Perfectly Inelastic collisions

  • Thread starter chamonix
  • Start date
  • #1
chamonix
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Homework Statement



In 1986, four high school students built an electric car that could reach a speed of 106 km/h. The mass of the car was just 60.0 kg. Imagine two of these cars performing in a stunt show. One car travels east with a speed of 106.0 km/h and the other travels west with a speed of 75.0 km/h. If each car is driven by a person with mass of 50.0 kg, how much kinetic energy is dissipated in a head on, perfectly inelastic collision?


Homework Equations


I used these three equations:
m1*v1i+m2*v2i=(m1+m2)vf
KEi=1/2*m1*v1i^2+1/2*m2*v2i^2
KEf=1/2*(m1+m2)vf^2
KE=KEi-KEf

The Attempt at a Solution


I've tried many attempts, but I think this one is most relevant...
m1*v1i+m2*v2i=(m1+m2)vf
(110kg)*(29.4m/s)+(110kg)*(-20.8m/s)=(220kg)vf
vf=4.3m/s
KEi=1/2*m1*v1i^2+1/2*m2*v2i^2
KEi=1/2*110kg*29.4m/s^2+1/2*110kg*-20.8^2
KEi=71335J
KEf=1/2*(m1+m2)vf^2
KEf=1/2*(220)4.3^2
KEf=2033.9J
KE=KEi-KEf
KE=71335-2033.9
KE=69301.1J
KE=69300 Joules
I think that I used all the right equations, but the answer is wrong.:confused: Any help is most appreciated. Thanks.
 

Answers and Replies

  • #2
Gokul43201
Staff Emeritus
Science Advisor
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Haven't checked your numbers, but the equations you've used are correct.

To find the KEi, did you add or subtract the 2 numbers to get the total?
 
  • #3
daguy
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Well i dont see any problems
 
  • #4
chamonix
21
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I subtracted them. But I think the final answer should be negative.
 
  • #5
xiankai
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I subtracted them. But I think the final answer should be negative.

if it was negative, that would mean initial energy is LESS than final energy, which would be a no-no :tongue2:
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
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If the problem had asked for the "change in kinetic energy" then the answer would be negative. If the problem asked for "kinetic energy dissapated" then the answer is positive.
 
  • #7
chamonix
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Oh, I see. So the answer should be positive then.
 
  • #8
daguy
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Challenge problem projectile

Mongolian Projectile

You are the artillery expert for the Mongolian army. You have just fired a shell from the
top of a hill 53.6 m high at a tank moving away from you at a constant speed of 14.0 m/s.
The muzzle velocity of your gun is 148 m/s and your launch angle was 12.0°. You
observed that the shell landed 73.5 metres behind the tank. At the moment of impact, the
tank started to accelerate at 1.17 m/s2 over a distance of 187 metres, then it resumed
contant speed. If it takes you 24.0 s from the time you observe the shell land to recalculate
a trajectory and fire your second shell, what is the launch angle that you should fire at in
order to hit the moving tank?
Sketch of first shot:
12.0º
148 m/s
73.5 m
53.6 m Tank.

I dont have the least i dea of what supposed to be done any help is welcome.:mad:
 
  • #9
cristo
Staff Emeritus
Science Advisor
8,122
74
Mongolian Projectile

You are the artillery expert for the Mongolian army. You have just fired a shell from the
top of a hill 53.6 m high at a tank moving away from you at a constant speed of 14.0 m/s.
The muzzle velocity of your gun is 148 m/s and your launch angle was 12.0°. You
observed that the shell landed 73.5 metres behind the tank. At the moment of impact, the
tank started to accelerate at 1.17 m/s2 over a distance of 187 metres, then it resumed
contant speed. If it takes you 24.0 s from the time you observe the shell land to recalculate
a trajectory and fire your second shell, what is the launch angle that you should fire at in
order to hit the moving tank?
Sketch of first shot:
12.0º
148 m/s
73.5 m
53.6 m Tank.

I dont have the least i dea of what supposed to be done any help is welcome.:mad:

Please don't hijack someone else's thread! Firstly, draw a diagram, this may help. Do you know any equations that can help in questions like this?
 

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