Collisions in two dimensions

In summary, using the principles of conservation of momentum and solving for the unknown variables using trigonometry, the speed and angle of the billiard balls after the collision were found to be v'B=0.661 m/s and Θ'B=-29.9°, respectively.
  • #1
rvnt
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0

Homework Statement


Billiard ball A of mass mA= 0.400kg moving with speed vA=1.80m/s strikes ball B initially at erst, of mass mB=0.500kg.As a result of the collision,ball A is deflected off at an angle of 30.0degrees with a speed v'A=1.10m/s. Taking the x-axis as the original direction of motion of ball A, solve for the speed v'B and angle Θ'B of ball B. Do not assume the collision is elastic.


Homework Equations


px conserved: mAvA=mAv'ACosΘ'A+mBv'BCosΘ'B
py conserved: mAvA=mAv'AsinΘ'A+mBv'BSinΘ'B



The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
-1.5407=CosΘ'BSinΘ'B
 
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  • #2
To solve for the angle, use a double angle trig identity:
sin(2θ) = 2 sinθ cosθ
 
  • #3
Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
 
  • #4
rvnt said:
Could you help me out with where I have went wrong please? or are my whole calculations incorrect and I should start over? Also, should I try figuring out the angle first or the speed?
My bad. I didn't read your work carefully and just looked at the last line. You were on the right track until that point. That last line was an error.

rvnt said:

The Attempt at a Solution


0.72=0.38105+0.500v'BCosΘ'B
0.338948/0.500=v'BCosΘ'B
0.677897=v'BCosΘ'B
0.677897/CosΘ'B= v'B

0=mAv'AsinΘ'A+mBv'BSinΘ'B
0=0.22+0.5v'BSinΘ'B
-0.22/0.5=v'BSinΘ'B
-0.44=v'BSinΘ'B

-0.44=(0.677897/CosΘ'B)SinΘ'B
That looks OK.
-1.5407=CosΘ'BSinΘ'B
Not sure how you got this from the previous line. Note:
-0.44=(0.677897/CosΘ'B)SinΘ'B = (0.677897)(SinΘ'B/CosΘ'B) = (0.677897)(TanΘ'B)

That will give you the angle; then you can find the speed.
 
  • #5
Oh right! k now I got it thank-you
 

1. What is meant by a collision in two dimensions?

A collision in two dimensions refers to the interaction between two objects in a two-dimensional space, where both objects have mass and are capable of moving in different directions.

2. How is momentum conserved in a two-dimensional collision?

Momentum is conserved in a two-dimensional collision when the total momentum of the system before the collision is equal to the total momentum after the collision. This means that the sum of the momenta of all objects involved in the collision remains the same.

3. What is the difference between an elastic and an inelastic collision?

In an elastic collision, both kinetic energy and momentum are conserved, meaning that the objects involved bounce off each other without any loss of energy. In an inelastic collision, kinetic energy is not conserved, and some energy is lost in the form of heat, sound, or deformation of the objects.

4. How do you calculate the final velocities of objects after a collision in two dimensions?

The final velocities of objects after a collision in two dimensions can be calculated using the conservation of momentum and the conservation of kinetic energy equations. These equations take into account the mass and initial velocities of the objects involved in the collision.

5. Can you have a perfectly elastic collision in two dimensions?

Yes, it is possible to have a perfectly elastic collision in two dimensions, where both momentum and kinetic energy are conserved. This type of collision is often seen in billiards or other similar games.

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