How fast are these particles sliding when they pass?

How fast are these particles sliding when they pass?

Question

Two particles $A$ and $B$ are placed side by side on rough horizontal ground, and are simultaneously projected along the ground in the same direction with speeds $14\,{\mathrm{m\,s^{-1}}}$ and $21\,{\mathrm{m\,s^{-1}}}$ respectively. $A$ comes to rest after travelling $70\,\text{m}$, and $B$ after travelling $35\,\text{m}$. Calculate the coefficient of friction between each particle and the ground, and show that the times for which the particles are in motion are in the ratio $3:1$.

Show also that $A$ passes $B$ before either particle comes to rest, and calculate the speed of each particle when this happens.

Take $g$ to be $9.8\,{\mathrm{m\,s^{-2}}}$.

Solution

Two particles $A$ and $B$ are placed side by side on rough horizontal ground, and are simultaneously projected along the ground in the same direction with speeds $14\,{\mathrm{m\,s^{-1}}}$ and $21\,{\mathrm{m\,s^{-1}}}$ respectively. $A$ comes to rest after travelling $70\,\text{m}$, and $B$ after travelling $35\,\text{m}$. Calculate the coefficient of friction between each particle and the ground, and show that the times for which the particles are in motion are in the ratio $3:1$.

Take $g$ to be $9.8\,{\mathrm{m\,s^{-2}}}$.

Let’s start by drawing a diagram.

The particles are moving so friction is a maximum and we can use the relationship $F=\mu R$, where $R$ is the normal reaction and $\mu$ is the coefficient of friction. $F$ is the only horizontal force on the object.

We know $R=mg$ and Newton’s 3rd law tells us that $F=ma$ leading to $\mu=a/g$ (where $a$ is directed to the left, in the same direction as the frictional force).

To find the coefficient of friction for each particle we must first find its acceleration. For this we can use the equation $v^2=u^2+2as$.

$a_A = \dfrac{0-14^2}{2\times70} =\dfrac{-196}{140} = -\dfrac{7}{5}$ so $\mu_A=\dfrac{7}{5g}=\dfrac{1}{7}$.

$a_B = \dfrac{0-21^2}{2\times35} =\dfrac{-441}{70} = -\dfrac{63}{10}$ so $\mu_B=\dfrac{63}{10g}=\dfrac{9}{14}$.

To find out the time of travel we can use $v=u+at$.

$t_A = \dfrac{0-14}{-7/5}=10$ and $t_B = \dfrac{0-21}{-63/10}=\dfrac{10}{3}$.

So the ratio $t_A:t_B=10:\frac{10}{3}=3:1$ as required.

Using

$s=ut+\frac{1}{2} at^2$

to find distances at time

$t$

,

$$\begin{align*} s_A &=14t-\frac{7}{10}t^2 \\ s_B &=21t-\frac{63}{20}t^2. \end{align*}$$

They pass when $s_A=s_B$, so subtracting the above equations, $7t-\frac{49}{20}t^2=0$ which means $t=0$ (i.e. the start) or $t=\frac{20}{7}$.

At this time $A$ has speed $14-\frac{7}{5} \times \frac{20}{7}= 10\,{\mathrm{m\,s^{-1}}}$ and $B$ has speed $21-\frac{63}{10} \times \frac{20}{7}=3\,{\mathrm{m\, s^{-1}}}$. This is clearly before either particle has come to rest.