Finding kinetic energy after an elastic collision

In summary: The problem was much easier than thought. Find velocity of the first object from kinetic energy. Use this velocity times the cosine of the angle. Use that velocity for finding the new kinetic energy. Giving an answer of 0.728J.
  • #1
Delta Sheets
10
0

Homework Statement


A ball with mass m = 0.230 kg and kinetic energy K1 = 1.16 J collides elastically with a second ball of thesame mass that is initially at rest. After the collision, the first ball moves away at an angle of θ1= 37.6° with respect to the horizontal. What is the kinetic energy of the first ball after the collision?


Homework Equations


KE=1/2mv^2
m1v1 + m2v2(initial)=m1v1 + m2v2(final)


The Attempt at a Solution


Since kinetic energy is conserved,I figured the kinetic energy of the two balls would be the same after the collision at the θ=0 position. So I then tried to solve for kinetic energy in the direction of θ=37.6. Seeing as the two kinetic energies of the masses should equal 1.16 J, I am lost on the problem and do not know where to start. I would like to be put in the right direction, not necessarily given an answer so I can figure the work out on my own.
 
Physics news on Phys.org
  • #2
Delta Sheets said:

Homework Statement


A ball with mass m = 0.230 kg and kinetic energy K1 = 1.16 J collides elastically with a second ball of the same mass that is initially at rest. After the collision, the first ball moves away at an angle of θ1= 37.6° with respect to the horizontal. What is the kinetic energy of the first ball after the collision?

Homework Equations


KE=1/2mv^2
m1v1 + m2v2(initial)=m1v1 + m2v2(final)

The Attempt at a Solution


Since kinetic energy is conserved,I figured the kinetic energy of the two balls would be the same after the collision at the θ=0 position.
That's not a valid assumption.
So I then tried to solve for kinetic energy in the direction of θ=37.6. Seeing as the two kinetic energies of the masses should equal 1.16 J, I am lost on the problem and do not know where to start. I would like to be put in the right direction, not necessarily given an answer so I can figure the work out on my own.
If the collision were such that there was no deflection to either side, then after the collision, the first ball would be at rest and the second ball would move with the same velocity that the first ball initially had.
 
  • #3
SammyS said:
That's not a valid assumption.

If the collision were such that there was no deflection to either side, then after the collision, the first ball would be at rest and the second ball would move with the same velocity that the first ball initially had.

Now I understand that part, but in what way can i calculate the kinetic energy of each?
 
  • #4
Delta Sheets said:
Now I understand that part, but in what way can i calculate the kinetic energy of each?
The conservation of momentum equation is a vector equation.

Therefore,
[itex]\displaystyle
\left(\,m_1\,(v_1)_x\ +\ m_2\,(v_2)_x\,\right)_\text{initial}
=\left(\,m_1\,(v_1)_x\ +\ m_2\,(v_2)_x\,\right)_\text{final}[/itex]

[itex]\displaystyle
\left(\,m_1\,(v_1)_y\ +\ m_2\,(v_2)_y\,\right)_\text{initial}
=\left(\,m_1\,(v_1)_y\ +\ m_2\,(v_2)_y\,\right)_\text{final}[/itex]​

Assuming that the initial velocity is in the x direction, what does that tell you about how the y-components of the final velocities of the two balls compare?
 
  • #5
SammyS said:
The conservation of momentum equation is a vector equation.

Therefore,
[itex]\displaystyle
\left(\,m_1\,(v_1)_x\ +\ m_2\,(v_2)_x\,\right)_\text{initial}
=\left(\,m_1\,(v_1)_x\ +\ m_2\,(v_2)_x\,\right)_\text{final}[/itex]

[itex]\displaystyle
\left(\,m_1\,(v_1)_y\ +\ m_2\,(v_2)_y\,\right)_\text{initial}
=\left(\,m_1\,(v_1)_y\ +\ m_2\,(v_2)_y\,\right)_\text{final}[/itex]​

Assuming that the initial velocity is in the x direction, what does that tell you about how the y-components of the final velocities of the two balls compare?

This shows the y components of velocity should equal 0 when added together
 
  • #6
Delta Sheets said:
This shows the y components of velocity should equal 0 when added together
Right.

For the first ball, you know how the x-component final velocity is related to the y-component final velocity.
 
  • #7
SammyS said:
Right.

For the first ball, you know how the x-component final velocity is related to the y-component final velocity.

The problem was much easier than thought. Find velocity of the first object from kinetic energy. Use this velocity times the cosine of the angle. Use that velocity for finding the new kinetic energy. Giving an answer of 0.728J.
 

Related to Finding kinetic energy after an elastic collision

1. What is kinetic energy?

Kinetic energy is the energy an object possesses due to its motion. It is a scalar quantity and is dependent on an object's mass and velocity.

2. What is an elastic collision?

An elastic collision is a type of collision in which both momentum and kinetic energy are conserved. This means that the total kinetic energy before and after the collision remains the same.

3. How do you find kinetic energy after an elastic collision?

To find the kinetic energy after an elastic collision, you can use the formula KE = 1/2 mv^2, where m is the object's mass and v is its velocity. You will need to calculate the velocities of each object after the collision and use those values in the formula.

4. What are some examples of elastic collisions?

Some examples of elastic collisions include billiard balls colliding, a rubber ball bouncing off a hard surface, or two cars colliding and bouncing off each other without any deformation.

5. Why is kinetic energy important in elastic collisions?

Kinetic energy plays a crucial role in elastic collisions because it helps us understand the behavior of objects after a collision. By calculating the kinetic energy before and after the collision, we can determine if the collision was elastic or inelastic and analyze the objects' velocities and directions of motion.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
544
Replies
1
Views
709
  • Introductory Physics Homework Help
Replies
22
Views
3K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
962
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top