A railroad car with a mass of 2.08* 10^4kg moving at 3.16 m/s joins with two railroad cars already joined together each with the same mass as the single car and initially moving in the same direction of 1.42 m/s. What will be the speed of the cars after the collision and what will be the change in kinetic energy of the system?
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This type of collision is an example of an inelastic collision. In this type of collision the momentum of the system is conserved, however kinetic energy is not.
To answer the question we must determine the total momentum before the collision. We can then set that equal to the total momentum after the collision and solve for the missing velocity.
MV1 + 2MV2 = 3MV
Where M is the mass of single car, V1= 3.16 m/s, V2 = 1.42 m/s ans is the two cars joined together, and V is the speed of all three cars after the collision.
We solve for V
V = (MV1+2MV2)/3M Factoring out the M and cancelling common factors
V = M(V1 + 2V2)/3M = (V1 + 2V2)/3
Plugging in the values for V1 and V2 gives
V = 2.00 m/s
So, the three cars together continue moving at 2.00 m/s
To get the loss in kinetic energy, we calculate the KE of all three cars before the collision and subtract the KE of the bound together cars after the collision:
Delta KE = 1/2 MV1^2 +1/2 (2M)V2^2 - 1/2 (3M)V^2
Delta KE = 1/2 M(V1^2 + 2V2^2 - 3V^2)
Replacing with the values for M, V1, V2, and V gives
Delta KE = 1.04 X 10^4(3.16^2 +2X1.42^2 - 12) joules
Delta KE = 2.10 X 10^4 joules
That is, the system has lost 2.10 X10^4 joules of kinetic energy into some other form of energy.
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