Hi Omar M.!
Step 1: Identify Collision Type
Just as we did in the previous question, we must first identify which type of collision we are considering so that we know which assumptions we may make.
In this case, we are told that the impact approximates an elastic collision.
In elastic collisions, both momentum and kinetic energy are conserved.
Step 2: Conservation of Momentum
Now that we have identified which collision type we are working with, we are once again prepared to set up our equations. In elastic, inelastic and perfectly inelastic collisions, we can assume that momentum is conserved. Only in elastic collisions can we assume that kinetic energy is conserved. This tells us that the initial kinetic energy must equal the final kinetic energy. We will make use of this information later in our solution.
Let us start with the conservation of momentum:
m1v1o + m2v2o = m1v1f + m2v2f
=> v1o = 0 because the car is initially at rest
m2v2o = m1v1f + m2v2f
Step 3: Conservation of Kinetic Energy
v1o + v1f = v2o + v2f
=> v1o = 0 because the car is initially at rest
v1f = v2o + v2f
Step 4: Solve
First, we will start with our first equation derived form the conservation of momentum. We will then substitute in our second equation derived from the conservation of kinetic energy. Finally, we will solve for both final velocities.
m2v2o = m1v1f + m2v2f
=> v1f = v2o + v2f
m2v2o = m1(v2o + v2f) + m2v2f = m1v2o + m1v2f + m2v2f
m2v2o - m1v2o = m1v2f + m2v2f = v2f(m1 + m2)
v2f = v2o(m2 - m1)/(m1 + m2)
<=> [(15.5 m*s-1)(1720 kg - 730 kg)]/(1720 kg + 730 kg)
v2f = 6.3 m*s-1 (final velocity of the truck)
v1f = v2o + v2f
<=> (15.5 m*s-1 + 6.3 m*s-1)
v1f = 22 m*s-1 (final velocity of the car)
Hope this helps!
Cheers
