- #1
jfnn
SOLVED
THANK YOU
THANK YOU
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Then keep going with..jfnn said:For the x direction: (the initial velocity of the particle is fully in the y-component, thus the initial velocity for particle one and two, which are attached, in the x-direction is 0).
(1/3m)*0 + (2/3m)*0 = (1/3m)*v1x' + (2/3m)*v2x'
0 = (1/3m)*v1x' + (2/3m)*v2x' where v1x' (the velocity of particle one in the x direction after collision is -48 m/s, negative because it is west)
0 = (1/3m)*-48 + (2/3m)*v2x'
-16m + (2/3m)*v2x' = 0 --> This is equation 1
Then I took equation 1 to find v2x'
-16m + (2/3m)*v2x' = 0 --> (2/3m)*v2x' = 16m --> Masses cancel and 16 - 2/3*v2x' --> 24 = v2x'
jfnn said:Then I did the same thing except for the y direction
(1/3m)*75 + (2/3m)*75 = (2/3m)v2y'
75m = (2/3m)v2y'
v2y' = 112.5 m/s
then I did the pathag theorem to find v2' = 115 m/s --> Is this the correct value/way to do this?
I have no clue how to find the direction
jfnn said:the velocity of particle one in the x direction after collision is -48 m/s, negative because it is west
jfnn said:I have no clue how to find the direction
I know how to do the relationships I think. So it would be theta = tan^-1(v2y'/v2x')?CWatters said:Yes conservation of momentum is correct approach. I haven't checked your working but you seemed to jump about a bit. The following would be better order..Then keep going with..
So now you have the x and y components of velocity.
Yes it's the correct way.
Make a drawing showing the components V2x, V2y and the resultant V2. You used Pythagoras so you know looks like a right angled triangle.
Do you know how to work out angles using Sin() , Cos() or Tan() ?
CWatters said:Re...
It would have been better to state at the beginning which directions you will define as positive (eg up and east).
In addition to using trig to find the angle with the horizontal... You know that one part went directly west. So what does conservation of momentum tell you about the compass direction of the other part? The V2x is positive so we know it's not got a westerly component.
jfnn said:OH! So since the V2x' is positive, it means that it is going east..
and since the y is positive as well (it is north) I can then say it is 78 deg N of E?
CWatters said:Correct. If it only splits into two parts and one goes exactly west the other must have a component that is exactly east. That way momentum in the Z axis is conserved.
No that's not correct. Y was the vertical axis. So your 78 degrees is the angle it makes with the horizontal. I have made a diagram. Everything happens in the XY plane..
View attachment 208313
CWatters said:PS I still haven't checked your calculations (just the method).
CWatters said:No "North of east" would mean it had a component in the Z axis. It must go exactly east and upwards.
The formula for finding the speed and direction of a particle after collision is: v2 = (m1v1 + m2v2) / (m1 + m2), where v2 is the final velocity, m1 and m2 are the masses of the colliding objects, and v1 is the initial velocity of one of the objects.
The direction of the particle after collision can be determined by using the angle of the initial velocity and the angle of the final velocity. If the two angles are the same, the particle will continue moving in the same direction. If the angles are different, the particle will change direction after the collision.
Yes, there is a difference in calculating the speed and direction for elastic and inelastic collisions. In elastic collisions, the total kinetic energy and momentum are conserved, while in inelastic collisions, some of the kinetic energy is converted into other forms of energy and is not conserved.
Yes, the speed and direction of a particle after collision can be negative. A negative velocity indicates that the particle is moving in the opposite direction of its initial velocity.
Yes, one of the assumptions made when using the formula is that there are no external forces acting on the colliding objects during the collision. Additionally, the objects are assumed to be rigid bodies and there is no loss of energy due to friction or air resistance.