2D inelastic collision

  • Thread starter Sally99
  • Start date
  • #1
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Homework Statement



You are moving due North in your car with velocity (0, 20m/s). You see another car travelling with a relative velocity of 15 m/s with an angle of 150 degrees. You have a mass of 30 kg, the other tricyclist has a mass of 40kg and the tricycles both have mass 10kg. You collide and stick together.

a) What is your velocity right after the collision?
b) Just after the collision, at what angle are you moving with respect to the x-axis?
c) What is your speed after the collision?
d) What is your change in kinetic energy from the collision?

Homework Equations



Vac = Vab + Vbc
m1v1 = (m1 + m2) v2
tan-1 = Vy/ Vx
pythagorean theorem




The Attempt at a Solution



First I would use the relative velocity equation to find the velocity of car 2 with respect to the ground. I'm not sure how to set this up though. Then I would use the x and y components to solve for the angle. After, I would use these two values and plug them into the pythagorean theorem. Finally, I can't seem to figure out what the equation is for change in kinetic energy!!

Any help from you guys would be greatly appreciated! I'm really stuck here :(

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
13
0
Ok so i noticed no one is replying to my thread so I'm going to elaborate a little.

to solve for initial velocity of car 2 i wrote
V21 = v2g - v1g
(-7.5, 13.0) = v2g - (0, 20)
v2g = (-7.5, -7)

Then to find the final velocity i did
m1v1 + m2v2 = (m1 + m2) v2
(40)(0,20) + (50)(-7.5,-7) = 90v2
(4.72, -3.89) = v2

theta = tan-1 3.89/ 4.72
= -39 +180
= 140

c) asquared =bsquared = csquared
(4.72)^2 + (-3.89)^2 = c^2
c= 6.12

Then change in KE = ???
 
  • #3
961
0
There may be an easier approach. Only relative velocities matter in mo conservation. So assume you are rest and collide with the other tricycle. What is your new velocity relative to the assumption of zero V initial velocity. Add this vectorially to the initial velocity by breaking down to x,y components. Compare the 1/2mv^2 values of the cars or tricycles. Same approach basically.
 
  • #4
13
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If I solved it using relative velocities would it give an incorrect answer?
 
  • #5
961
0
If I solved it using relative velocities would it give an incorrect answer?
Nope, so long as you do what I describe and consider the initial veocity as part of the kinetic energy. In other words if you were on a bike at 50 mph and hit a flea....

Here you are hitting an object of similar mo. But you had KE before and that needs to be considered. If you compute the changes in velocity from your assumption, then add these to the initial velocity of 20m/s and use Pythags to get a resultant velocity, square that and remember that the final mass is the sum of the two,....

Momenta conservation:

MOx=Cos(150)(15m/s)(40Kg)=(40kg+30Kg) Vx;
Vx is relative change in horizontal V, solve for it
MoY=Sin sin(150)(15m/s)*(40Kg)=(40+30)*Vy solve for it and subtract from initial Vy of 20m/s

Calculate the resultant velocity and using 1/2 mv'^2-1/2mv^2, compute the change in KE.
 
  • #6
13
0
Ahhhh ok I see, thanks so much for your input!
 

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