Question

A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s. (a) What is the speed of the three coupled cars after the collision? 1.27 Incorrect: Your answer is incorrect. m/s (b) How much kinetic energy is lost in the collision?

Answer 1

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

$m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}$

in this case:

$3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s$

the initital kinetic energy is:

$K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ$

and the final:

$K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ$

The energy lost is given by:

$E_l=|K_f-K_i|\\E_l=23kJ$