Question

A car of mass m moving at a speed v1 collides and couples with the back of a truck of mass 3m moving initially in the same direction as the car at a lower speed v2. (Use any variable or symbol stated above as necessary.) (a) What is the speed vf of the two vehicles immediately after the collision? vf = (b) What is the change in kinetic energy of the car❝truck system in the collision? ?K =_______.

Answer 1

Answer:

$v_{f} = \frac{mv_{1} +3mv_{2} }{4m}= \frac{v_{1} +3v_{2}}{4}$

$change in K = (\frac{1}{2} mv_{1}^{2} - \frac{1}{2}.3mv_{2} ^{2} ) - \frac{1}{2}(4m)v_{f}^{2}$

Explanation:

Using the principle of conservation of momentum

momentum before the collision  =  momentum after the collision

$mv_{1} + 3mv_{2}  = (m+3m)v_{f}\\mv_{1} + 3mv_{2}  =4mv_{f}\\v_{f} = \frac{mv_{1} +3mv_{2} }{4m}= \frac{v_{1} +3v_{2}}{4}$

Change in kinetic energy =  sum of kinetic energies after the collision -  sum of kinetic energies before the collision

$change in K = (\frac{1}{2} mv_{1}^{2} - \frac{1}{2}.3mv_{2} ^{2} ) - \frac{1}{2}(4m)v_{f}^{2}$

The simplification for kinetic energy can be made by substitution the answer for the first part into the second part