A $30$ton rail car and a $90$ton rail car, initially at rest, are connected together with a giant but massless compressed spring between them. When released, the $30$ton car is pushed away at a speed of $4.0m/s$relative to the $90$ ton car. What is the speed of the $30$ ton car relative to the ground?

We need to find the the speed of the car relative to the ground.

The velocity of $30ton$car relative to the ground. so this car is relative to the $90ton$is $4m/s$in opposite direction. So the velocity of $30ton$car that relative to the ground $v_{30}$is calculated by

role="math" localid="1649424236193" $v_{30}=-4m/s-v_{90}\left(1\right)$

$90ton$car's velocity is $v_{90}$that is relative to the ground. we need to find the expression for $v_{90}$and both cars are at the rest, if using the conversation law of momentum,

$$\begin{gathered} \stackrel{\rightharpoonup }{P_i}=\stackrel{\rightharpoonup }{P_f} \\ {m}_{30}{v}_{30}=m_{90}{v}_{90} \\ {v}_{90}=\frac{m_{30}{v}_{30}}{m_{90}} \end{gathered}$$

Using expression of $v_{90}$into equation (1) to get $v_{30}$by

$v_{30}=(4m/s)-\frac{\left(m_{30}{v}_{30}\right)}{\left(m_{90}\right)}\left(2\right)$

Putting the values for $m_{30}and{m}_{90}$in equation (2) to get $v_{30}$

$$\begin{gathered} v_{30}=(4m/s)-\frac{\left(m_{30}{v}_{30}\right)}{\left(m_{90}\right)} \\ {v}_{30}=(4m/s)-\left(\frac{30ton}{90ton}\right)v_{30} \\ {v}_{30}=-3m/s \end{gathered}$$