Chapter 4, Problem 173P

(a)

To determine

The distance from block B’s initial position do the two blocks meet and the time elapsed from $t=0$ until the block meet.

Answer to Problem 173P

The distance from block B’s initial position is $1.65\text{m}$ and the time elapsed is $2.02\text{s}$.

Explanation of Solution

Write the expression for the acceleration of each block.

$\begin{array}{l}{a}_A=\frac{F_A}{m_A}\\ a_B=\frac{F_B}{m_B}\end{array}$ (I)

Here, $a_A$ is the acceleration of the block A, $a_B$ is the acceleration of the block B, $F_A$ is the force acting on A and $F_B$ is the force acting on B, $m_A$ is the mass of the block A, $m_B$ is the mass of the block B.

Write the expression for displacement after a time $\Delta t$ when the objects are at rest initially.

$\begin{array}{l}\Delta x_{A1}=\frac{1}{2}{a}_A{\left(\Delta t\right)}^2\\ \Delta x_{B1}=\frac{1}{2}{a}_B{\left(\Delta t\right)}^2\end{array}$ (II)

Here, $\Delta x_{A1}$ is the displacement for the block A, $\Delta x_{B1}$ is the displacement for the block B, $\Delta t$ is the time interval.

Substitute equation I in equation II,

$\begin{array}{l}\Delta x_{A1}=\frac{1}{2}\left(\frac{F_A}{m_A}\right){\left(\Delta t\right)}^2\\ \Delta x_{B1}=\frac{1}{2}\left(\frac{F_B}{m_B}\right){\left(\Delta t\right)}^2\end{array}$ (III)

Write the expression for the velocity of the blocks.

$\begin{array}{l}{v}_A=a_A\Delta t\\ v_B=a_B\Delta t\end{array}$ (IV)

Here, $v_A$ is the velocity of the block A, $v_B$ is the velocity of the block B.

Substitute equation I in equation IV,

$\begin{array}{l}{v}_A=\frac{F_A}{m_A}\Delta t\\ v_B=\frac{F_B}{m_B}\Delta t\end{array}$ . (V)

At the end of the $\Delta t$,  the blocks are moving with constant velocity until they meet.

Write the expression for the displacement of each block during $\Delta t_2$.

$\begin{array}{l}\Delta x_{A2}=v_A\Delta t_2\\ \Delta x_{B2}=v_B\Delta t_2\end{array}$ (VI)

Here, $\Delta x_{A2}$ is the displacement of the block A during $\Delta t_2$, $\Delta x_{B2}$ is the displacement of the block B during $\Delta t_2$.

Substitute equation IV and equation I in equation VI.

$\begin{array}{l}\Delta x_{A2}=\frac{F_A}{m_A}\Delta t\Delta t_2\\ \Delta x_{B2}=\frac{F_B}{m_B}\Delta t\Delta t_2\end{array}$ (VII)

Write the expression for the distance.

$\begin{array}{c}d=\Delta x_{A1}+\Delta x_{A2}-\Delta x_{B1}-\Delta x_{B2}\\ =\frac{1}{2}\left(\frac{F_A}{m_A}\right){\left(\Delta t\right)}^2+\frac{F_A}{m_A}\Delta t\Delta t_2-\frac{1}{2}\left(\frac{F_B}{m_B}\right){\left(\Delta t\right)}^2-\frac{F_B}{m_B}\Delta t\Delta t_2\end{array}$ (VIII)

The displacements of the block B are subtracted because they are intrinsically negative.

Write the expression for the displacement of the block B.

$d_B=-\frac{1}{2}\left(\frac{F_B}{m_B}\right){\left(\Delta t\right)}^2-\frac{F_B}{m_B}\Delta t\Delta t_2$ (IX)

Conclusion:

Rearrange equation VIII to find $\Delta t_2$.

$\Delta t_2=\frac{d-\frac{{\left(\Delta t\right)}^2}{2}\left(\frac{F_A}{m_A}-\frac{F_B}{m_B}\right)}{\left(\frac{F_A}{m_A}-\frac{F_B}{m_B}\right)\Delta t}$ (X)

Substitute $3.40\text{m}$ for $d$, $0.100\text{s}$ for $\Delta t$, $2.00\text{N}$ for $F_A$, $-5.00\text{N}$ for $F_B$, $0.225\text{kg}$ for $m_A$ and $0.600\text{kg}$ for $m_B$ in the equation IX to find $\Delta t_2$.

$\begin{array}{c}\Delta t_2=\frac{3.40\text{m-}\frac{{\left(0.100\text{s}\right)}^2}{2}\left(\frac{2.00\text{N}}{0.225\text{kg}}-\frac{-5.00\text{N}}{0.600\text{kg}}\right)}{\left(\frac{2.00\text{N}}{0.225\text{kg}}-\frac{-5.00\text{N}}{0.600\text{kg}}\right)\left(0.100\text{s}\right)}\\ =1.92\text{s}\end{array}$

Write the equation for elapsed time from $t=0$ until the blocks meet.

$t=\Delta t+\Delta t_2$

Substitute $0.100\text{s}$ for $\Delta t$ and $1.92\text{s}$ for $\Delta t_2$ in the above equation to find $t$.

$\begin{array}{c}t=0.100\text{s+1}\text{.92}\text{s}\\ \text{=2}\text{.02}\text{s}\end{array}$

The distance from the block $B\text{'}s$ initial position to where the two blocks meet is negative the displacement of $B$.

Substitute $-5.00\text{N}$ for $F_B$,$0.600\text{kg}$ for $m_B$ and $0.100\text{s}$ for $\Delta t$ in the equation X to find $d_B$.

$\begin{array}{c}{d}_B=-\frac{1}{2}\left(\frac{-5.00\text{N}}{0.600\text{kg}}\right){\left(0.100\text{s}\right)}^2-\frac{-5.00\text{N}}{0.600\text{kg}}\left(0.100\text{s}\right)\left(1.924\text{s}\right)\\ =1.65\text{m}\end{array}$

Therefore, The distance from block B’s initial position is $1.65\text{m}$ and the time elapsed is $2.02\text{s}$.