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#.