Question

A 1470-kilogram truck moving with a speed of 27.0 m/s runs into the rear end of a 1010-kilogram stationary car. If the collision is completely inelastic, how much kinetic energy is lost in the collision

Answer 1

Answer:

$\Delta KE = 218.375\ kJ$

Explanation:

Given,

Mass of the truck, M =  1470 Kg

initial speed of the truck, u = 27 m/s

mass of car, m = 1010 Kg

initial speed of car, u' = 0 m/s

Final speed, V = ?

Using conservation of momentum

$Mu = (M+m) V$

$1470\times 27 = (1470 + 1010 ) V$

$V = 16\ m/s$

Change in kinetic energy

$\Delta KE = \dfrac{1}{2} Mu^2 - \dfrac{1}{2}(M+m)V^2$

$\Delta KE = \dfrac{1}{2}\times 1470\times 27^2 - \dfrac{1}{2}\times (1470 + 1010)\times 16^2$

$\Delta KE = 218.375\ kJ$

Change in KE = $\Delta KE = 218.375\ kJ$