In this question, we're presented with a situation in which two balls of different densities but equal volumes are held underneath the surface of a container of water. Then, each ball is released and allowed to accelerate up to the surface. The question is to determine how the acceleration of each ball compares to the other. In order to answer this, let's start by imagining all of the forces acting on the submerged ball.

In the x-direction, the forces acting to the left of the ball are exactly equal to the forces acting on the right of the ball. Therefore, all of the forces acting in the x-direction cancel out, resulting in a net force of zero in the x-direction.

In the y-direction, we have to consider two things. The first is the upward buoyant force caused by the displacement of water. Since both balls have an identical volume, both of them displace the same amount of water. Consequently, both balls will experience the same upward buoyant force. However, we must also consider the downward force caused by the weight of the ball itself. In this scenario, the downward weight of Ball B is greater than that of Ball A.

We can write the net force acting on the ball in the y-direction as the difference between the upward buoyant force and the downward weight as follows:

\(\displaystyle \Sigma F_{y}=B-W\)

Due to the fact that Ball A has a smaller weight (a smaller \(\displaystyle W\) component in the above equation), the result is that the net upward force is greater than that of Ball B. Thus, we would expect Ball A to have a greater acceleration.

Consider all the forces on the gold marble:

\(\displaystyle F_{net}=F_1+F_2+...\)

\(\displaystyle F_{net}=F_{buoyant}+F_{gravity}\)

Recall the equation for buoyant force:

\(\displaystyle F_{net}=V_{object}\rho_{object} g+m_{object}g\)

Substitute:

\(\displaystyle F_{net}=V_{object}\rho_{medium} g-mg\)

Find the mass of the marble:

\(\displaystyle D=\frac{m}{V}\)

\(\displaystyle m=19.32*\left(\frac{4}{3}\pi*.95^3*10 \right )\)

Plug in values:

\(\displaystyle F_{net}=\frac{4}{3}\pi*.95^3*13.59*10-\left(19.32*\left(\frac{4}{3}\pi*.95^3 \right )*10 \right )\)

\(\displaystyle F_{net}= 487.819-693.5\)

\(\displaystyle F_{net}=-205.68\frac{g\cdot m}{s^2}\)

\(\displaystyle F_{net}=-0.20568N\)

Note how the buoyant force points up while the gravitational force points down.

Use the equation for buoyant force:

\(\displaystyle F_{buoyant}=\rho g V\)

Where \(\displaystyle \rho\) is the density of the medium around the object in question

\(\displaystyle g\) is the gravitational acceleration near the surface of the earth

\(\displaystyle V\) is the volume of the object in question

Convert \(\displaystyle \frac{g}{mL}\) into \(\displaystyle \frac{kg}{L}\)

\(\displaystyle 1.5\frac{g}{mL}*\frac{1000ml}{1L}*\frac{1kg}{1000ml}=\frac{1.5kg}{L}\)

Plug in values:

\(\displaystyle F_{buoyant}=1.5*9.8*3.5\)

\(\displaystyle F_{buoyant}=51N\)

Use the equation for buoyant force:

\(\displaystyle F_b=\rho*|g|*V\)

Where

\(\displaystyle \rho\) is the density of the medium

\(\displaystyle g\) is the acceleration due to gravity

\(\displaystyle V\) is the volume

Plugging in values:

\(\displaystyle F_b=\frac{1kg}{L}*|-9.8\frac{m}{s^2}|*2L\)

\(\displaystyle F_b=19.6N\)

Determining density:

\(\displaystyle d=\frac{m}{v}\)

Volume of a sphere:

\(\displaystyle v=\frac{4}{3}\pi r^3\)

Combining equations:

\(\displaystyle d=\frac{3m}{4\pi r^3}\)

Converting \(\displaystyle kg\) to \(\displaystyle g\) and plugging in values:

\(\displaystyle d=\frac{3*10,000}{4\pi 10^3}\)

\(\displaystyle d=2.38\frac{g}{cm^3}\)

It is denser than water, so it will sink.

According to Archimedes's principle, when the block is placed under water, 3.5L of water are displaced. We can then calculate the buoyant force provided by the water:

\(\displaystyle F_{buoyant} = m_{displaced}g\)

Where:

\(\displaystyle m= \rho_w V = \left ( 1.0\frac{kg}{L}\right )(3.5L) = 3.5kg\)

Plugging in our values to the first expression:

\(\displaystyle F_b = (3.5kg)\left ( 10\frac{m}{s^2}\right )\)

\(\displaystyle F_b = 35N\)

Then we can use Newton's 2nd law to determine the acceleration of the block:

\(\displaystyle F_{net}=ma\)

There are two forces acting on the block, gravity and buoyancy, and they are in opposite directions. If we designate a downward force as positive, we get:

\(\displaystyle F_g -F_b = ma\)

Substituting in the expression for force due to gravity:

\(\displaystyle mg-F_b = ma\)

Rearranging for acceleration:

\(\displaystyle a = \frac{mg-F_b}{m}\)

Substituting in values:

\(\displaystyle a = \frac{ (5kg)\left ( 10\frac{m}{s^2}\right )-35N}{5kg}\)

\(\displaystyle a = 3\frac{m}{s^2}\)

We will start with Newton's 2nd law for this problem:

\(\displaystyle F_{net} = ma\)

