Impulse and momentum in an inelastic collision

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SUMMARY

The discussion focuses on the principles of impulse and momentum in an inelastic collision involving a cement truck and a Smart car. The cement truck, with a mass of 10,000 kg, collides with a stationary Smart car weighing 750 kg, traveling at 15 mph (6.68 m/s). The final velocity after the collision is calculated to be 6.21 m/s, with an average acceleration of 445.28 m/s² for the Smart car. The analysis concludes that riding in the cement truck is safer due to its lower acceleration during the crash, which is 33.4 m/s².

PREREQUISITES
  • Understanding of momentum conservation principles
  • Knowledge of inelastic collisions
  • Familiarity with basic physics equations (F = ma, p = mv)
  • Ability to convert units (e.g., mph to m/s)
NEXT STEPS
  • Study the concept of impulse and its relation to momentum changes
  • Learn about the differences between elastic and inelastic collisions
  • Explore real-world applications of momentum conservation in vehicle safety
  • Investigate how different masses affect collision outcomes in physics simulations
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Students studying physics, automotive safety engineers, and anyone interested in understanding the dynamics of vehicle collisions and their implications for safety.

runner2392
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Please check my work for the first question. The second and third questions, I'm not 100% sure how to solve them.
A ssume that a cement truck with a mass of 10,000 kg hits you while you were stopped at a traffic light, and that you are driving a Smart car with a mass of 750 kg.

1. If the truck was traveling at 15 mph and the crash makes you stick together (you and the truck have the same final velocity), what is your final velocity? Ignore friction with the road.

Momentum is conserved so Pinitial = Pfinal
15mi/h(1609m/mi)(hr/3600s) = 6.68 m/s
Pinitial = (10,000kg)(6.68 m/s) + 750kg*0 = 66,800
Pfinal = (10,000kg + 750kg)vfinal = 66,800 --> vfinal = 6.21 m/s2. If the crash impact took 0.2 s, what was your average acceleration? Convert it to units of g, the acceleration of gravity.

Since F = delta p/delta t, and delta p = 66,800
For the car,
F = 66,800 / 0.2 s = 333,400 = ma --> 333,400/750 = a = 445.28 ms^-2.

3. Solve the previous problem assuming you are in the cement truck instead of the Smart car. In which vehicle would you rather ride out the crash? Why?
For the truck,
F = 66,800 / 0.2 s = 333,400 = ma --> 333,400/10,000= a = 33.4 ms^-2.
So it would be more preferable to be in the truck because the truck accelerates less?

I'm really not sure whether to use 66800 for the truck's delta p since p doesn't change for the truck. does it? please help!
 
Last edited:
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Delta p means the change in momentum, not the total momentum. You need to calculate how much the momentum changed for either vehicle. Calculate the p before impact and after impact for either vehicle.
 

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