Problem 174
Easy Translational Dynamics

A railway wagon of mass m_1 = 8 t , standing still, carries…

A railway wagon of mass $m_1 = 8\ \mathrm{t}$, standing still, carries a rocket launcher. Two rockets of mass $m_2 = 80\ \mathrm{kg}$ each are launched at the same time with speed $v_0 = 1000\ \mathrm{m/s}$, horizontally along the track. How far does the wagon move because of the launch, if the total friction coefficient during its motion is $\mu = 0.06$?

Explanation of Method

A wagon recoiling from two rockets.

Step 1: Recoil speed

The total momentum stays zero ($m_1$ is the wagon's mass after the launch): $m_1 v_1 = 2 m_2 v_0$.

\[ v_1 = \frac{2 \cdot 80 \cdot 1000}{8000} = 20\ \mathrm{m/s}\ \text{(backward)}. \]

Step 2: Braking by friction

\[ a = \frac{\mu m_1 g}{m_1} = \mu g = 0.589\ \mathrm{m/s^2}, \]

\[ s = \frac{v_1^2}{2\mu g} = \frac{400}{1.177}. \]

\[s \approx 340\ \mathrm{m}\]

$2 \times 80\ \mathrm{kg},\ 1000\ \mathrm{m/s}$ $v = 20\ \mathrm{m/s}$ $8000\ \mathrm{kg},\ \mu = 0.06$ $\text{then brakes: } s = \frac{v^2}{2\mu g} \approx 340\ \mathrm{m}$
The rockets carry momentum $2 \cdot 80 \cdot 1000$ forward, so the wagon gets the same momentum backward. Friction then stops it.

Result

\[\boxed{\,s \approx 340\ \mathrm{m}\,}\]

Remark

Check. The original result (339.8 m) is correct. If the 8 t already included the rockets, the wagon alone would have 7840 kg, $v_1 = 20.4\ \mathrm{m/s}$ and $s \approx 354\ \mathrm{m}$.