Where there are 4 total forces acting on the scuba diver and ball: gravity and buoyancy acting on both the scuba diver and the ball. However, the diver has the same density as the water, so the gravitational and buoyancy forces acting on the diver will cancel out. If we designate an upward force as being positive, we can say:

\(\displaystyle F_{b_{ball}}-F_{g_{ball}}=m_Ta\) (1)

Where:

\(\displaystyle F_{b_{ball}}=m_{w,displaced}g\)

\(\displaystyle F_{g_{ball}}=m_bg\)

\(\displaystyle m_T = m_d+m_b\)

Where:

\(\displaystyle m_w = \rho_w V_b\)

\(\displaystyle m_b = \rho_b V_b\)

So,

\(\displaystyle F_b = \rho_w V_bg\)

\(\displaystyle F_{g}=\rho_bV_bg\)

\(\displaystyle m_T = m_d+\rho_bV_b\)

Plugging these expressions back in to expression (1), we get:

\(\displaystyle \rho_w V_bg-\rho_b V_bg = (m_d + \rho_b V_b)a\)

Now let's start rearranging for the density of the ball:

\(\displaystyle \rho_wV_bg - m_da = \rho_bV_b(a+g)\)

\(\displaystyle \rho_b = \frac{\rho_wV_bg-m_da}{V_b(a+g)}\)

\(\displaystyle \rho_b = \frac{\left ( 1000\frac{kg}{m^3}\right )(0.2m^3)\left ( 10\frac{m}{s^2}\right )-(80kg)\left ( 1.5\frac{m}{s^2}\right )}{(0.2m^s)(1.5\frac{m}{s^2}+10\frac{m}{s^2})}\)

\(\displaystyle \rho_b = 817\frac{kg}{m^3}\)

We will start with Newton's 2nd law for this problem:

\(\displaystyle F_{net}=ma\)

Since the block is traveling at a constant velocity we can say:

\(\displaystyle F_{net}=0\)

There are 3 forces acting on the block: gravitational, buoyant, and drag force. If we denote a downward force being positive, the expression becomes:

\(\displaystyle F_g -F_b -F_d = 0\)

Where:

\(\displaystyle F_g = m_{block}g\)

\(\displaystyle F_b = m_{w,displaced}g\)

\(\displaystyle F_d = 3N\)

Substituting these in:

\(\displaystyle m_{block}g - m_{w,displaced}g - 3N = 0\)

Where according to Archimedes's principle:

\(\displaystyle m_w = \rho_wV_b\)

Plugging this in:

\(\displaystyle m_{block}g - \rho_w V_bg - 3N = 0\)

Rearranging for volume:

\(\displaystyle m_{block}g - 3N = \rho_w V_bg\)

\(\displaystyle V_b = \frac{m_{block}g - 3N }{\rho_w g}\)

Plugging in values:

\(\displaystyle V_b =\left (\frac{(2kg)\left ( 10\frac{m}{s^2}\right )-3N}{\left ( 1000\frac{kg}{m^3}\right )\left ( 10\frac{m}{s^2}\right )} \right )\left ( \frac{1000L}{1m^3}\right )\)

\(\displaystyle V_b = 1.7L\)

Since the ball is held in place, we know that:

\(\displaystyle F_{net}=0\)

There are three forces acting on the ball: gravitational, buoyant, and tension. If we denote a downward force as positive, we get:

\(\displaystyle F_g -F_b - F_t=0\)     (1)

\(\displaystyle F_g = mg\)

\(\displaystyle F_b = m_{w,displaced}g\)

And using Archimedes's principal:

\(\displaystyle m_w = \rho_w V_b\)

Where:

\(\displaystyle V_b = \frac{4}{3} \pi r^3\)

Plugging all of these into expression (1), we get:

\(\displaystyle m_bg - \frac{4}{3}\pi r^3 \rho_wg = F_t\)

Plugging in our values, we get:

\(\displaystyle F_t =(6kg)\left ( 10\frac{m}{s^2}\right ) - \frac{4}{3}\pi (0.10m)^3\left ( 1000\frac{kg}{m^3}\right )\left ( 10\frac{m}{s^2}\right )\)

\(\displaystyle F_t= 18N\)

We are asked when:

\(\displaystyle F_g = F_b\)

\(\displaystyle m_bg = m_{w,displaced}g\)

\(\displaystyle m_b = m_{w,displaced}\)

\(\displaystyle m_b = \rho_wV_b\)

Now we need to develop an expression for the mass in the ball using the rate at which water enters the ball:

\(\displaystyle m_b = m_i + \dot{m}t\)

Where:

\(\displaystyle \dot{m} = \dot{V}\rho_wt\)

\(\displaystyle m_b = m_i + \dot{V}\rho_wt\)

Plugging this into expression (1):

\(\displaystyle m_i + \dot{V}\rho_wt= \rho_wV_b\)

Rearranging for time, we get:

\(\displaystyle t= \frac{\left ( \rho_wV_b- m_i \right )}{\dot{V}\rho_w}\)

Plugging in our values, we get:

\(\displaystyle t= \frac{\left ( 1\frac{kg}{L}\right )(0.8L)- (0.4kg)}{\left ( 0.6\frac{L}{min}\right )\left ( 1\frac{kg}{L}\right )}\)

\(\displaystyle t = \frac{2}{3}min = 40s\)