# Question #e2720

Jun 13, 2015

$v = 23.19 \frac{m}{s} = 83.49 \frac{K m}{h}$.

#### Explanation:

For the linear conservation law:

${m}_{c} {v}_{i c} + {m}_{t} {v}_{i t} = {m}_{c} {v}_{f c} + {m}_{t} {v}_{f t}$

(where $i$=initial, $f$=final, $c$=car, $t$=truck).

But

${v}_{i t} = 0$ because the truck is initially still

and

${v}_{f c} = {v}_{f t} = {v}_{f}$ because they will proceed together,

so:

${m}_{c} {v}_{i c} = \left({m}_{c} + {m}_{t}\right) {v}_{f}$.

To know $\left(1\right) = {v}_{i c} = \frac{{m}_{c} + {m}_{t}}{m} _ c {v}_{f}$, we have to know ${v}_{f}$.

The motion after the impact is a decelerated one, so:

the two laws of the accelerated motion are:

$s = {s}_{0} + {v}_{0} t + \frac{1}{2} a {t}^{2}$

and

$v = {v}_{0} + a t$.

But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.

$\left(2\right) = {v}^{2} = {v}_{0}^{2} + 2 a \Delta s$ in which $\Delta s$ is the space run.

But we need to find $a$.

The resistence force (of the friction) is:

${F}_{f} = \mu P = \mu \left({m}_{c} + {m}_{t}\right) g$ where $\mu$ is the coefficient of friction.

For the second law of Newton:

${F}_{f} = \left({m}_{c} + {m}_{t}\right) a \Rightarrow$

$\mu \left({m}_{c} + {m}_{t}\right) g = \left({m}_{c} + {m}_{t}\right) a \Rightarrow a = \mu g$.

Now we can use the $\left(2\right)$ law written before:

${v}_{0}^{2} = {v}^{2} - 2 a \Delta s = {0}^{2} + 2 \mu g \Delta s$

($a$ is a deceleration! Remember that ${v}_{0}$ is the initial velocity of this new motion that is the same of the final of the first motion!).

${v}_{0} = \sqrt{2 \mu g \Delta s} = \left({v}_{f}\right)$.

Now, using the $\left(1\right)$ equation:

${v}_{i c} = \frac{{m}_{c} + {m}_{t}}{m} _ c \sqrt{2 \mu g \Delta s} =$

$= \frac{\left(920 + 2300\right) K g}{920 K g} \cdot \sqrt{2 \cdot 0.8 \cdot 9.8 \frac{m}{s} ^ 2 \cdot 2.8 m} = 23.19 \frac{m}{s}$.