Perfectly Elastic Collision Problem

In summary, the problem involves a perfectly elastic collision between a 110 g ball and a 400 g ball moving in the same direction. The collision leads to two unknown final velocities, but conservation of momentum and energy can be used to solve the problem. By using the Galilean transformation of velocities, the second object's velocity can be made into zero and the final velocity of the 110 g ball can be found.
  • #1
jonnejon
27
0

Homework Statement


A 110 g ball moving to the right at 4.0 m/s catches up and collides with a 400 g ball that is moving to the right at 1.2 m/s.
If the collision is perfectly elastic, what is the speed of the 110 g ball after the collision?

Homework Equations


KEi=KEf
1/2m1v1i^2 + 1/2m2v2i = 1/2m1v1f^2 + 1/2m2v2f^2


The Attempt at a Solution


Tried plugging it in but I don't really understand the concept so I am let with 2 final velocity. Please help.

110g < 400g so 110g should move left and 400g should move right, so does that make v2f=0?
 
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  • #2
You have two unknowns and therefore need two simultaneous equations. What else is conserved besides energy?
 
  • #3
Momentum and kinetic energy?
 
  • #4
jonnejon said:
Momentum and kinetic energy?
Correct! So you already have one equation (conservation of energy), can you now write a second equation using conservation of momentum? You should then have a system of two simultaneous equations.
 
  • #5
Nope, I don't understand how. There is an equation my textbook derived for perfectly elastic collisions but it is only if the second mass is at rest.
 
  • #6
jonnejon said:
Nope, I don't understand how. There is an equation my textbook derived for perfectly elastic collisions but it is only if the second mass is at rest.
What does the principle of conservation of momentum state?
 
  • #7
Pi=Pf mivi=mfvf
 
  • #8
jonnejon said:
Pi=Pf mivi=mfvf
Correct, so can you now apply conservation of momentum to your problem?
 
  • #9
So, I solved for m2vf2 in the momentum equation and plug that into the energy equation and solve for vf1 but I got it wrong.
 
  • #10
jonnejon said:
So, I solved for m2vf2 in the momentum equation and plug that into the energy equation and solve for vf1 but I got it wrong.
Could you detail your steps?
 
  • #11
solve for v2f in momentum equation:
v2f = (m1v1i + m2v2i - m1v1f) / m2
v2f = (v1i/m2 + v2i - v1f/m2)
plug it in the energy equation:
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2(v1i/m2 + v2i - v1f/m2)^2
solve for v1f?
 
  • #12
jonnejon said:
solve for v2f in momentum equation:
v2f = (m1v1i + m2v2i - m1v1f) / m2
v2f = (v1i/m2 + v2i - v1f/m2)
Your step from the first line to the second is incorrect. Where did the m1 go?
 
  • #13
Opps, I thought it cancels out because of the subtraction sign.

1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1v1i/m2 + v2i - m1v1f/m2)^2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + 1/2m2 (m1^2v1i^2/m2^2 + v2i^2 - m1^2v1f^2/m2^2)
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2v1i^2 - m1^2v1f^2)/2m2 - m2v2i^2/2
1/2m1v1i^2 + 1/2m2v2i^2 = 1/2m1v1f^2 + (m1^2 (v1i^2 - v1f^2)/2m2) - m2v2i^2/2
(1/2m1v1i^2 + 1/2m2v2i^2)(2m2)/(m1^2) + m2v2i^2/2 = 1/2m1v1f^2 + (v1i^2 - v1f^2)

Man, this equation just getting to complicated. Am I even doing it right?
 
  • #14
Yes it is getting rather complicated. I really can't decipher what you've written, you could use latex to make things easier to read,

Conservation of Energy
[tex]m_1v_{1i}^2+m_2v_{2i}^2 = m_1v_{1f}^2+m_2v_{2f}^2[/tex]

Conservation of Momentum
[tex]m_1v_{1i}+}m_2v_{2i} = m_1v_{1f}+m_2v_{2f}[/tex]
 
  • #15
I tried doing that but it was just too messy for me. I actually found out how to do it. There are perfectly elastic collision equations in my textbook for velocities at rest. But they said you can use the Galilean transformation of velocities to make the 2nd object's velocity into zero and then find the final velocity.

Used:
v' = v - V

Thanks for your help though.
 
  • #16
in elastic collision momentum and energy is conserved
 

1. What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which 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.

2. How is momentum conserved in a perfectly elastic collision?

In a perfectly elastic collision, both momentum and kinetic energy are conserved. This means that the total momentum of the system before the collision is equal to the total momentum after the collision.

3. What is the equation for calculating the velocities of two objects after a perfectly elastic collision?

The equation for calculating the velocities of two objects after a perfectly elastic collision is:
v1f = (m1-m2)/(m1+m2) * v1i + (2*m2)/(m1+m2) * v2i
v2f = (2*m1)/(m1+m2) * v1i + (m2-m1)/(m1+m2) * v2i
where v1f and v2f are the final velocities of objects 1 and 2, m1 and m2 are the masses of the objects, and v1i and v2i are the initial velocities of the objects.

4. Can a perfectly elastic collision occur in real life?

In theory, a perfectly elastic collision can occur in real life, but it is very rare. In most real-life situations, there is some loss of kinetic energy due to factors such as friction, deformation of the objects, and sound waves. However, in certain cases, such as collisions between subatomic particles, perfectly elastic collisions can occur.

5. How does a perfectly elastic collision differ from an inelastic collision?

In an inelastic collision, there is a loss of kinetic energy and the two objects stick together after the collision. In a perfectly elastic collision, there is no loss of kinetic energy and the two objects bounce off each other after the collision. Additionally, in an inelastic collision, momentum is conserved but kinetic energy is not, while in a perfectly elastic collision, both momentum and kinetic energy are conserved.

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