Inelastic collision problem - question doesn't make sense

[randomkent]

Homework Statement

Q5. A 1500kg car is traveling east @ 15.0m/s, when it crashes into a 2500kg truck traveling in the same direction @ 12m/s. The car has a velocity of 13m/s affter the collision. Assuming this is an inelastic collision, determine:
a) speed of the truck just after the collision
b) how much kinetic energy is lost during the collision. Into what is this KE converted?

Homework Equations

m1u1 + m1u2 = m1v2 + m1v2
m1u1 + m1u2 = Vf(m1 + m2)
0.5mu^2 + 0.5mu^2 ≠ 0.5mv^2 + 0.5mv^2

The Attempt at a Solution

This is a question from a past exam paper, and I need to know this for my upcoming exam. The solutions page uses m1u1 + m1u2 = m1v2 + m1v2 for part a to get 13.2. They then used this info to get 4200J (lost) as heat and sound for part b.

From my understanding, the answer is already given (13); because aren't inelastic collisions meant to move at the same speed? Since they are moving at different speeds it must actually be ELASTIC, but then the answer is already there and part b would be invalid because no KE would be lost... The only thing I can think of is the fact that it says JUST AFTER the collision, but still the car and truck should stick together and move at the same speed, right? m1u1 + m1u2 = Vf(m1 + m2) couldn't be used because it says the car after is 13m/s; and if they stick together they would both be @ 13.125m/s to conserve momentum...

 

[Infinitum]

Homework Statement

Q5. A 1500kg car is traveling east @ 15.0m/s, when it crashes into a 2500kg truck traveling in the same direction @ 12m/s. The car has a velocity of 13m/s affter the collision. Assuming this is an inelastic collision, determine:
a) speed of the truck just after the collision
b) how much kinetic energy is lost during the collision. Into what is this KE converted?

Homework Equations

m1u1 + m1u2 = m1v2 + m1v2
m1u1 + m1u2 = Vf(m1 + m2)
0.5mu^2 + 0.5mu^2 ≠ 0.5mv^2 + 0.5mv^2

The Attempt at a Solution

This is a question from a past exam paper, and I need to know this for my upcoming exam. The solutions page uses m1u1 + m1u2 = m1v2 + m1v2 for part a to get 13.2. They then used this info to get 4200J (lost) as heat and sound for part b.

From my understanding, the answer is already given (13); because aren't inelastic collisions meant to move at the same speed? Since they are moving at different speeds it must actually be ELASTIC, but then the answer is already there and part b would be invalid because no KE would be lost... The only thing I can think of is the fact that it says JUST AFTER the collision, but still the car and truck should stick together and move at the same speed, right? m1u1 + m1u2 = Vf(m1 + m2) couldn't be used because it says the car after is 13m/s; and if they stick together they would both be @ 13.125m/s to conserve momentum...

If it were a completely inelastic collision your analysis would be correct. Both would move with the speed 13.125m/s, stuck together. The question has specified an 'inelastic' collision, but it has that the car finally moves with 13m/s. This means, the collision is partially elastic, which should have been mentioned in the question.

 

[randomkent]

Yeah, I guess that is the only way for it to be correct; if it was partially elastic. If we ignored "the car has a velocity of 13m/s affter the collision" and said they both moved at 13.125m/s, 4218J would be lost which is not that much more than 4200J, but that's irrelevant anyway... However, this should actually prove that it is partially elastic shouldn't it; less energy is lost. Thanks for your help!

 

[Infinitum]

Yeah, I guess that is the only way for it to be correct; if it was partially elastic. If we ignored "the car has a velocity of 13m/s affter the collision" and said they both moved at 13.125m/s, 4218J would be lost which is not that much more than 4200J, but that's irrelevant anyway... However, this should actually prove that it is partially elastic shouldn't it; less energy is lost. Thanks for your help!

Yep, you got it!