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Perfectly Inelastic collisions

  • Thread starter chamonix
  • Start date
  • #1
21
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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
Gold Member
7,051
17
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
2
0
Well i dont see any problems
 
  • #4
21
0
I subtracted them. But I think the final answer should be negative.
 
  • #5
31
0
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
41,808
933
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
21
0
Oh, I see. So the answer should be positive then.
 
  • #8
2
0
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
8,107
73
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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