MCAT Passage: Inelastic Collision

In summary, the problem involves a collision between a car and a truck on a track, with both vehicles accelerating at the same rate before the collision. The collision is perfectly inelastic and the vehicles skid for 7m before coming to rest. The total kinetic energy of the vehicles just after the collision can be solved using either momentum conservation or an energy approach, with the given coefficient of kinetic friction being used in the latter method. The correct answer is B. 147 kJ.
  • #1
mrlucky0
69
1
[SOLVED] MCAT Passage: Inelastic Collision

Homework Statement



A car (1000 kg) and a truck (2000kg) start from rest on a long, straight track. At time t=0, the truck is at position x=0 and the car is at position x=100m. Both vehicles then accelerate toward each other and collide.

Both the truck and the car accelerate uniformly - and at the same rate - until the speed of the each vehicle is 21 m/s, just as they collide. The resulting collision is perfectly inelastic, and the vehicles skid for 7 m before coming to rest. The coefficient of kinetic friction between the track and the vehicle's tires is 0.35.

What was the total kinetic energy of the vehicles just after the collision?

A. 73.5 kJ
B. 147 kJ
C. 330.8 kJ
D. 661.5 kJ

Homework Equations



KE = 1/2mv^2

The Attempt at a Solution



Can I apply the conservation of momentum here? I'm also not sure how to use the coefficient of kinetic friction although I'm sure the solution involves it.

I've determined that the initial total KE of the system is 661.5 kJ and was tempted to choose that as the answer but it's not right.

I'm also confused about the wording of the problem (if someone can clarify): does "just after the collision" imply after the cars have crashed but before the combined mass starts to skid? If this is the case, why bother with kinetic friction?
 
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  • #2
Hi mrlucky0,

Yes, I believe "just after the collision" means after the crash and before any appreciable skidding has occurred. I think the reason they give the coefficient of friction is so you can solve the problem in two ways: either use momentum conservation during the collision, or use an energy approach after the collision.
 
  • #3
alphysicist said:
Hi mrlucky0,

Yes, I believe "just after the collision" means after the crash and before any appreciable skidding has occurred. I think the reason they give the coefficient of friction is so you can solve the problem in two ways: either use momentum conservation during the collision, or use an energy approach after the collision.

Thanks. Your tip helped me solve the problem!
 

1. What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. In other words, the total kinetic energy of the system before and after the collision is not the same.

2. How is an inelastic collision different from an elastic collision?

In an elastic collision, kinetic energy is conserved, meaning the total kinetic energy of the system before and after the collision remains the same. In an inelastic collision, kinetic energy is lost due to factors such as deformation or heat.

3. What is the formula for calculating the final velocity in an inelastic collision?

The formula for calculating the final velocity in an inelastic collision is: vf = (m1v1 + m2v2) / (m1 + m2), where vf is the final velocity, m1 and m2 are the masses of the objects involved, and v1 and v2 are the initial velocities of the objects.

4. Can an inelastic collision result in an increase in kinetic energy?

No, an inelastic collision cannot result in an increase in kinetic energy. Kinetic energy is always lost in an inelastic collision due to factors such as deformation or heat.

5. How does the concept of momentum apply to inelastic collisions?

Inelastic collisions still follow the law of conservation of momentum, which states that the total momentum of a closed system remains constant. However, in an inelastic collision, the total momentum may be conserved while kinetic energy is not, leading to a decrease in the velocity of the objects involved.

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