Chapter 3, Problem 2SP

(a)

To determine

The velocities of the two balls $1.8\text{seconds }$ after they are released.

Answer to Problem 2SP

The velocity of the ball $\text{A}$ is $17.6\text{m}/\text{s}$ and velocity of the ball $\text{B}$ is $31.6\text{m}/\text{s}$.

Explanation of Solution

Given info: The time after which velocity is to be find is $1.8\text{s}$, the acceleration due y gravity is $9.8\text{m}/{\text{s}}^{\text{2}}$, initial velocity of the ball A is $0\text{m}/\text{s}$ and initial velocity of the ball B is $14\text{m}/\text{s}$.

Consider the motion of the ball $\text{A}$.

Write the equation of motion of the ball A.

$v_{\text{A}}=v_{0\text{A}}+at$

Here,

$v_{\text{A}}$ is the velocity of the ball A

$v_{0\text{A}}$ is the initial velocity of the ball A

$a$ is the acceleration of the ball A

$t$ is the time

Substitute $0\text{m}/\text{s}$ for $v_{0\text{A}}$ ,$9.8{\text{m}/\text{s}}^2$ for $a$ and $1.8\text{s}$ for $t$ in the above equation to get $v_{\text{A}}$.

$\begin{array}{c}{v}_{\text{A}}=0\text{m}/\text{s}+\left(9.8\text{m}/\text{s}\right)\left(1.8\text{m}/\text{s}\right)\\ v_{\text{A}}=17.64\text{m}/\text{s}\end{array}$

Consider the motion of the ball B.

Write the equation of motion of the ball B.

Write the equation of motion of the ball B.

$v_{\text{B}}=v_{0\text{B}}+at$

Here,

$v_{\text{B}}$ is the velocity of the ball B

$v_{0\text{B}}$ is the initial velocity of the ball B

$a$ is the acceleration of the ball B

$t$ is the time

Substitute $14\text{m}/\text{s}$ for $v_{0\text{B}}$ ,$9.8{\text{m}/\text{s}}^2$ for $a$ and $1.8\text{s}$ for $t$ in the above equation to get $v_{\text{B}}$.

$\begin{array}{c}{v}_{\text{B}}=14\text{m}/\text{s}+\left(9.8\text{m}/\text{s}\right)\left(1.8\text{m}/\text{s}\right)\\ v_{\text{B}}=14\text{m}/\text{s}+17.64\text{m}/\text{s}\\ =31.6\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity of the ball $\text{A}$ is $17.6\text{m}/\text{s}$ and velocity of the ball $\text{B}$ is $31.6\text{m}/\text{s}$.

(b)

To determine

The distance dropped by each ball after $1.8\text{seconds}$.

Answer to Problem 2SP

The distance dropped by ball $\text{A}$ in $1.8\text{s}$ is $16.2\text{m}$ and the distance dropped by ball B is $41.4\text{m}$.

Explanation of Solution

Given info: The time after which velocity is to be find is $1.8\text{s}$, the acceleration due to gravity is $10{\text{m}/\text{s}}^{\text{2}}$, initial velocity of the ball A is $0\text{m}/\text{s}$ and initial velocity of the ball B is $14\text{m}/\text{s}$.

Consider the motion of the ball $\text{A}$.

Write the equation of distance of the ball A.

$d_{\text{A}}=v_{0\text{A}}t+\frac{1}{2}a{t}^2$

Here,

$d_{\text{A}}$ is the distance travelled by the ball A

Substitute $0\text{m}/\text{s}$ for $v_{0\text{A}}$ ,$9.8{\text{m}/\text{s}}^2$ for $a$ and $1.8\text{s}$ for $t$ in the above equation to get $d$.

$\begin{array}{r}d=\left(0\text{m}/\text{s}\right)\left(1.8\text{s}\right)+\frac{1}{2}\left(10\text{m}/{\text{s}}^{\text{2}}\right){\left(1.8\text{s}\right)}^2\\ d=16.2\text{m}\end{array}$

Consider the motion of the ball B.

Write the equation of motion of the ball B.

$d=v_{\text{0B}}t+\frac{1}{2}a{t}^2$

Here,

$d_{\text{B}}$ is the distance travelled by the ball B

Substitute $14\text{m}/\text{s}$ for $v_0$ ,$9.8{\text{m}/\text{s}}^2$ for $v$ and $1.8\text{s}$ for $t$ in the above equation to get $a$.

$\begin{array}{c}d=\left(14\text{m}/\text{s}\right)\left(1.8\text{s}\right)+\frac{1}{2}\left(10{\text{m}/\text{s}}^2\right){\left(1.8\text{s}\right)}^2\\ =41.4\text{m}\end{array}$

Conclusion:

Thus, the distance dropped by ball A in $1.8\text{s}$ is $16.2\text{m}$ and the distance dropped by ball B is $41.4\text{m}$.

(c)

To determine

Whether, the difference in the velocities of the two balls changes at any time after their release.

Answer to Problem 2SP

Since the acceleration of the both the balls are same as acceleration due to gravity, the difference in the velocities of the balls never changes until they become rest at the ground.

Explanation of Solution

Even though the balls are dropped at two different velocities they are accelerated at the acceleration due to gravity.

For both the balls time at which they dropped are same and also acceleration are same.

Write the equation of motion of the ball A.

$v_{\text{A}}=v_{0\text{A}}+at$ (1)

Write the equation of motion of the ball B.

$v_{\text{B}}=v_{0\text{B}}+at$ (2)

Take the difference of the equation (1) by (2) to get the difference in their position.

$v_{\text{B}}-v_{\text{A}}=v_{0\text{B}}-v_{\text{0A}}$

Substitute $0\text{m}/\text{s}$ for $v_{0\text{A}}$ and $14\text{m}/\text{s}$ for $v_{\text{0B}}$ in the above equation to get $v_{\text{B}}-v_{\text{A}}$.

$\begin{array}{c}{v}_{\text{B}}-v_{\text{A}}=14\text{m}/\text{s}-0\text{m}/\text{s}\\ =14\text{m}/\text{s}\end{array}$

Therefore the difference in the velocities of the both the ball do not change in time it is a constant equal to $14\text{m}/\text{s}$.

Conclusion:

Thus, the difference between the velocities of the both the balls are always constant and is equal to $14\text{m}/\text{s}$.