Chapter 14, Problem 14.64AP

Review. Assume a certain liquid, with density 1 230 kg/m³, exerts no friction force on spherical objects. A ball of mass 2.10 kg and radius 9.00 cm is dropped from rest into a deep tank of this liquid from a height of 3.30 m above the surface. (a) Find the speed at which the hall enters the liquid. (b) Evaluate the magnitudes of the two forces that are exerted on the ball as it moves through the liquid. (c) Explain why the ball moves down only a limited distance into the liquid and calculate this distance. (d) With what speed will the ball pop up out of the liquid? (c) How does the time interval ∆t_down, during which the ball moves from the surface down to its lowest point, compare with the lime interval ∆t_up for the return trip between the same two points? (f) What If? Now modify the model to suppose the liquid exerts a small friction force on the ball, opposite in direction to its motion. In this case, how do the time intervals ∆t_down and ∆t_up compare? Explain your answer with a conceptual argument rather than a numerical calculation.

To determine

The speed of the ball which enters the liquid.

Answer to Problem 14.64AP

The speed of the ball which enters the liquid is $8.05\text{m}/\text{s}$.

Explanation of Solution

The density of the liquid is $1230\text{kg}/{\text{m}}^3$, the radius of the ball is $9.00\text{cm}$, the mass of the ball is $2.10\text{kg}$ and the height from which the ball dropped is $3.30\text{m}$.

By the conservation of energy,

    $mgh=\frac{1}{2}m{v}_{\text{i}}{}^2$

Here, $m$ is the mass of the ball, $g$ is the acceleration due to gravity, $h$ is the height of the ball from the liquid level and $v_{\text{i}}$ is the initial velocity of the ball.

Substitute $3.30\text{m}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $v_{\text{i}}$.

    $\begin{array}{c}9.81\text{m}/{\text{s}}^{\text{2}}\times 3.30\text{m}=\frac{1}{2}{v}_{\text{i}}{}^2\\ v_{\text{i}}{}^2=2\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 3.30\text{m}\\ v_{\text{i}}=\sqrt{2\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 3.30\text{m}}\\ =8.05\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed of the ball which enters the liquid is $8.05\text{m}/\text{s}$.

To determine

The magnitudes of the two forces that are exerted on the ball as move through liquid.

Answer to Problem 14.64AP

The magnitude of the gravitational force that is exerted on the ball as move through liquid is $20.6\text{N}$ and the magnitude of the buoyant force that is exerted on the ball as move through liquid is $36.84\text{N}$.

Explanation of Solution

Formula to calculate the gravitational force or weight of the ball is,

    $F_{\text{g}}=mg$

Here, $F_{\text{g}}$ is the gravitational force exerted on the ball.

Substitute $2.10\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $F_{\text{g}}$.

    $\begin{array}{c}{F}_{\text{g}}=2.10\text{kg}\times 9.81\text{m}/{\text{s}}^{\text{2}}\\ =20.6\text{N}\end{array}$

Thus, the gravitational force exerted on the ball is $20.6\text{N}$.

The buoyant force exerted on the ball is equal to the volume of water displaced by the ball.

Formula to calculate the buoyant force exerted on the ball is,

    $B=\rho g\times V$        (1)

Here, $\rho$ is the density of the fluid and $B$ is the buoyant force exerted on the ball.

Formula to calculate the volume of the spherical ball is,

    $V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius of spherical ball.

Substitute $\frac{4}{3}\pi r^3$ for $V$ in equation (1).

    $B=\rho g\times \frac{4}{3}\pi r^3$

Substitute $9.00\text{cm}$ for $r$, $1230\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $B$.

    $\begin{array}{c}B=1230\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times \frac{4}{3}\pi {\left(9.00\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3\\ =36.84\text{N}\end{array}$

Conclusion:

Therefore, the magnitudes of the gravitational force that are exerted on the ball as move through liquid is $20.6\text{N}$ and the magnitudes of the buoyant force that are exerted on the ball as move through liquid is $36.84\text{N}$.

To determine

The distance covered by the ball in water.

Answer to Problem 14.64AP

The distance covered by the ball in water is $4.19\text{m}$.

Explanation of Solution

The buoyant force exerted on the ball is greater than the weight of the ball, therefore the ball certain distance covered inside the water because it changes the direction of motion.

From third law of motion,

    $v_{\text{f}}{}^2-v_{\text{i}}{}^2=2ad$        (2)

Here, $v_{\text{f}}$ is the final velocity of the ball, $d$ is the distance covered by the ball inside theater, $u$ is the initial; velocity of the ball and $a$ is the acceleration of the ball.

Formula to calculate the acceleration of the ball is,

    $a=\frac{F}{m}$        (3)

Formula to calculate the net force acting on a ball is,

    $F=B-F_{\text{g}}$

Substitute $B-F_{\text{g}}$ for $F$ in equation (3).

    $a=\frac{B-F_{\text{g}}}{m}$

Substitute $\frac{B-F_{\text{g}}}{m}$ for $a$ in equation (I).

    $v_{\text{f}}{}^2-v_{\text{i}}{}^2=2\left(\frac{B-F_{\text{g}}}{m}\right)d$

Substitute $0$ for $v_{\text{f}}$, $8.05\text{m}/\text{s}$ for $v_{\text{i}}$, $36.84\text{N}$ for $B$, $20.6\text{N}$ for $F_{\text{g}}$ and $2.10\text{kg}$ for $m$ to find $d$.

    $\begin{array}{c}{0}^2-{\left(8.05\text{m}/\text{s}\right)}^2=2\left(\frac{36.84\text{N}-20.6\text{N}}{2.10\text{kg}}\right)d\\ d=\frac{0^2-{\left(8.05\text{m}/\text{s}\right)}^2}{2\left(\frac{36.84\text{N}-20.6\text{N}}{2.10\text{kg}}\right)}\\ =-4.19\text{m}\end{array}$

The negative sign shows direction of the ball in downward direction.

Conclusion:

Therefore, the distance covered by the ball in water is $4.19\text{m}$ in the downward direction.

To determine

The speed of the ball pop up out of the liquid.

Answer to Problem 14.64AP

The speed of the ball which enters the liquid is $8.05\text{m}/\text{s}$.

Explanation of Solution

The speed of the ball which enters the liquid is equal to the speed of the ball pop up out of the liquid because absence of friction, no energy losses occur in this system. Hence the speed of the ball pop up out of the liquid is $8.05\text{m}/\text{s}$.

Conclusion:

Therefore, the speed will the ball pop up out of the liquid is $8.05\text{m}/\text{s}$.

To determine

The result of comparison the time interval during which the ball moves from the surface to its lowest point with the time interval for return trip at the same point.

Answer to Problem 14.64AP

The time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point.

Explanation of Solution

The time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point because ball going down and up acceleration of the ball and distance covered by the ball is same.

Conclusion:

Therefore, the time interval during which the ball moves from the surface to its lowest point is identical to the time interval for return trip at the same point.

To determine

Compare the time interval during which the ball moves from the surface to its lowest point with the time interval for return trip at the same point.

Answer to Problem 14.64AP

The time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point.

Explanation of Solution

The time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point when friction is present because energy losses by the system.

Conclusion:

Therefore, the time interval during which the ball moves from the surface to its lowest point is not equal to the time interval for return trip at the same point.