Chapter 3, Problem 3SP

(a)

To determine

The time taken by each ball to reach the floor.

Answer to Problem 3SP

The time taken by each ball to reach the floor is $0.4\text{s}$.

Explanation of Solution

Given info: The height of the table top is equal to $0.8\text{m}$ , vertical component of initial velocity is $0\text{m}/\text{s}$ and acceleration due to gravity is $10\text{m}/{\text{s}}^{\text{2}}$ along the downward direction.

Consider the motion of the ball $\text{A}$ and B. There is only change in their horizontal velocity and all other parameters are same.

Therefore, Write the equation of vertical distance travelled by the both the balls.

$d_{\text{vertical}}=v_0{}_{\text{vertical}}t+\frac{1}{2}{a}_{\text{vertical}}{t}^2$

Here,

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

$v_{\text{0vertical}}$ is the vertical component of initial velocity

$a_{\text{vertical}}$ is the vertical ;component of acceleration

In the case of ball the only vertical acceleration is the acceleration due to gravity and distance travelled by the ball is equal to the height of the table.

Substitute $0\text{m}/\text{s}$ for $v_0{}_{\text{vertical}}$, $10\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $0.8\text{m}$ for $d_{\text{vertical}}$ in the above equation to get $t$.

$\begin{array}{c}0.8\text{m}=\left(0\text{m}/\text{s}\right)t+\frac{1}{2}\left(10\text{m}/{\text{s}}^{\text{2}}\right)t^2\\ t=\sqrt{\frac{1.6\text{m}}{10\text{m}/{\text{s}}^{\text{2}}}}\\ =0.4\text{s}\end{array}$

Therefore, the time taken by both the balls are equal to $0.4\text{s}$.

Conclusion:

Thus, time taken by both the balls to reach the ground is $0.4\text{s}$.

(b)

To determine

The horizontal distance travelled by the balls before hitting the floor.

Answer to Problem 3SP

Before hitting the ground ball A travelled a horizontal distance of $1.2\text{m}$ and ball B travelled a horizontal distance of $2.0\text{m}$.

Explanation of Solution

Given info: the horizontal velocity of ball A is $3\text{m}/\text{s}$, the horizontal velocity of ball B is $5\text{m}/\text{s}$ and there is no horizontal acceleration for both the balls.

Consider the motion of the ball $\text{A}$. In horizontal direction there is no acceleration.

Therefore, write the equation of horizontal distance of the ball A.

$d_{\text{horizontal-A}}=v_{0\text{horizontal-A}}t$

Here,

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

$v_{0\text{horizontal-A}}$ is the horizontal component of velocity of ball A

$t$ is the time

Substitute $3\text{m}/\text{s}$ for $v_{0\text{horizontal-A}}$ and $0.4\text{s}$ for $t$ in the above equation to get $d_{\text{horizontal-A}}$.

$\begin{array}{c}{d}_{\text{horizontal-A}}=\left(3\text{m}/\text{s}\right)\left(0.4\text{s}\right)\\ =1.2\text{m}\end{array}$

Consider the motion of the ball B. In horizontal direction there is no acceleration.

Therefore, write the equation of horizontal distance of the ball B.

$d_{\text{horizontal-B}}=\left(v_{0\text{horizontal-B}}\right)t$

Here,

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

$v_{0\text{horizontal-B}}$ is the horizontal component of velocity of ball B

Substitute $5\text{m}/\text{s}$ for $v_{0\text{horizontal-B}}$ and $0.4\text{s}$ for $t$ in the above equation to get $d_{\text{horizontal-B}}$.

$\begin{array}{c}{d}_{\text{horizontal-B}}=\left(5\text{m}/\text{s}\right)\left(0.4\text{s}\right)\\ =2.0\text{m}\end{array}$

Conclusion:

Thus, before hitting the ground ball A travelled a horizontal distance of $1.2\text{m}$ and ball B travelled a horizontal distance of $2.0\text{m}$.

(c)

To determine

Whether the two balls will reach the floor at same time, if the two balls started at the same time at a point $1.2\text{m}$ behind the edge of the table.

Answer to Problem 3SP

Since the ball B has larger velocity than ball A, ball B will reach the edge of the table first and therefore A will reach the floor first.

Explanation of Solution

Given info: The distance behind the edge at which the balls are started rolling is $1.2\text{m}$

In part a, it is showed that both the ball will reach the ground at the same time if they are allowed to roll off the balls at the table top.

But in this case they are started rolling at same time at a point $1.2\text{m}$ behind the edge of the table.

Therefore, write the equation of horizontal distance of the ball A.

$d_{\text{horizontal-A}}=v_{0\text{horizontal-A}}t$

Rearrange the above equation to get $t$.

$t=\frac{d_{\text{horizontal-A}}}{v_{0\text{horizontal-A}}}$

Substitute $3\text{m}/\text{s}$ for $v_{0\text{horizontal-A}}$ and $1.2\text{m}$ for $d_{\text{horizontal-A}}$ in the above equation to get $t$.

$\begin{array}{c}t=\frac{1.2\text{m}}{3\text{m}/\text{s}}\\ =0.4\text{s}\end{array}$

Write the equation of horizontal distance of the ball B.

$d_{\text{horizontal-B}}=v_{0\text{horizontal-B}}t$

Rearrange the above equation to get $t$.

$t=\frac{d_{\text{horizontal-A}}}{v_{0\text{horizontal-A}}}$

Substitute $5\text{m}/\text{s}$ for $v_{0\text{horizontal-A}}$ and $1.2\text{m}$ for $d_{\text{horizontal-A}}$ in the above equation to get $t$.

$\begin{array}{c}t=\frac{1.2\text{m}}{5\text{m}/\text{s}}\\ =0.24\text{s}\end{array}$

Therefore ball B has taken less time than ball A to cover $1.2\text{m}$. Since for the rest of the distance both of them will take equal time, ball B will reach the ground compared to the ball A.

Conclusion:

Since the ball B has larger velocity than ball A, ball B will reach the edge of the table first and therefore A will reach the floor first.