You are a professional witness in a lawsuit. What do you tell the judge? What do
ID: 2120208 • Letter: Y
Question
You are a professional witness in a lawsuit. What do you tell the judge? What do you tell the Physicist (Read the following scenario.) You are on an RTD Stretch Bus (the long busses with the accordion in the middle. (This is called an articulating bus.)) The bus is fully loaded with fuel and people. You are traveling at 30 [mph] when a mini-Van in front of the bus pulls in front of the bus as it travels in a similar direction; the van is traveling 20 [mph]. The bus driver slams on the brakes, but there is black ice so the bus gracefully slides right into the van sending the van flying ahead. Three passengers, upon realizing their opportunity to make some quick cash, start crying of back and neck pain. Mass of bus + passengers + fuel is about 66,600 [lb] or 30,200 [kg]. Mass of van is about 2,000 [kg]. Please Note: A few students have contacted me and said that I made an error converting from pounds to kilograms. Here are my conversion notes: 66600*4.4492454 /9.8 = 30236.709 #[lb] * 4.4492454[N] / 1 [lb] / 9.8[m/s/s]---> kg 66600*0.454 = 30236.400 #[lb] * 0.454...[kg] / 1 [lb] ---> kg See Wolfram Alpha for more information. 1 pound is about 4.45 Newton. Forensics estimates from the bending of the light-pole that the van was moving at an uninhibited speed of about 35 [mph] when it hit the pole after sliding on ice. You are a professional witness; what is your statement to the judge and why? (Three other passengers tell the judge they didn't even feel the collision. They heard it and saw the van fly forward; but they claim they felt nothing.) You are a professional witness; what is your statement to the Physicist and why?
Explanation / Answer
Mass of the bus m1 = 30,200 kg
Mass of the van m2 = 2,000 kg
Initial speed of the bus u1 = 30 mph
Initial speed of the van u2 = 20 mph
The final speed of the van is v2 = 35 mph
As there is no froction acting while this collision happend, the collision be considered as elastic and according to the
principle of conservation of linear momentum,
m1u1+m2u2 = m1v1+m2v2
Where v1 is the final velocity of the bus after the collision
30,200 kg*30 mph + 2,000 kg*20 mph = 30,200 kg*v1 + 2000 kg*35 mph
v1 = 29 mph
The change in momentum of the bus is
m1u1-m1v1 = m1(u1-v1)
= 30,200 kg*(30 mph - 29 mph)
= 30,200 kg*0.447 m/s
= 13499.4 Ns
As the change in momentum and thus the force(Newton's second law) acting on the bus is very small compared to the
initial momentum of the bus,
the passangers in the bus do not feel any force acting them.
The passengers who are claimed to be injured are actuall do not fell any pain actually.
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