Totally Inelastic Collisions question

[CaptainOfSmug]

Homework Statement

A 2000-kg truck is sitting at rest (in neutral) when it is rear-ended by a 1000-kg car going 28m/s .After the collision, the two vehicles stick together.
#1What is the final speed of the car-truck combination?
#2What is the kinetic energy of the two-vehicle system before the collision?
#3What is the kinetic energy of the system after the collision?
#4Based on the results of the previous parts, what can you conclude about which type of collision this is?
#5Calculate the coefficient of restitution for this collision.

Homework Equations

The Attempt at a Solution

Okay, so for the record I got all these answers correct (masteringphysics) if you've had to use this horrendous program :p

#1So for the final speed of the car-truck combination:
1000(28)+2000(0)=m₁₂(V_12xf
=9.33m/s
#2 The kinetic energy of the two vehicle system before the collision:
K_car=0.5(mv²)
.5(1000)(28)^2
=392000J
#3 KE after the collision:
K_truckandcar=.5(1000+2000)(9.33)²
#4 what type of collision is it?
I know the answer is totally inelastic but I cannot figure out why? By my books definition the final velocity of the truck and car must equal 0? Am I wrong here or can someone explain this for me?
#5 Since the answer to the previous question was totally inelastic I can assume e=0, but when I calculate it out this is not the case. I need an explanation on this to help my understanding of this! Thanks in advance!

 

[NascentOxygen]

Hi CoS. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif

from Wikipedia:-
In a perfectly inelastic collision, i.e., a zero coefficient of restitution, the colliding particles stick together. In such a collision, kinetic energy is lost by bonding the two bodies together. This bonding energy usually results in a maximum kinetic energy loss of the system.

 

[CaptainOfSmug]

Thank you for your reply, but I'm wondering when calculating the coefficient of restitution which is e=(V_12xf)/(v_12xi) isn't zero or am I missing something? Isn't the velocity of the two cars 9.33m/s? Shouldn't it actually be zero for e to =0?

 

[NascentOxygen]

coefficient of restitution involves relative velocities. When the bodies stick together their relative velocity is zero. After the collision they are often moving but locked together, so while some K.E. may be lost, it is not all lost.

 

[HalfLostAndSearching]

In a totally inelastic collision, the two objects stick together after the collision and move with a common final velocity. In this case, the final speed of the car-truck combination is indeed 9.33 m/s, which is not equal to 0. This is because in a totally inelastic collision, some kinetic energy is lost due to the deformation of the objects involved. In this case, the kinetic energy of the system after the collision is less than the kinetic energy before the collision, which is consistent with a totally inelastic collision.

To calculate the coefficient of restitution for this collision, we can use the formula e = (V12xf-V12xi)/(V21xi-V12xi), where V12xf is the final velocity of the car-truck combination, V12xi is the initial velocity of the car, and V21xi is the initial velocity of the truck. Plugging in the values, we get e = (9.33-28)/(0-28) = 0.33. This means that the coefficient of restitution for this collision is 0.33, which is less than 1 and consistent with a totally inelastic collision.