1350Kg tuned up Civic moving at 15m/s hits a 1450Kg car moving at 12m/s

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In summary, the conversation is about collisions and how to determine the final velocity of two cars in an elastic and inelastic collision. The participants discuss the equations and assumptions needed for each type of collision, and one person asks for help solving the problem.
  • #1
DrVirus
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Im not really good at elastic and inelastic collisions,, can someone help me with collisions please.

In the movie the fast and the furious, a 1350Kg tuned up Civic moving at 15m/s hits a 1450Kg car moving at 12m/s. Determine the final velocity of the cars if (a) the collision was elastic AND (b) the collision was inelastic.

THank you for any help you can give.
 
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  • #2
DrVirus,
you have 2 unknowns (the final velocities).

In an elastic collision, both momentum and kinetic energy are conserved. So 2 equations, fine.

In an inelastic collision, only momentum is conserved. To get the 2nd equation, you usually assume that the bodies stick together after collision.
 
  • #3
Well, its been an hour since i posted the question. I got the first part for elastic collision right.. but the inelastic i can't solve it, can someone show me how to solve it, because then tomorrow i have a lab that has to do with collisions and i have to understand this. Thanks alot
 
  • #4
As I said:
Momentum: pfinal=pinitial
Sticking: v1,final=v2,final=:vfinal
=>
(m1+m2)vfinal=pinitial

You have calculated pinitial in part (a), haven't you?
 
  • #5
No, i didnt solve for Pi. i used an equation that works with elastic collision only and has to have V2 as 0. so i change the frame of reference to be 0, and changed it back after i got the answer.
 
  • #6
pinitial=m1v1,initial+m2v2,initial

Just plug in.
 

1. What is the total momentum of the two cars after the collision?

The total momentum of the two cars after the collision is equal to the sum of their individual momentums, which can be calculated by multiplying the mass of each car by its velocity. In this case, the total momentum would be 1350kg x 15m/s + 1450kg x 12m/s = 37,950 kg*m/s.

2. What is the kinetic energy of the two cars before and after the collision?

The kinetic energy of a moving object is equal to half of its mass multiplied by the square of its velocity. Before the collision, the kinetic energy of the 1350kg Civic would be (1/2)(1350kg)(15m/s)^2 = 151,875 Joules. The kinetic energy of the 1450kg car would be (1/2)(1450kg)(12m/s)^2 = 104,400 Joules. After the collision, the total kinetic energy of the two cars would be (1/2)(1350kg + 1450kg)(15m/s)^2 = 202,500 Joules.

3. What is the final velocity of the two cars after the collision?

To calculate the final velocity of the two cars after the collision, we can use the conservation of momentum principle, which states that the total momentum of a closed system remains constant. This means that the total momentum before the collision must be equal to the total momentum after the collision. Using this principle, we can set up the following equation: (1350kg x 15m/s) + (1450kg x 12m/s) = (1350kg + 1450kg)v, where v is the final velocity of the two cars. Solving for v, we get a final velocity of 13.5 m/s.

4. What type of collision is this?

This is an inelastic collision, where the two cars stick together after the collision and move with a common final velocity. This can be seen by the fact that the final velocity of the two cars after the collision is less than the initial velocity of the Civic (15m/s).

5. What is the change in momentum of the 1350kg Civic?

The change in momentum of an object can be calculated by subtracting its initial momentum from its final momentum. In this case, the initial momentum of the Civic is (1350kg x 15m/s) = 20,250 kg*m/s, and the final momentum is (2800kg x 13.5m/s) = 37,800 kg*m/s. Therefore, the change in momentum of the 1350kg Civic is 37,800 kg*m/s - 20,250 kg*m/s = 17,550 kg*m/s.

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