Solving a 2kg Ball Collision Problem

In summary, the conversation discusses the collision of two identical balls, with one ball initially moving with a speed of 3.0m/s and hitting the stationary ball elastically. The conversation then goes on to determine the speed and direction of both balls after the collision, using conservation of momentum and conservation of energy equations. The final solution involves combining the squared equations and eliminating theta to find the final velocities of both balls after the collision.
  • #1
kehwon
7
0

Homework Statement



A 2.0 kg ball moving with a speed of 3.0m/s hits, elastically, an identical stationary ball as shown. If the first ball moves away with angle 30 to the original path, determine
a) the speed of the first ball after the collision.
b) the speed and direction of the second ball after the collision

Homework Equations


m1v1 + m2v2 = m1v1' + m2v2'
1/2 m1v1^2 + 1/2m2v2^2 = 1/2 m1v1'^2 + 1/2 m2v2'^2


The Attempt at a Solution


X component:
v1 = v1' + v2' (since v2 is 0, and m's are equal)

Y component:
v1' = v2' (since v1 and v2 are 0 and m's are equal)

Because of this, does this mean that the angles should be equal as well?
I'm very confused about what I can deduce and assume from the identical mass of the balls..

It'd be great if someone could point me in the right direction. Thanks!
 
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  • #2


Vx= Vcos theta and simailry Vy= Vsint heta can be used here.
 
  • #3


So then do I divide the two components?
Cuz I tried that,

using
X: m1(v1 - v1'cos30) = m2(v2'costheta - 0)
v1 - v1'cos30 = v2costheta

Y: m1(0 -v1'sin30) = m2(v2'sintheta - 0)
- v1'sin30 = v2'sin theta

IF I divide Y by X, I get
v1-v1'sin30 / -v1'cos30 = tantheta

Then I'm stuck again :S
 
  • #4


You have 3 unknowns but only 2 equations. You need to use conservation of energy as well.
 
  • #5


Using conservation of energy,
v1 = v1'+v2'

with the combined X and Y component equations I've been trying to solve this and came up with several crazy dead-end solutions, using different trig identities..
 
  • #6


kehwon said:
Using conservation of energy,
v1 = v1'+v2'
Try squares of velocities.
 
  • #7


Okay,

so I have
(v1 - v1')(v1+v1') = v2'^2

which I would liek to incorporate into the x component: v1 - v1'cos30 = v2'costheta, but the cos30 is getting in the way.

I also tried putting Y component, with v2' = v1'sin30 / sin theta into the X component...
 
  • #8


Your two momentum equations have in them sin theta and cos theta. Can you see how to combine them in a way that eliminates theta?
 
  • #9


can I square the two momentum equations? then somehow combine them into sin^2theta + cos^2theta = 1?
 
  • #11


I'm having troubles linking the two. If I just add the two squard equations together, that makes sense since it would be like pythagorean theorem.

However, since theX componenti s v1 - v1'cos30 on the left side, square of that would be v1^2 - 2v1v1'cos30 + v1'^2...
 
  • #12


Sounds right. Why does that bother you? Now combine that with your energy equation and eliminate v2'
 
  • #13


great! thanks for all your help!
 

1. What is a 2kg ball collision problem?

A 2kg ball collision problem is a physics problem that involves two balls of equal mass colliding with each other and calculating the resulting velocities and momenta.

2. How do you solve a 2kg ball collision problem?

To solve a 2kg ball collision problem, you need to use the principles of conservation of momentum and conservation of kinetic energy. These principles state that the total momentum and kinetic energy of a system before and after a collision must remain the same.

3. What information is needed to solve a 2kg ball collision problem?

The mass and initial velocities of both balls are needed to solve a 2kg ball collision problem. The direction and type of collision (elastic or inelastic) may also be given.

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

In an elastic collision, kinetic energy is conserved and the two objects bounce off each other without any loss of energy. In an inelastic collision, kinetic energy is not conserved and some energy is lost, usually in the form of heat or sound.

5. Are there any real-life applications of solving a 2kg ball collision problem?

Yes, there are many real-life applications of solving a 2kg ball collision problem. It can be used to understand and predict the behavior of billiard balls, car crashes, and even the motion of molecules in a gas. It is also important in sports, such as calculating the trajectory of a golf ball after it hits the ground.

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