We can assume this, since no information tells us that the road
was wet.
W = M*g; where g = 9.81m/s^2, the acceleration of gravity
The frictional force F decelerates the two entangled cars to a
complete stop in 18 m, so:
F = [1.07)*M*g = M*A; where A is the deceleration magnitude.
A = (1.07)*g
and (1/2)*A*t^2 = 18 m, so
t^2 = (36 m)/A = (36 m)/(1.07*g)
t = sqrt[36/(1.07*g)] = 1.852 s
Calculate t and then the average speed after the collision was
(18 m)/t = 9.72 m/s
Average speed is half the initial speed, Vi, so
Vi = (36 m)/t = 19.44 m/s {answer to first question}
Driver of the blue car was 0.5 m above the ground and went 8 m horizontally in the time that it took him to reach the ground.
Let T be his time of fall, so that
(1/2)*g*T^2 = 0.5 m
T^2 = (2*0.5 m)/g = 0.102 s, so
T = 0.319 s
His average horizontal speed was
(8 m)/0.319s = 25.08 m/s This is Vb, how fast the blue car was traveling at the time of impact.
Because the perpendicular components of momentum for the two
cars was equal, their "tangential" components had to also be
equal: Mr*Vr*cos(15 deg) = Mb*Vb*cos(15 deg). The total
tangential component of momentum before the crash was thus
(Mr*Vr + Mb*Vb)cos(15 deg), and the momentum after the crash
and in this same direction must equal this amount
(Mr + Mb)*Vi = (Mr*Vr + Mb*Vb)cos(15 deg)
(19.44 m/s)*(Mr + Mb) = (Mr*Vr + Mb*25.08 m/s)*(0.966)
There is not enough information given to solve for other values unless we make the assumption that the masses of the two cars are equal: that is, Mr = Mb.
(Mr + Mb) = (Mr*Vr + Mb*25.08 m/s)*(0.966)/(19.44 m/s) Now divide by Mr to get:
2 = 0.*[Vr + 25.08 m/s]
Vr + 25.08 m/s = 2/(0.0497) = 40.24 m/s
So that Vr = (40.24 -25.08) = 15.16 m/s . This is the speed of the red car just before the crash.
For the other questions, I see no information that lets you calculate what the mass common to both cars might be. However, you can find the momenta and the loss of kinetic energy as a function of the common mass (Mr or Mb), you just cannot get a final numerical answer without more information about the mass.
For the kinetic energy lost, you could find it as a fraction or a percentage of the initial energy, since then the mass will cancel out of the calculation.
And that's all I have to offer on this problem.