The gravitational force is $\vec{F}_g = Mg \hat{g} = -Mg \hat{j}$.

The net force on the ball in the non-inertial frame is $\vec{F}_{net} = \vec{F}_g + \vec{F}_{pseudo} + \vec{F}_B = -Mg \hat{j} - Ma \hat{i} + 4M (g \hat{j} + a \hat{i}) = -Mg \hat{j} - Ma \hat{i} + 4Mg \hat{j} + 4Ma \hat{i} = 3Ma \hat{i} + 3Mg \hat{j}$.

The acceleration of the ball relative to the container is $\vec{a}_{ball/container} = \frac{\vec{F}_{net}}{M} = \frac{3Ma \hat{i} + 3Mg \hat{j}}{M} = 3a \hat{i} + 3g \hat{j}$.

So, the horizontal acceleration relative to the container is $a_x = 3a$ and the vertical acceleration relative to the container is $a_y = 3g$.

The ball is released from point O, which is on the left wall at a height $d/2$ from the bottom. The initial velocity of the ball relative to the container is zero.

Let the origin be at point P (bottom left corner). The initial position of the ball is $(0, d/2)$.

The horizontal distance to the right wall is $d$. The vertical position of the top wall is $d$.

The horizontal motion is $x(t) = x_0 + v_{0x} t + \frac{1}{2} a_x t^2 = 0 + 0 \cdot t + \frac{1}{2} (3a) t^2 = \frac{3}{2} a t^2$.

The vertical motion is $y(t) = y_0 + v_{0y} t + \frac{1}{2} a_y t^2 = \frac{d}{2} + 0 \cdot t + \frac{1}{2} (3g) t^2 = \frac{d}{2} + \frac{3}{2} g t^2$.

For the ball to strike the right wall, $x(t) = d$. Let the time taken be $t_R$.

$d = \frac{3}{2} a t_R^2 \implies t_R^2 = \frac{2d}{3a} \implies t_R = \sqrt{\frac{2d}{3a}}$.

The vertical position when it strikes the right wall is $y(t_R) = \frac{d}{2} + \frac{3}{2} g t_R^2 = \frac{d}{2} + \frac{3}{2} g \left(\frac{2d}{3a}\right) = \frac{d}{2} + \frac{gd}{a}$.

Point R is the top right corner with coordinates $(d, d)$. For the ball to strike at R, we must have $y(t_R) = d$ when $x(t_R) = d$.

$d = \frac{d}{2} + \frac{gd}{a}$ $\frac{d}{2} = \frac{gd}{a}$ $\frac{1}{2} = \frac{g}{a} \implies a = 2g$.

So, the ball will strike at point R if $a = 2g$. This confirms option (A).

Let's check the time taken when $a = 2g$.

$t_R = \sqrt{\frac{2d}{3a}} = \sqrt{\frac{2d}{3(2g)}} = \sqrt{\frac{2d}{6g}} = \sqrt{\frac{d}{3g}}$.

So, if the ball strikes at point R (which happens when $a=2g$), the time taken is $t = \sqrt{\frac{d}{3g}}$. This confirms option (D).