Assuming that there R no other nonconservative forces involved wut impulse was givin 2 each truck at collision

Question

A.) Truck 1=
B.) Truck 2=

You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass(3,000 kg each) were involved in a rear-end accident at a stop sign. here is what you know:
Truck 1 approched the intersection from the top of a 22-meter hill.
Truck 2 was on a flat stretch of road directly in front of truck 1.
At the bottom of the hill, before braking for the stop sign, Truck 1 was going 20 m/s and truck 2 was going 35 m/s.
From the skid marks on the road you can see that truck 1 applied force on its brakes for 2 seconds, 80 meters before the stop sign.
There were no skid marks left by truck 2. The collision occured at the stop sign, where truck 2 had stopped.
After the collision, both trucks were moving together in the same direction at 10 m/s, before slowly rolling to a stop.
You must now push truck 2 using 1,000 N of force, 8 meters off the side of the road so no one else gets hurt.

Answer 1

First let's throw out the irrelevant data. Truck 2 was stopped and Truck 1 rear-ended it. How high the hill, how long and over what distance the brakes were hit, and the junk about pushing Truck 2 off the road don't apply. By "no other nonconservative forces" I assume the problem means no energy losses other than that of the collision, which conserves momentum but not energy. Momentum before the collision Mi = momentum after collision Mf. Mf = vf*(m1+m2) = 10*6000 = 60,000 kg-m/s. Mi = 60,000 = m1*vi1+m2*0, so Truck 1's initial velocity vi1 = 20 m/s. The collision added an impulse of +30,000 kg-m/s to Truck 1 and -30,000 kg-m/s to Truck 2.