Hi Omar M.!
Step 1: Identify Collision Type
We must first identify which type of collision we are considering so that we know which assumptions we can make.
Because the two objects stick together following the collision, this must be a perfectly inelastic collision.
In perfectly inelastic collisions, momentum is conserved, but kinetic energy is not.
Step 2: Conservation of Momentum
Now that we have identified which collision type we are considering, we are 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 suggests that the system will lose kinetic energy due to this collision. So, we should expect a negative value for our change in kinetic energy.
Let us start with the conservation of momentum:
m1v1o + m2v2o = m1v1f + m2v2f
=> m2v2o falls out because v2o = 0 (the second cart is at rest)
m1v1o = m1v1f + m2v2f
=> v1f + v2f = vf (because the two objects stick together following the collision, they will travel together with the same velocity)
m1v1o = m1vf + m2vf = (m1 + m2)vf => vf = m1v1o/(m1 + m2)
<=> (0.35kg*0.95m*s-1)/(0.35kg + 0.55kg)
vf = 0.37 m*s-1
Step 3: Change in Kinetic Energy
Now that have ascertained the final velocity of the system, we are ready to find the change in the kinetic energy due to the collision between these two objects.
ΔKE = KEf - KE0 = [(1/2)(m1 + m2)vf2] - [(1/2)m1v1o2 + (1/2)m2v2o2]
=> v2o = 0 m*s-1 because it is at rest
ΔKE = KEf - KE0 = [(1/2)(m1 + m2)vf2] - [(1/2)m1v1o2]
<=> [(1/2)(0.35kg + 0.55kg)(0.37m*s-1)2] - [(1/2)(0.35kg)(0.95m*s-1)2]
ΔKE = -0.097J
Hope that helps!
Cheers
