Answer to Question #164642 in Mechanics | Relativity for Lucy green

Question

A car with a mass of 900.0 kg is moving at 26.0 m/s [E] when it crashes into a truck with a mass of 1500.0 kg moving at 30.0 m/s [W]. After the 0.4 s collision, both vehicles are coupled together and have the same velocity. Find the final velocity and average net force of each vehicle after the collision.

Answer 1

(a) Let the east be the positive direction. Then, according to the law of conservation of momentum, we have:

$$m_1v_1-m_2v_2=(m_1+m_2)v_f,$$ $$v_f=\dfrac{m_1v_1-m_2v_2}{(m_1+m_2)},$$ $$v_f=\dfrac{900\ kg\cdot26\ \dfrac{m}{s}-1500\ kg\cdot30\ \dfrac{m}{s}}{(900\ kg+1500\ kg)}=-9\ \dfrac{m}{s}.$$

The sign minus means that the couple of the car and truck moves to the west.

(b) Let's find the average net force of each vehicle after the collision:

$$F_{avg}\Delta t=m|(v_f-v_i)|,$$ $$F_{avg}=\dfrac{m|(v_f-v_i)|}{\Delta t},$$ $$F_{avg, car}=\dfrac{900\ kg\cdot|(-9\ \dfrac{m}{s}-26\ \dfrac{m}{s})|}{0.4\ s}=78750\ N,$$ $$F_{avg, truck}=\dfrac{1500\ kg\cdot|(-9\ \dfrac{m}{s}-30\ \dfrac{m}{s})|}{0.4\ s}=146250\ N.$$