How Do You Calculate Vehicle Speeds After a Collision?

[mandos]

Homework Statement

The diagram below shows a skethc drawn by an accident investigator following a head-on collision between two vehicles.

Direction of Travel A (speed 12.5 m/s) and 2400kg -------------> <---------------- Direction of Travel of B, 3600kg, unknown velocity.Final Position of vehicles:

The diagram shows two boxes, repsenting the vehicles, attached together. One box is A, and is 2400 kg. The second is 3600 kg. Also, is X (where the vehicles collided) and an arrow to the left with 8.4m, which is the distance the vehicles travelled.

The Question: Determine the speed of the interlocked vehicles immediately after impact.

The second question: Vehicle A was known to be moving at 12.5m/s just before the impact. Calculate the speed of vehicle B just before impact.

Homework Equations

Right, for the first question I assume the equation Kinetic Energy = 1/2 mv^2 is relevant. The answer is a combined speed of 4.6 m/s but unfortunately, there is little point in me knowing the answer unless I can figure out how to get there.

For the second question, I am clueless what sort of equation would be used. However, I know the answer is 16 m/s.

The Attempt at a Solution

For the first question, I tried to do Kinetic Energy = 1/2 mv^2 = 1/2 mv^2 to say that the total energy when the two vehicles collide is the same and rearrange it to find the second velocity. But in retrospect, that seems to be going down the wrong avenue?

Other than that, I can't figure out what else to use or try.

I'd appreciate anyone who can shed some insight or help. If you don't understand the question, I'll elaborate as best I can.

Thanks.

 

[tiny-tim]

For the first question, I tried to do Kinetic Energy = 1/2 mv^2 = 1/2 mv^2 to say that the total energy when the two vehicles collide is the same and rearrange it to find the second velocity. But in retrospect, that seems to be going down the wrong avenue?

Hi mandos!

Yes, this is a totally inelastic collision, so KE isn't conserved.

What is the whole question?

There isn't enough information to solve this.

 

[thomate1]

Your approach for the first question is on the right track. The conservation of energy principle states that the total energy of a closed system remains constant. In this case, the two vehicles are considered a closed system as they collide and stick together. Therefore, the initial kinetic energy of the system (when the two vehicles are moving with their respective velocities) is equal to the final kinetic energy of the system (when the vehicles are interlocked and moving with a combined velocity).

So, you can set up the equation as follows:

Initial Kinetic Energy = Final Kinetic Energy
1/2 (2400 kg) (12.5 m/s)^2 + 1/2 (3600 kg) (unknown velocity)^2 = 1/2 (6000 kg) (combined velocity)^2

Solving for the unknown velocity, we get:

(Unknown velocity)^2 = (1/2 (2400 kg) (12.5 m/s)^2 + 1/2 (3600 kg) (unknown velocity)^2) / (1/2 (6000 kg))

(Unknown velocity)^2 = 2.5 (2400 kg) (12.5 m/s)^2 + 2.5 (3600 kg) (unknown velocity)^2

(Unknown velocity)^2 = 60000 kg m^2/s^2 + 90000 kg m^2/s^2

(Unknown velocity)^2 = 150000 kg m^2/s^2

(Unknown velocity) = √(150000 kg m^2/s^2)

(Unknown velocity) = 122.5 m/s

Therefore, the combined velocity of the two vehicles immediately after impact is 122.5 m/s.

For the second question, you can use the same approach but this time, you know the velocity of one of the vehicles (12.5 m/s) and you need to find the velocity of the other vehicle.

So, the equation becomes:

Initial Kinetic Energy = Final Kinetic Energy
1/2 (2400 kg) (12.5 m/s)^2 + 1/2 (3600 kg) (unknown velocity)^2 = 1/2 (6000 kg) (combined velocity)^2

Solving for the unknown velocity, we get:

(Unknown velocity)^2 = (1/2 (2400 kg) (12.5 m/s)^2 + 1/2 (3600 kg) (unknown velocity