Answer to Question #243008 in Physics for Sheyni

Question

A 1,000 kg car running at 50 km/h eastward a head-on collision with a 2,000 kg truck running at 30 km/h. If the two vehicles are locked together after collision, calculate the final velocity of the wreckage. What is the loss in kinetic energy of the system?

Answer 1

We can find the final velocity of the wreckage from the law of conservation of energy:

$$m_1v_{1i}-m_2v_{2i}=(m_1+m_2)v_f,$$ $$v_f=\dfrac{m_1v_{1i}-m_2v_{2i}}{m_1+m_2},$$ $$v_f=\dfrac{1000\ kg\cdot13.89\ \dfrac{m}{s}-2000\ kg\cdot8.33\ \dfrac{m}{s}}{1000\ kg+2000\ kg}=-0.92\ \dfrac{m}{s}.$$

The sign minus means that the final velocity of the wreckage directed to the west.

We can find the loss in kinetic energy of the system as follows:

$$\Delta KE=KE_i-KE_f,$$ $$\Delta KE=\dfrac{1}{2}m_1v_{1i}^2+\dfrac{1}{2}m_2v_{2i}^2-\dfrac{1}{2}(m_1+m_2)v_{f}^2,$$

$\Delta KE=\dfrac{1}{2}\cdot1000\ kg\cdot(13.89\ \dfrac{m}{s})^2+\dfrac{1}{2}\cdot2000\ kg\cdot(8.33\ \dfrac{m}{s})^2-\dfrac{1}{2}\cdot(1000\ kg+2000\ kg)\cdot(0.92\ \dfrac{m}{s})^2=165\ kJ.$

Answer:

$v_f=0.92\ \dfrac{m}{s},$ to the west.

$\Delta KE=165\ kJ.$