For the linear conservation law:
#m_cv_(ic)+m_tv_(it)=m_cv_(fc)+m_tv_(ft)#
(where #i#=initial, #f#=final, #c#=car, #t#=truck).
But
#v_(it)=0# because the truck is initially still
and
#v_(fc)=v_(ft)=v_f# because they will proceed together,
so:
#m_cv_(ic)=(m_c+m_t)v_f#.
To know #(1)=v_(ic)=(m_c+m_t)/m_cv_f#, we have to know #v_f#.
The motion after the impact is a decelerated one, so:
the two laws of the accelerated motion are:
#s=s_0+v_0t+1/2at^2#
and
#v=v_0+at#.
But if you solve the system of the two equations, you can find a third law really useful in those cases in which you haven't the time, or you haven't to find it.
#(2)=v^2=v_0^2+2aDeltas# in which #Deltas# is the space run.
But we need to find #a#.
The resistence force (of the friction) is:
#F_f=muP=mu(m_c+m_t)g# where #mu# is the coefficient of friction.
For the second law of Newton:
#F_f=(m_c+m_t)arArr#
#mu(m_c+m_t)g=(m_c+m_t)arArra=mug#.
Now we can use the #(2)# law written before:
#v_0^2=v^2-2aDeltas=0^2+2mugDeltas#
(#a# is a deceleration! Remember that #v_0# is the initial velocity of this new motion that is the same of the final of the first motion!).
#v_0=sqrt(2mugDeltas)=(v_f)#.
Now, using the #(1)# equation:
#v_(ic)=(m_c+m_t)/m_csqrt(2mugDeltas)=#
#=((920+2300)Kg)/(920Kg)*sqrt(2*0.8*9.8m/s^2*2.8m)=23.19m/s#.
