Momentum and Kinetic Energy, Elastic Collision

In summary, when a 2.0 kg ball moving at 3.0 m/s with an angle of 30° hits an identical stationary ball elastically, the first ball will move away with a speed of 2.598075 m/s at an angle of 30° and the second ball will move with a speed of 1.5 m/s at an angle of 60°. This can be solved using conservation of momentum and kinetic energy equations.
  • #1
cassie123
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Homework Statement


[/B]
A 2.0 kg ball moving with a speed of 3.0 m/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


px: m1v1=m1v'1cosΘ1+m2v'2cosΘ2
py: 0=m1v'1sinΘ1+m2v'2sinΘ2
KE: 1/2m1v1^2=1/2m1v'1+1/2m2v'2

where I am using 1 and 2 to denote the first and second balls. and the "prime's" denoting final speed.
Θ1=30°
Since the balls have identical mass, I believe mass cancels out of the above equations.

The Attempt at a Solution


[/B]
What I tried to do was to find the components of the vectors of final motion, use the conservation of momentum and kinetic energy to create three equations to solve for the three unknowns (Θ2, v'1, v'2).

I canceled the mass out of all three equations, rearranged so that the Θ2 terms are on the same side of the momentum equations. I then squared the momentum equations and added them so that I could use the identity cos^2Θ+sin^2Θ=1 to get rid of the Θ2 terms and only have to solve for v'1 and v'2 using the KE equation and the added momentum equations. This is where I am getting stuck with the algebra:

(v1-v'1cos(30))^2+(-v'1sin(30))^2=(v'2)^2
and v'2=(v1)^2-(v'1)^2
so, subbing in for v'2:
(v1-v'1cos(30))^2+(-v'1sin(30))^2=(v1)^2-(v'1)^2

and at this point I should be able to solve for v'1, but i can't figure it out
 
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  • #2
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  • #3
Just expand the terms and simplify (a lot).
 
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  • #4
haruspex said:
Just expand the terms and simplify (a lot).

I gave that shot and ended up with v'1=2.598075 by using the quadratic equation.
This makes v'2=1.5 and Θ2=60°.

Does that make sense?
 
  • #5
cassie123 said:
I gave that shot and ended up with v'1=2.598075 by using the quadratic equation.
This makes v'2=1.5 and Θ2=60°.

Does that make sense?
Looks right. You'll get more insight (and a heap of other benefits) if you resist the temptation to plug in numbers until the final step. Writing the original angle as theta, instead of 30 degrees, you find v1'=v1cos(theta), v2'=v1sin(theta).
 
  • #6
Yes.
 

FAQ: Momentum and Kinetic Energy, Elastic Collision

What is momentum?

Momentum is a measure of an object's motion and is calculated by multiplying its mass by its velocity. It is a vector quantity, meaning it has both magnitude and direction.

How is momentum related to kinetic energy?

Momentum and kinetic energy are both measures of an object's motion, but they are not the same. Kinetic energy is the energy an object has due to its motion, while momentum is a measure of the object's motion itself.

What is an elastic collision?

An elastic collision is a type of collision where there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

How is momentum conserved in an elastic collision?

In an elastic collision, the total momentum of the system before and after the collision remains the same. This means that the sum of the momenta of all the objects involved in the collision is equal before and after the collision.

Can an elastic collision occur between objects of different masses?

Yes, an elastic collision can occur between objects of different masses. In this type of collision, the momentum and kinetic energy of each individual object may change, but the total momentum and kinetic energy of the system remains constant.

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